# Percolation of terraces,
and enhancements for the orthant model

Mark Holmes111University of Melbourne. Email<EMAIL_ADDRESS>and
Thomas S. Salisbury222York University. Email<EMAIL_ADDRESS>
###### Abstract

We study a model of an i.i.d. random environment in general dimensions $d\geq
2$, where each site is equipped with one of two environments. The model comes
with a parameter $p$ which governs the frequency of the first environment, and
for each dimension $d$ there is a critical parameter $p_{c}(d)$ at which there
is a phase transition for the _geometry_ of a particular connected cluster
(the cluster is infinite for all $p$). We use the celebrated methodology of
enhancements in this novel setting to prove that $p_{c}(d)$ is strictly
monotone in $d$ for this model. To do so we study the discrete geometry and
percolation theory of higher-dimensional structures called terraces.

Keywords: percolation, enhancement, random environment, orthant model,
critical point, terraces, discrete geometry.

MSC2020: 60K35, 52C99

###### Contents

  1. 1 Introduction
    1. 1.1 Notation and Results
  2. 2 Terraces
    1. 2.1 Facts about terraces
    2. 2.2 Corners
    3. 2.3 Local terraces
    4. 2.4 Relation with $\mathcal{C}_{o}$
  3. 3 Enhancement
    1. 3.1 Disturbed orthant model
    2. 3.2 Local modification
    3. 3.3 Orthant model for slabs

##  1 Introduction

In this paper, we study certain random directed subgraphs of the lattice
$\mathbb{Z}^{d}$ (called the _orthant model_ and the _half-orthant_ model -
see Examples 1 and 2 below) in which (with probability $p$) a vertex will
assume a configuration that blocks immediate passage to sites below it. This
leads to a phase transition in which passage downwards is feasible for small
$p$ but not for large $p$. These random directed graphs have been studied
earlier in the case $d=2$, in [9] and [10]. More recently, [13] defined a
critical point $p_{c}(d)$ for these (and other) models, while [14] examined
the asymptotic shape of the set of sites in $\mathbb{Z}^{d}$ that can be
reached from the origin $o$ when $p$ is large. Beekenkamp [3] extended the
validity of the shape theorem for $p$ down to a (potentially different)
critical value $p^{\prime}_{c}(d)$. Random walks in various directed graphs
including these kinds were studied in [11] and [12].

The fact that $p_{c}(d)$ is monotone increasing in $d$ is straightforward. The
main result of this paper is that $p_{c}(d)$ is _strictly_ monotone in $d\geq
2$ \- see Theorem 2. This turns out to be highly non-trivial and requires a
detailed excursion into discrete geometry, because the argument centres on the
study of percolation for higher dimensional structures called terraces.

###  1.1 Notation and Results

Let $d\geq 2$. Let $[d]=\\{1,2,\dots,d\\}$, and let $(e_{i})_{i\in[d]}$ denote
the canonical basis vectors for $\mathbb{Z}^{d}$. Set
$\mathcal{E}_{+}(d)=\\{e_{i}:i\in[d]\\}$,
$\mathcal{E}_{-}(d)=\\{-e_{i}:i\in[d]\\}$, and
$\mathcal{E}(d)=\mathcal{E}_{+}(d)\cup\mathcal{E}_{-}(d)$. When there is no
chance of confusion, we will drop $d$ from the above notation.

Let $\mu$ be a probability measure on the measurable space
$\big{(}\mathcal{P}(\mathcal{E}),\mathcal{P}(\mathcal{P}(\mathcal{E}))\big{)}$,
where $\mathcal{P}(\mathcal{A})$ denotes the set of subsets of $\mathcal{A}$.
We will abuse notation and write $\mu(\mathcal{G})$ instead of
$\mu(\\{\mathcal{G}\\})$ where $\mathcal{G}\subset\mathcal{E}$ (i.e.
$\mathcal{G}\in\mathcal{P}(\mathcal{E})$). At each $x\in\mathbb{Z}^{d}$
independently choose a $\mathcal{G}_{x}\subset\mathcal{E}$ with law $\mu$. We
refer to $\boldsymbol{\mathcal{G}}=(\mathcal{G}_{x})_{x\in\mathbb{Z}^{d}}$ as
a _degenerate random environment_. Inserting arrows from $x$ to each of the
vertices $\\{x+e:e\in\mathcal{G}_{x}\\}$, yields a random directed graph with
vertex set $\mathbb{Z}^{d}$ and directed edge set
$\\{(x,x+e):x\in\mathbb{Z}^{d},e\in\mathcal{G}_{x}\\}$. The two examples that
are relevant to the present work are as follows.

######  Example 1 (Orthant model).

$\mu(\mathcal{E}_{+})=p=1-\mu(\mathcal{E}_{-})$.

######  Example 2 (Half-orthant model).

$\mu(\mathcal{E}_{+})=p=1-\mu(\mathcal{E})$.

Hereafter we are interested in the set of vertices
$\mathcal{C}_{x}\subset\mathbb{Z}^{d}$ that can be reached from $x$ by
following the arrows. In both examples above $\mathcal{C}_{x}$ _is always
infinite_. To see this note that $e_{1}\in\mathcal{E}_{+}$ and
$-e_{2}\in\mathcal{E}_{-}$ so that from $x$ we can follow an infinite self-
avoiding path in $\mathcal{C}_{x}$ consisting of just $e_{1}$ and $-e_{2}$
steps. Degenerate random environments for which the forward clusters
$(\mathcal{C}_{x})_{x\in\mathbb{Z}^{d}}$ are a.s. infinite are the natural
random directed graphs to study in order to understand random walks in non-
elliptic random environments. In this context $\mathcal{G}_{x}$ represents the
set of possible steps that the walk can take from $x$ and the fact that
forward clusters are a.s. infinite ensures that the random walker does not get
stuck on a finite set of sites (see e.g. [11, Lemma 2.2]).

Consider a random walk in random environment (RWRE) in which the walker
chooses uniformly at random from the available arrows in a degenerate random
environment $\mathcal{G}_{x}$. In [12] it is shown that this RWRE, for
$\mathcal{G}_{x}$ given by either Example 1 or Example 2, is ballistic in
direction $1:=\sum_{e\in\mathcal{E}_{+}}e$ when $d=2$ and $p>p_{c}(d)$, or
when $d\geq 3$ and $p$ is large. In [3] the latter statement is extended to
$p>p^{\prime}_{c}(d)$. For the former example, this should in fact hold for
all $p>1/2$ (but we don’t know how to prove this), while for the latter it
should not be hard to prove that this (and a CLT) holds for all $p>0$ (see
e.g. [4]).

We can couple the environments together for all $p$ and all pairs of subsets
of $\mathcal{E}$ in the natural way, using independent uniform random
variables $(U_{x})_{x\in\mathbb{Z}^{d}}$. In other words, for $p\in[0,1]$ and
$E,F\subset\mathcal{E}$ define

$\mathcal{G}_{x}=\mathcal{G}_{x}(p)=\begin{cases}E,&\text{if $U_{x}\leq p$}\\\
F,&\text{if $U_{x}>p$.}\end{cases}$

The sets $\mathcal{C}_{x}=\mathcal{C}_{x}(p)$ are clearly increasing in $E$
and $F$ (so $\mathcal{C}_{x}$ for the orthant model is a subset of
$\mathcal{C}_{x}$ for the half-orthant model). When $E\subset F$ (e.g. for the
half-orthant model $E=\mathcal{E}_{+}$ and $F=\mathcal{E}$), they are also
decreasing in $p$. Let

$\displaystyle\Omega_{1}$ $\displaystyle=\\{x\in\mathbb{Z}^{d}:U_{x}\leq
p\\}=\\{x\in\mathbb{Z}^{d}:\mathcal{G}_{x}=E\\}$ (1) $\displaystyle\Omega_{0}$
$\displaystyle=\\{x\in\mathbb{Z}^{d}:U_{x}>p\\}=\\{x\in\mathbb{Z}^{d}:\mathcal{G}_{x}=F\\}.$
(2)

The orthant model and its phase transition in 2 dimensions are studied in [9,
10]. Let $o=(0,\dots,0)$ denote the origin in $\mathbb{Z}^{d}$. Using a
duality with oriented site percolation on the triangular lattice, those papers
proved a shape theorem for a particular boundary of $\mathcal{C}_{o}$ for this
model, and established improved estimates for the critical value of the dual
percolation model. Similar arguments apply to the half-orthant model with
$d=2$.

To formulate the phase transitions in general dimensions, define for
$x\in\mathbb{Z}^{d}$,

$L^{(i)}_{x}(p):=\inf\left\\{k\in\mathbb{Z}:x+ke_{i}\in\mathcal{C}_{o}(p)\right\\}.$
(3)

Also, for $z\in\mathbb{Z}^{d}$ and $i\in[d]$ define
$z_{\\{+i\\}}=\\{z+ke_{i}:k\geq 0\\}$. For a set $A\subset\mathbb{Z}^{d}$
define $A_{\\{+i\\}}=\cup_{z\in A}z_{\\{+i\\}}$. The following result
concerning the half-orthant model is proved in [13]. Related results can be
stated for the orthant model, but since e.g. the orthant model is a non-
monotone model (in $p$), it is cleaner to state results for the half-orthant
model.

######  Theorem 0 ([13]).

For the half-orthant model (Example 2) there exists $p_{c}(d)\in(0,1)$ such
that:

  1. (I)

for $p<p_{c}(d)$, $\mathcal{C}_{o}(p)=\mathbb{Z}^{d}$ a.s. and;

  2. (II)

when $p\in(p_{c}(d),1)$, for each $i\in[d]$

    1. (a)

$L^{(i)}_{x}(p)$ is a.s. finite for every $x\in\mathbb{Z}^{d}$, and

    2. (b)

$\mathcal{C}_{o}(p)=(\bigcup_{x\in\mathbb{Z}^{d}}\\{x+L^{(i)}_{x}(p)\\})_{\\{+i\\}}$,
and

  3. (III)

when $p=p_{c}(d)$, either $\mathcal{C}_{o}(p)=\mathbb{Z}^{d}$ a.s., or (II)(a)
and (b) hold a.s. for each $i\in[d]$.

It is shown in [13] that in any dimension the orthant and half-orthant models
share the same $\mathcal{C}_{o}(p)$ “outer”-boundary (for $p\geq 1/2$), in the
sense that $L^{(i)}_{x}(p)$ are equal for the two models for each $x,i$. Thus,
interpreted as a critical point for finiteness of the $L_{x}^{(i)}$, the
orthant and half-orthant models have the same critical value.

Beekenkamp’s $p^{\prime}_{c}(d)$ is defined in [3] as the infimum of $p$’s
such that $\mathcal{C}_{o}(p)$ is contained in certain convex cones. It is
known that $p_{c}(d)\leq p^{\prime}_{c}(d)$, and it is conjectured that they
are equal, but this is only known when $d=2$.

In the next section we will define a geometric class of subsets of
$\mathbb{Z}^{d}$ that we call _terraces_. We can then express the above phase
transition in terms of percolation of terraces as follows.

######  Theorem 1.

The critical point $p_{c}(d)$ of the orthant or half-orthant model is the
infimum of those $p$ such that there a.s. exists a terrace $\Lambda$ with
$\Lambda\subset\Omega_{1}$.

This paper is therefore a contribution to the general program of extending
standard percolation techniques to such higher-dimensional structures. For
other examples of high-dimensional percolation structures, see [6, 7]. Our
main result is the following.

######  Theorem 2.

For the orthant or half-orthant model $p_{c}(d)<p_{c}(d+1)$.

We will prove Theorem 2 via the methodology of enhancements, introduced by
Aizenman and Grimmett in [1]. This is the “standard” methodology for proving
strict inequality results for critical points in percolation models (see e.g.
[15, 2]). Application of this methodology usually requires verifying some
combinatorial statement. Sometimes the statement in question is
straightforward (or even obvious), while elsewhere it can be extremely
challenging (see e.g. [8], where strict inequality between the critical values
of bond and site percolation is established in various lattices). The full
story of enhancements of site percolation on $\mathbb{Z}^{d}$ is complicated
and currently incomplete (see e.g. [2]).

Our use of enhancements involves the deformation of terraces, which poses a
significant challenge. Sections 2.1–2.3 are entirely geometric, establishing
deterministic facts about terraces that will be required in Section 3, where
the main theorem is proved. The approach in Section 3 is to use a slab to
interpolate between $\mathbb{Z}^{d}$ and $\mathbb{Z}^{d+1}$. We analyze this
slab by perturbing the $\mathbb{Z}^{d}$ model at points $V_{d}$ of a sub-
lattice, which is taken to be sparse in order to ensure sufficient
independence. The enhancement approach then requires us to estimate the number
of pivotal sites in $V_{d}$.

We will do that by modifying a given terrace locally in order to pass through
some point of $V_{d}$. The obstacle to achieving this is that our terraces can
be tightly constrained by boundary conditions, which can in some cases impose
a high degree of rigidity on the terraces allowed. A consequence of the ideas
of Section 2, that illustrates this issue, is that the kinds of boundary
configurations that make the terrace rigid must already force it to go through
points of $V_{d}$. Our approach to this problem is to show that the terraces
already contain (or can be deformed locally in order to contain) long linear
segments.

######  Remark 3.

An alternate approach which would also establish our main result (while
bypassing the interpolation via slabs) would be to fix the dimension $d$ and
to consider two models, one with sets $E_{0},F_{0}$ and the other with sets
$E_{1},F_{1}$, satisfying the following:

  * •

$F_{0}=\mathcal{E}$ and $\\{e_{1}\\}\subset E_{0}\subsetneq\mathcal{E}_{+}$,
and

  * •

$F_{1}=\mathcal{E}$ and $E_{0}\subsetneq E_{1}\subset\mathcal{E}_{+}$,

Strict monotonicity would follow from establishing that the critical points
for the above two models are not equal. However, this approach seems to
require an understanding of a more general class of geometric objects than the
terraces considered here.

##  2 Terraces

As indicated earlier, we will require a detailed study of the geometry of
certain sets that we call _terraces_.

For $z\in\mathbb{Z}^{d}$ we let $z^{\scriptscriptstyle{[}i]}=z\cdot e_{i}$
(note that when used in a _superscript_ , the notation $[i]$ refers to the
$i$-th coordinate of $z$ not $z^{\\{1,\dots,d\\}}$) and define

$\displaystyle z_{+}$
$\displaystyle=\\{y\in\mathbb{Z}^{d}:y^{\scriptscriptstyle{[}i]}\geq
z^{\scriptscriptstyle{[}i]}\text{ for each }i\in[d]\\},\text{ and }$
$\displaystyle z_{-}$
$\displaystyle=\\{y\in\mathbb{Z}^{d}:y^{\scriptscriptstyle{[}i]}\leq
z^{\scriptscriptstyle{[}i]}\text{ for each }i\in[d]\\}.$

For $C\subset\mathbb{Z}^{d}$ we define $-C=\\{-x:x\in C\\}$,

$\displaystyle C_{+}=\bigcup_{z\in C}z_{+},\quad\text{ and }\quad
C_{-}=\bigcup_{z\in C}z_{-}.$ (4)

Trivially $\varnothing_{+}=\varnothing$ and

if $A\subset B$ then $A_{+}\subset B_{+}$. (5)

  * •

A set $C\subset\mathbb{Z}^{d}$ is saturated if for every $y\in\mathbb{Z}^{d}$
and every $i\in[d]$, there exist $z\in C$ and $k\in\mathbb{Z}$ such that
$z=y+ke_{i}$.

  * •

A set $C\subset\mathbb{Z}^{d}$ is solid above (resp.  _below_) if $C_{+}=C$
(resp. $C_{-}=C$).

  * •

For $i\in[d]$, a set $C$ is $i$-bounded below (resp.  _above_) if for every
$y\in\mathbb{Z}^{d}$, the set $\\{k\in\mathbb{Z}:y+ke_{i}\in C\\}$ is bounded
below (resp. above).

  * •

We say that $C$ is bounded below (resp.  _above_) if it is $i$-bounded below
(resp. above) for every $i\in[d]$.

  * •

If $C$ is saturated and bounded below, we may define
$\gamma_{\scriptscriptstyle{C}}:\mathbb{Z}^{d}\times[d]\to C$ by
$\gamma_{\scriptscriptstyle{C}}(y,i)=z$, where $z=y+ke_{i}$ for $k$ the
smallest integer making this point in $C$. For such a $C$, let
$\lambda_{C}=\\{\gamma_{\scriptscriptstyle{C}}(y,i):y\in\mathbb{Z}^{d},i\in[d]\\}$.

It is trivial that $B_{+}$ is solid above and $B_{-}$ is solid below for any
set $B\subset\mathbb{Z}^{d}$. The following are also trivial observations:

if $G$ and $G^{\prime}$ are solid above then so is $G\cup G^{\prime}$,  (6) if
$G$ and $G^{\prime}$ are bounded below then so is $G\cup G^{\prime}$,  (7)

######  Definition 4 (Good set).

A set $C\subset\mathbb{Z}^{d}$ is said to be good  if $C$ is saturated, solid
above, and bounded below.

It follows from the above that

if $G$ and $G^{\prime}$ are good then so is $G\cup G^{\prime}$.  (8)

######  Definition 5 (Terrace).

A set $\Lambda\subset\mathbb{Z}^{d}$ is a terrace if $\Lambda=\lambda_{C}$ for
some good set $C$.

The following is essentially a reformulation of Theorem  ‣ 1.1, in the
language of terraces, though we will not prove this until Section 2.4.

######  Proposition 6.

For the half-orthant model, Example 2:

  1. (i)

$\mathcal{C}_{o}$ is solid above for each $p\in[0,1]$ and is saturated for
$p<1$;

  2. (ii)

If $p\in(p_{c},1)$ then $\mathcal{C}_{o}$ is good,
$\gamma_{\scriptscriptstyle{\mathcal{C}_{o}}}(x,i)=x+L_{x}^{(i)}$, and
$\lambda_{\mathcal{C}_{o}}=\\{\gamma_{\scriptscriptstyle{\mathcal{C}_{o}}}(x,i):x\in\mathbb{Z}^{d},i\in[d]\\}\subset\mathcal{C}_{o}$
is a terrace with $\lambda_{\mathcal{C}_{o}}\subset\Omega_{1}$ and
$(\lambda_{\mathcal{C}_{o}})_{+}=\mathcal{C}_{o}$.

  3. (iii)

If $p<p_{c}$ then $\mathcal{C}_{o}(p)=\mathbb{Z}^{d}$, and $\Omega_{1}$
contains no terrace.

###### Proof of Theorem 1.

The statement of Theorem 1 for the half-orthant model follows immediately from
Proposition 6. The corresponding statement for the orthant model follows in
turn, using results of [13] which establish that $(\mathcal{C}_{o}(p))_{+}$
for the orthant model coincides with $\mathcal{C}_{o}(p)$ for the half-orthant
model. ∎

Figure 1: A simulation of a portion of $\mathcal{C}_{o}$ for the half-orthant
model of Example 2 in 3 dimensions when $p=0.95$. The structure of the terrace
$\lambda_{\mathcal{C}_{o}}$ is clearly visible.

Figure 1 depicts a realisation of part of the boundary of $\mathcal{C}_{o}$
for the orthant model in 3 dimensions with $p=0.95$.

###  2.1 Facts about terraces

This section consists of deterministic geometric properties about terraces.
For $a,b\in\mathbb{Z}^{d}$ with $b\in a_{+}$ define the box $[a,b]$ with
corners $a$ and $b$ by

$[a,b]=\\{z\in\mathbb{Z}^{d}:a^{\scriptscriptstyle{[}i]}\leq
z^{\scriptscriptstyle{[}i]}\leq b^{\scriptscriptstyle{[}i]}\text{ for each
}i\in[d]\\}.$ (9)

Note that $[a,b]$ is non-empty, but consists of a single point in the case
$a=b$.

######  Lemma 7.

If $\Lambda\subset\mathbb{Z}^{d}$ is a terrace then:

  1. (i)

$\Lambda$ is saturated, bounded below, and for each $z\in\Lambda$ there is an
$i\in[d]$ with $z-e_{i}\notin\Lambda_{+}$;

  2. (ii)

If $C$ is good and such that $\Lambda=\lambda_{C}$, then $C=\Lambda_{+}$ (and
hence $\Lambda_{+}$ is good);

  3. (iii)

For any line $L$ parallel to an axis, $\Lambda\cap L$ is finite;

  4. (iv)

$\Lambda_{+}\cap\Lambda_{-}=\Lambda$;

  5. (v)

$(\Lambda_{+})^{c}$ is solid below, and $-((\Lambda_{+})^{c}\cup\Lambda)$ is
good;

  6. (vi)

$\Lambda_{+}\setminus\Lambda$ is solid above;

  7. (vii)

If a nearest neighbour path starts in $\Lambda_{+}$ and ends in
$(\Lambda_{+})^{c}$, and if $k$ is the first time that the path is in
$(\Lambda_{+})^{c}$ then the path is in $\Lambda$ at time $k-1$;

  8. (viii)

$\Lambda_{+}\cap(x_{-})$ is finite for every $x$.

  9. (ix)

If $x\in\Lambda_{+}$, $y\in\Lambda$, and $y\in x_{+}$ then
$[x,y]\subset\Lambda$.

###### Proof of Lemma 7.

Suppose that $\Lambda$ is a terrace. Then $\Lambda=\lambda_{C}$ for some good
set $C$.

Let $y\in\mathbb{Z}^{d}$ and $i\in[d]$. Then
$\gamma_{\scriptscriptstyle{C}}(y,i)\in\Lambda$, so $\Lambda$ is saturated.
Next, $\Lambda\subset C$ and $C$ is bounded below so $\Lambda$ is also bounded
below. Now let $z\in\Lambda$. Then $z=\gamma_{\scriptscriptstyle{C}}(y,i)$ for
some $y\in\mathbb{Z}^{d}$ and $i\in[d]$, so $z-e_{i}\notin C$. Since $C=C_{+}$
(solid above), $z-e_{i}\notin C_{+}$ and therefore $z-e_{i}\notin\Lambda_{+}$
by (5). This verifies (i).

For (ii), note that by (5), $\Lambda_{+}\subset C_{+}$. Since $C$ is solid
above, $\Lambda_{+}\subset C$. Next, if $z\in C$ and $i\in[d]$ then
$z^{\prime}:=\gamma_{\scriptscriptstyle{C}}(z,i)\in\Lambda$ and $z\in
z^{\prime}_{+}$, so $z\in\Lambda_{+}$. Thus, $C\subset\Lambda_{+}$.

For (iii), suppose without loss of generality that $L$ is the line
$\mathbb{Z}e_{1}$, and suppose that this line contains infinitely many points
in $\Lambda$. Then (since $\Lambda$ is bounded below) there exists some
$j\in[d]$ and an increasing sequence $(n_{r})_{r\in\mathbb{N}}$ such that
$n_{r}e_{1}=\gamma_{\scriptscriptstyle{\Lambda_{+}}}(n_{r}e_{1},j)$. Now
consider the line $L^{\prime}=\mathbb{Z}e_{1}-e_{j}$. Since $\Lambda$ is
saturated, there exists $x\in L^{\prime}\cap\Lambda\subset\Lambda_{+}$, and
since $\Lambda_{+}$ is solid above, $x+ne_{1}\in\Lambda_{+}$ for every
$n\in\mathbb{N}$. Choose $n_{r}$ such that $n_{r}>x\cdot e_{1}$. Then
$n_{r}e_{1}=\gamma_{\scriptscriptstyle{\Lambda_{+}}}(n_{r}e_{1},j)$ but
$n_{r}e_{1}-e_{j}\in\Lambda_{+}$, which is a contradiction.

To prove (iv), it is trivial that $\Lambda\subset\Lambda_{+}\cap\Lambda_{-}$.
To prove the reverse inclusion, suppose that $x\in\Lambda_{-}$, so that $x\in
z_{-}$ for some $z\in\Lambda$. Then by the definition of a terrace and (ii),
for some $j\in[d]$ we have $z-e_{j}\notin\Lambda_{+}$, and thus
$x-e_{j}\notin\Lambda_{+}$. Suppose also that
$x\in\Lambda_{+}\setminus\Lambda$. Then for each $i\in[d]$,
$\gamma_{\scriptscriptstyle{\Lambda_{+}}}(x,i)\neq x$, so there exists
$n_{i}\in\mathbb{N}$ such that $x-n_{i}e_{i}\in\Lambda_{+}$, and since
$\Lambda_{+}$ is solid above we have that $x-e_{i}\in\Lambda_{+}$ for each
$i$. Thus $x-e_{j}\in\Lambda_{+}$, which is a contradiction.

For (v), let $z\in(\Lambda_{+})^{c}$, and let $x\in z_{-}$. Then $z\in x_{+}$,
so if $x\in\Lambda_{+}$ then $z\in\Lambda_{+}$ as well. This is a
contradiction, hence $x\in(\Lambda_{+})^{c}$, so $(\Lambda_{+})^{c}$ is solid
below. For the second claim it suffices to show that
$(\Lambda_{+})^{c}\cup\Lambda$ is saturated, bounded above and solid below.
Note that $(\Lambda_{+})^{c}\cup\Lambda$ is saturated since $\Lambda$ is (by
(i)). Let $x\in\mathbb{Z}^{d}$ and $i\in[d]$. Then
$\gamma_{\scriptscriptstyle{\Lambda_{+}}}(x,i)\in\Lambda$ and
$\\{\gamma_{\scriptscriptstyle{\Lambda_{+}}}(x,i)+ke_{i}:k\geq
0\\}\subset\Lambda_{+}$ since $\Lambda_{+}$ is solid above. Thus,
$\\{k\in\mathbb{Z}:x+ke_{i}\in(\Lambda_{+})^{c}\\}$ is bounded above. By (iii)
we also have that $\\{k\in\mathbb{Z}:x+ke_{i}\in\Lambda\\}$ is bounded above.
Thus $(\Lambda_{+})^{c}\cup\Lambda$ is bounded above. Suppose that
$(\Lambda_{+})^{c}\cup\Lambda$ is not solid below. Then there exist
$x\in(\Lambda_{+})^{c}\cup\Lambda$ and $y\in x_{-}$ such that
$y\notin(\Lambda_{+})^{c}\cup\Lambda$ (i.e.
$y\in\Lambda_{+}\setminus\Lambda$). Since $(\Lambda_{+})^{c}$ is solid below
we must have that $x\in\Lambda$. Since $y\in x_{-}$ this means that
$y\in\Lambda_{-}$. Thus $y\in(\Lambda_{+}\setminus\Lambda)\cap(\Lambda_{-})$,
which contradicts (iv).

For (vi), let $x\in\Lambda_{+}\setminus\Lambda$. Then $x-e_{i}\in\Lambda_{+}$
for every $i\in[d]$. Let $y\in x_{+}$, so $y\in\Lambda_{+}$. If $y\in\Lambda$
then there is a $j$ with $y-e_{j}\notin\Lambda_{+}$. But then
$x-e_{j}\notin\Lambda_{+}$ as well (as $(x-e_{j})^{[i]}\leq(y-e_{j})^{[i]}$
for every $i$). This gives a contradiction, so
$y\in\Lambda_{+}\setminus\Lambda$ for every $y\in x_{+}$, as required.

For (vii), there is a first time $k$ the path enters $(\Lambda_{+})^{c}$, so
at $k-1$ it is at a point $x\in\Lambda_{+}$. Since $x+e_{i}\in\Lambda_{+}$ for
each $i\in[d]$ the $k$th step must be from $x$ to some
$x-e_{j}\in(\Lambda_{+})^{c}$. Since $x-e_{j}\in\Lambda_{+}^{c}$ we have that
$\gamma_{\scriptscriptstyle{\Lambda_{+}}}(x,j)=x$, so
$x\in\lambda_{\Lambda_{+}}=\Lambda$.

Turning to (viii), if $\Lambda_{+}\cap x_{-}$ is infinite then there is an $i$
such that the set of reals $e_{i}\cdot(\Lambda_{+}\cap x_{-})$ fails to be
bounded below. Let $n\in\mathbb{N}$. Then there exists $y\in\Lambda_{+}\cap
x_{-}$ with $y^{\scriptscriptstyle{[}i]}<x^{\scriptscriptstyle{[}i]}-n$, and
since $\Lambda_{+}$ is solid above it follows that $x-ne_{i}\in
y_{+}\subset\Lambda_{+}$. Having this for every $n\in\mathbb{N}$ contradicts
the assumption that $\Lambda_{+}$ is bounded below.

To see (ix), note first that since $y\in x_{+}$, $[x,y]\subset x_{+}$ by
definition. Let $z\in[x,y]$. Then $z\in x_{+}\subset\Lambda_{+}$, and $y\in
z_{+}$. If $z\in\Lambda_{+}\setminus\Lambda$, then (vi) would imply that $y\in
z_{+}\subset\Lambda_{+}\setminus\Lambda$, which is impossible. So
$z\in\Lambda$. ∎

######  Remark 8.

It follows from Lemma 7(i),(vii) that if $\Lambda$ is a terrace then

$x\in\Lambda\iff x\in\Lambda_{+}\text{ and }x-e_{i}\notin\Lambda_{+}\text{ for
some }i.$

######  Lemma 9.

If
$\Lambda^{\scriptscriptstyle{(}0)}\supset\Lambda^{\scriptscriptstyle{(}1)}\supset\dots$
is a decreasing sequence of terraces, then
$\hat{\Lambda}:=\cap_{s=0}^{\infty}\Lambda^{\scriptscriptstyle{(}s)}$
satisfies $\hat{\Lambda}=\lambda_{G}$, where
$G:=\cap_{s=0}^{\infty}\Lambda^{\scriptscriptstyle{(}s)}_{+}$ is good.

###### Proof.

Clearly $G$ is bounded below since $\Lambda^{\scriptscriptstyle{(}0)}_{+}$ is.
If $x\in G$ and $y\in x_{+}$ then $x\in\Lambda^{\scriptscriptstyle{(}s)}_{+}$
for every $s$, and since $\Lambda^{\scriptscriptstyle{(}s)}_{+}$ is solid
above we have $y\in\Lambda^{\scriptscriptstyle{(}s)}_{+}$ (for each $s$). Thus
$G$ is solid above. To show that $G$ is saturated, it is sufficient to show
that $\hat{\Lambda}$ is saturated (since $G\supset\hat{\Lambda}$). Let $L$ be
a line parallel to one of the axes. Then
$L\cap\Lambda^{\scriptscriptstyle{(}s)}$ is finite, non-empty, and the sets
$L\cap\Lambda^{\scriptscriptstyle{(}s)}$ are decreasing in $s$. Hence
$\cap_{s=0}^{\infty}(L\cap\Lambda^{\scriptscriptstyle{(}s)})=L\cap\hat{\Lambda}$
is non-empty as required. We have shown that $G$ is good.

To show that $\hat{\Lambda}=\lambda_{G}$, note that the
$\Lambda^{\scriptscriptstyle{(}s)}_{+}$ are decreasing as well. Suppose that
$x\in\cap_{s=0}^{\infty}\Lambda^{\scriptscriptstyle{(}s)}$. Then $x\in G$ and
$x=\gamma_{\Lambda^{\scriptscriptstyle{(}0)}_{+}}(x,i)$ for some $i\in[d]$, so
$x-e_{i}\notin\Lambda^{\scriptscriptstyle{(}0)}_{+}$ and hence $x-e_{i}\notin
G$. Therefore $x=\gamma_{G}(x,i)\in\lambda_{G}$.

Conversely, suppose that $x\in\lambda_{G}$. Then $x\in G$ and
$x=\gamma_{G}(x,i)$ for some $i\in[d]$, so $x-e_{i}\notin G$. Therefore
$x\in\Lambda^{\scriptscriptstyle{(}s)}_{+}$ for every $s$, and there exists
$s_{x}$ such that for every $s>s_{x}$,
$x-e_{i}\notin\Lambda^{\scriptscriptstyle{(}s)}_{+}$. Therefore
$x\in\Lambda^{\scriptscriptstyle{(}s)}$ for every $s>s_{x}$. But the
$\Lambda^{\scriptscriptstyle{(}s)}$ are decreasing, so
$x\in\Lambda^{\scriptscriptstyle{(}s)}$ for every $s$, so
$x\in\cap_{s=0}^{\infty}\Lambda^{\scriptscriptstyle{(}s)}$. ∎

If $\Lambda$ is a terrace then define

$H_{\Lambda}=\\{x\in\Lambda:x+e_{i}\in\Lambda\text{ for every }i\in[d]\\}.$

A terrace $\Lambda\subset\mathbb{Z}^{d}$ is said to be irreducible if
$H_{\Lambda}$ is empty.

######  Lemma 10.

Let $\Lambda$ be a terrace. Then $-\Lambda$ is a terrace if and only if
$H_{\Lambda}=\varnothing$.

###### Proof.

Suppose that $x\in H_{\Lambda}$. Then $x\in\Lambda$ and $x+e_{i}\in\Lambda$
for every $i\in[d]$. Thus $-x\in-\Lambda$ and $-x-e_{i}\in-\Lambda$ for every
$i\in[d]$. Thus by Lemma 7(i), $-\Lambda$ is not a terrace.

Now suppose that $H_{\Lambda}$ is empty. By Lemma 7(v) we have that
$K:=-((\Lambda_{+})^{c}\cup\Lambda)$ is good. We claim that
$-\Lambda=\lambda_{K}$. Let $u\in-\Lambda\subset K$. Then $v:=-u\in\Lambda$
and since $H_{\Lambda}$ is empty there exists $i\in[d]$ such that
$v+e_{i}\in\Lambda_{+}\setminus\Lambda$. Therefore
$v+e_{i}\notin(\Lambda_{+})^{c}\cup\Lambda=-K$ and thus $u-e_{i}\notin K$.
Therefore $u=\gamma_{\scriptscriptstyle{K}}(u,i)\in\lambda_{K}$. Thus,
$-\Lambda\subset\lambda_{K}$. Now let $u\in\lambda_{K}$. Then $u\in K$ and
there exists $i\in[d]$ such that $u-e_{i}\notin K$. Thus,
$v=-u\in(\Lambda_{+})^{c}\cup\Lambda$ and
$v+e_{i}\notin(\Lambda_{+})^{c}\cup\Lambda$. Then
$v+e_{i}\in\Lambda_{+}\setminus\Lambda$ which implies that
$v=(v+e_{i})-e_{i}\in\Lambda_{+}$. Since also
$v=\in(\Lambda_{+})^{c}\cup\Lambda$ we conclude that $v\in\Lambda$ and
therefore $u\in-\Lambda$. This confirms that $\lambda_{K}\subset-\Lambda$. ∎

We’ll only require a special case of the following construction (see Lemma
27), but give the general version here for completeness.

######  Lemma 11.

Let $\Lambda$ be a terrace, and suppose $x,y\in\Lambda$. There is a path
$z_{\scriptscriptstyle{(}0)},\dots,z_{\scriptscriptstyle{(}n)}$ of points in
$\Lambda$ such that

  1. (a)

$z_{\scriptscriptstyle{(}0)}=x$, $z_{\scriptscriptstyle{(}n)}=y$;

  2. (b)

the $\ell_{1}$ distance from $x$ to $z_{\scriptscriptstyle{(}k)}$ is strictly
increasing in $k$, and the $\ell_{1}$ distance from
$z_{\scriptscriptstyle{(}k)}$ to $y$ is strictly decreasing;

  3. (c)

each increment $z_{\scriptscriptstyle{(}j+1)}-z_{\scriptscriptstyle{(}j)}$ has
the form $e_{i}$, $-e_{k}$, or $e_{i}-e_{k}$ for some $i\neq k$;

  4. (d)

each coordinate $z_{\scriptscriptstyle{(}j)}^{\scriptscriptstyle{[}i]}$ is
monotone in $j$.

###### Proof.

Property (d) follows from the others, as they together imply that the $e_{i}$
or $e_{i}-e_{k}$ are only used when
$x^{\scriptscriptstyle{[}i]}<z^{\scriptscriptstyle{[}i]}<y^{\scriptscriptstyle{[}i]}$
and the $-e_{k}$ or $e_{i}-e_{k}$ are only used when
$x^{\scriptscriptstyle{[}k]}>z^{\scriptscriptstyle{[}k]}>y^{\scriptscriptstyle{[}k]}$.
We therefore will ignore (d) in the rest of the proof.

By Lemma 7(i) we cannot have
$x^{\scriptscriptstyle{[}i]}<y^{\scriptscriptstyle{[}i]}$ for every $i$, nor
can we have $x^{\scriptscriptstyle{[}i]}>y^{\scriptscriptstyle{[}i]}$ for
every $i$. If $x^{\scriptscriptstyle{[}i]}\leq y^{\scriptscriptstyle{[}i]}$
for every $i$, fix an $i$ with
$x^{\scriptscriptstyle{[}i]}<y^{\scriptscriptstyle{[}i]}$ (if no such $i$
exists then $x=y$ and we are done) and let
$z_{\scriptscriptstyle{(}1)}=x+e_{i}\in\Lambda_{+}$. Then
$z^{\scriptscriptstyle{[}j]}_{\scriptscriptstyle{(}1)}\leq
y^{\scriptscriptstyle{[}j]}$ for every $j$, so by Lemma 7(vi) we see that
$z_{\scriptscriptstyle{(}1)}\in\Lambda$. Iterating this argument, we keep
stepping in direction $e_{i}$ till the $i$-coordinates agree. Then pick
another coordinate to increase, etc. Eventually all coordinates will agree,
and we’ve reached $y$.

If $x^{\scriptscriptstyle{[}i]}\geq y^{\scriptscriptstyle{[}i]}$ for every
$i$, construct a path from $y$ to $x$ as above, and traverse it backwards.

Now assume that $x^{\scriptscriptstyle{[}i]}<y^{\scriptscriptstyle{[}i]}$ and
$x^{\scriptscriptstyle{[}k]}>y^{\scriptscriptstyle{[}k]}$. Define

$z_{\scriptscriptstyle{(}1)}=\begin{cases}x-e_{k},&\text{ if
}x-e_{k}\in\Lambda_{+}\\\ x-e_{k}+e_{i},&\text{ if
}x-e_{k}\notin\Lambda_{+},\,x-e_{k}+e_{i}\in\Lambda_{+}\\\ x+e_{i},&\text{
otherwise}.\end{cases}$ (10)

If $x-e_{k}\in\Lambda_{+}$ then
$z_{\scriptscriptstyle{(}1)}:=x-e_{k}\in\Lambda$ by Lemma 7(v). If
$x-e_{k}\notin\Lambda_{+}$ and $x-e_{k}+e_{i}\in\Lambda_{+}$, then
$z_{\scriptscriptstyle{(}1)}:=x-e_{k}+e_{i}=\gamma_{\scriptscriptstyle{\Lambda_{+}}}(x-e_{k},i)\in\Lambda$.
Finally, if $x-e_{k}+e_{i}\notin\Lambda_{+}$ (and $x\in\Lambda_{+}$), then
$x+e_{i}\in\Lambda_{+}$ and
$z_{\scriptscriptstyle{(}1)}:=x+e_{i}=\gamma_{\scriptscriptstyle{\Lambda_{+}}}(x+e_{i},k)\in\Lambda$,
as required. Either this reaches $y$, or we can repeat one or the other of the
above arguments, to find $z_{\scriptscriptstyle{(}2)}$, and then
$z_{\scriptscriptstyle{(}3)}$, etc. ∎

######  Lemma 12.

Let $\Lambda$ be a terrace. If $x,y\in\mathbb{Z}^{d}\setminus\Lambda_{+}$ or
$x,y\in\Lambda_{+}\setminus\Lambda$ then there is a self avoiding nearest
neighbour path connecting $x$ and $y$ which is disjoint from $\Lambda$.

###### Proof.

Suppose $x,y\in\mathbb{Z}^{d}\setminus\Lambda_{+}$. Set
$z_{\scriptscriptstyle{(}0)}=x$. Given $z_{\scriptscriptstyle{(}k)}$, take
$z_{\scriptscriptstyle{(}k+1)}=z_{\scriptscriptstyle{(}k)}-e_{i}$ as long as
we can find an $i$ with
$z_{\scriptscriptstyle{(}k)}^{\scriptscriptstyle{[}i]}>y^{\scriptscriptstyle{[}i]}$.
Because $x\in(z_{\scriptscriptstyle{(}k)})_{+}$ we have
$z_{\scriptscriptstyle{(}k)}\in\mathbb{Z}^{d}\setminus\Lambda_{+}$. Once we’ve
run out of such $i$, start taking
$z_{\scriptscriptstyle{(}k+1)}=z_{\scriptscriptstyle{(}k)}+e_{j}$ where
$z_{\scriptscriptstyle{(}k)}^{\scriptscriptstyle{[}j]}<y^{\scriptscriptstyle{[}j]}$.
Because $y\in(z_{\scriptscriptstyle{(}k)})_{+}$ we still have
$z_{\scriptscriptstyle{(}k)}\in\mathbb{Z}^{d}\setminus\Lambda_{+}$. Eventually
this sequence reaches $y$.

The same argument applies if $x,y\in\Lambda_{+}\setminus\Lambda$, except first
we increase coordinates and then we decrease them. ∎

###  2.2 Corners

A point $z$ is a corner of $C\subset\mathbb{Z}^{d}$ if $z\in C$ but for each
$i$, $z-e_{i}\notin C$. In this section, we examine corners of terraces. We
will really use local versions of these results (stated in the next section),
but we will prove the terrace versions first, as they are both simpler and of
more intrinsic interest.

######  Lemma 13.

Let $\Lambda$ be a terrace, with a corner at $z$. Then:

  * (i)

$\Lambda_{+}\setminus\\{z\\}$ is good;

  * (ii)

$\Lambda\setminus\\{z\\}\subset\lambda_{\Lambda_{+}\setminus\\{z\\}}\subset(\Lambda\setminus\\{z\\})\cup\\{z+e_{j}:j\in[d]\\}$.

###### Proof.

$\Lambda_{+}\setminus\\{z\\}$ is bounded below because $\Lambda_{+}$ is, and
it is saturated because $\Lambda_{+}$ was saturated and
$\Lambda_{+}\setminus\\{z\\}$ contains $z+e_{j}$ for each $j\in[d]$. So let us
prove that $\Lambda_{+}\setminus\\{z\\}$ is solid above.

Let $x\in\Lambda_{+}\setminus\\{z\\}$ and $y\in x_{+}$. Then
$y\in\Lambda_{+}$, and we need only show that $y\neq z$. So assume $z=y$. Then
$z\in x_{+}$ and $x\neq z$, so we have $z\in(x+e_{i})_{+}$ for some $i$. Hence
$z-e_{i}\in x_{+}\subset\Lambda_{+}$. But $z-e_{i}\notin\Lambda$, because $z$
is a corner, so $z-e_{i}\in\Lambda_{+}\setminus\Lambda$. By Lemma 7 (vi) we
conclude that $z\in\Lambda_{+}\setminus\Lambda$ which is a contradiction. So
$z\neq y$ as required.

To prove (ii), note that if $x\in\Lambda\setminus\\{z\\}$ then
$x=\gamma_{\scriptscriptstyle{\Lambda_{+}}}(x,i)$ for some $i$. Thus
$x=\gamma_{\Lambda_{+}\setminus\\{z\\}}(x,i)$ as well, so
$x\in\lambda_{\Lambda_{+}\setminus\\{z\\}}$. This proves the first half of
(ii). For the second half, let $x\in\lambda_{\Lambda_{+}\setminus\\{z\\}}$.
Then $x\in\Lambda_{+}\setminus\\{z\\}$ and
$x-e_{j}\notin\Lambda_{+}\setminus\\{z\\}$ for some $j$. The latter implies
that either $x-e_{j}\notin\Lambda_{+}$ or $x-e_{j}=z$. In the first case,
$x\in\Lambda\setminus\\{z\\}$ and in the second, $x=z+e_{j}$. Either way, (ii)
is shown. ∎

For a terrace $\Lambda$ containing a corner $z\in\Lambda$, define

$\Lambda_{(\uparrow z)}=\lambda_{\Lambda_{+}\setminus\\{z\\}}.$ (11)

We describe $\Lambda_{(\uparrow z)}$ as the terrace resulting from pushing up
the corner $z$ of the terrace $\Lambda$. See e.g. Figure 2.

$x$$z$ Figure 2: Part of a terrace $\Lambda$ with $d=2$ in which $x\in
H_{\Lambda}$ and $z$ are corners. The dashed lines show the terrace with both
$x$ and $z$ pushed up.

######  Lemma 14.

Let $\Lambda$ be a terrace and $x\in H_{\Lambda}$. Then either $x$ is a corner
of $\Lambda$ or there exists $i\in[d]$ such that $x-e_{i}\in H_{\Lambda}$.

###### Proof.

If $x$ is not a corner of $\Lambda$ then there exists $i\in[d]$ such that
$x-e_{i}\in\Lambda$. Let $j\in[d]$. Then $x+e_{j}\in\Lambda$ since $x\in
H_{\Lambda}$. Also $x-e_{i}+e_{j}\in\Lambda_{+}$ since $x-e_{i}\in\Lambda$.
These two facts together with Lemma 7(vi) show that $x-e_{i}+e_{j}\in\Lambda$.
Since this holds for every $j\in[d]$ it follows that $x-e_{i}\in H_{\Lambda}$.
∎

######  Lemma 15.

If $\Lambda$ is a terrace then:

  1. (i)

if $x\in H_{\Lambda}$ and $x$ is a corner of $\Lambda$ then
$\Lambda_{(\uparrow x)}=\Lambda\setminus\\{x\\}$;

  2. (ii)

there exists an irreducible terrace $\Lambda^{\prime}\subset\Lambda$.

###### Proof.

From Lemma 13 we know that $\Lambda\setminus\\{x\\}\subset\Lambda_{(\uparrow
x)}\subset(\Lambda\setminus\\{x\\})\cup\\{x+e_{j}:j\in[d]\\}$. But the fact
that $x\in H_{\Lambda}$ means that each $x+e_{j}$ already
$\in\Lambda\setminus\\{x\\}$. Therefore (i) holds.

To prove (ii), let $\Lambda^{\scriptscriptstyle{(}0)}:=\Lambda$. If
$H_{\Lambda^{\scriptscriptstyle{(}0)}}$ is empty then
$\Lambda^{\scriptscriptstyle{(}0)}=\Lambda$ is irreducible and we are done.
Otherwise we may enumerate the vertices of
$H_{\Lambda^{\scriptscriptstyle{(}0)}}$ as
$(x^{\scriptscriptstyle{(}m)})_{m\in\mathbb{N}}$.

We proceed recursively. Given
$\Lambda^{\scriptscriptstyle{(}n)}\subset\Lambda^{\scriptscriptstyle{(}0)}$
with $H_{\Lambda^{\scriptscriptstyle{(}n)}}$ nonempty, we let
$M_{n}=\min\\{m:x^{\scriptscriptstyle{(}m)}\in
H_{\Lambda^{\scriptscriptstyle{(}n)}}\\}$ (which is finite, since
$H_{\Lambda^{\scriptscriptstyle{(}n)}}\subset
H_{\Lambda^{\scriptscriptstyle{(}0)}}$). If
$y^{\scriptscriptstyle{(}n,0)}:=x^{\scriptscriptstyle{(}M_{n})}$ is a corner
of $\Lambda^{\scriptscriptstyle{(}n)}$ then define
$\Lambda^{\scriptscriptstyle{(}n+1)}=\Lambda^{\scriptscriptstyle{(}n)}\setminus\\{x^{\scriptscriptstyle{(}M_{n})}\\}$.
Otherwise, by Lemma 14, there exists $i_{1}\in[d]$ such that
$y^{\scriptscriptstyle{(}n,1)}:=x^{\scriptscriptstyle{(}M_{n})}-e_{i_{1}}\in
H_{\Lambda^{\scriptscriptstyle{(}n)}}$. If $y^{\scriptscriptstyle{(}n,1)}$ is
a corner of $\Lambda^{\scriptscriptstyle{(}n)}$ then define
$\Lambda^{\scriptscriptstyle{(}n+1)}=\Lambda^{\scriptscriptstyle{(}n)}\setminus\\{y^{\scriptscriptstyle{(}n,1)}\\}$.
Otherwise we continue to find points $y^{\scriptscriptstyle{(}n,s)}$ for
$s\in\mathbb{N}$ such that $y^{\scriptscriptstyle{(}n,s)}\in
H_{\Lambda^{\scriptscriptstyle{(}n)}}$ and
$y^{\scriptscriptstyle{(}n,s)}=y^{\scriptscriptstyle{(}n,s-1)}-e_{i_{s}}$ for
some $i_{s}$. Let $s_{n}=\inf\\{s:y^{\scriptscriptstyle{(}n,s)}\text{ is a
corner of $\Lambda^{\scriptscriptstyle{(}n)}$}\\}$. Then $s_{n}$ is finite by
Lemma 7(viii), because each $y^{\scriptscriptstyle{(}n,s)}\in
x^{\scriptscriptstyle{(}M_{n})}_{-}$. We define
$\Lambda^{\scriptscriptstyle{(}n+1)}=\Lambda^{\scriptscriptstyle{(}n)}\setminus\\{y^{\scriptscriptstyle{(}n,s_{n})}\\}$.
By (i), $\Lambda^{\scriptscriptstyle{(}n+1)}$ is a terrace.

Thus we have produced a decreasing sequence
$(\Lambda^{\scriptscriptstyle{(}n)})_{n\in\mathbb{Z}_{+}}$ of terraces with
$\Lambda^{\scriptscriptstyle{(}0)}=\Lambda$. By Lemma 9, the intersection of
these, $\Lambda^{\prime}$ is a terrace. We claim that $\Lambda^{\prime}$ is
irreducible, i.e. that $H_{\Lambda^{\prime}}=\varnothing$. To see this, first
note that the sets $H_{\Lambda^{\scriptscriptstyle{(}n)}}$ are decreasing
since $\Lambda^{\scriptscriptstyle{(}n)}$ are, and similarly
$H_{\Lambda^{\prime}}\subset H_{\Lambda^{\scriptscriptstyle{(}n)}}$ for each
$n$ since $\Lambda^{\prime}\subset\Lambda^{\scriptscriptstyle{(}n)}$ for each
$n$. To complete the proof it is sufficient to show that for each $m$,
$x^{\scriptscriptstyle{(}m)}\notin H_{\Lambda^{\scriptscriptstyle{(}n)}}$ for
all $n$ sufficiently large (since then $H_{\Lambda^{\prime}}=\varnothing$). To
verify this, it is (without loss of generality) sufficient to show that
$x^{\scriptscriptstyle{(}1)}\notin H_{\Lambda^{\scriptscriptstyle{(}n)}}$ for
all $n$ sufficiently large. But
$x^{\scriptscriptstyle{(}1)}_{-}\cap\Lambda_{+}$ is finite by Lemma 7(viii).
It follows that $x^{\scriptscriptstyle{(}1)}\notin
H_{\Lambda^{\scriptscriptstyle{(}n)}}$ for $n>K_{1}+1$, where $K_{1}$ is the
cardinality of the set $x^{\scriptscriptstyle{(}1)}_{-}\cap\Lambda_{+}$ (since
by this time we have removed
$\\{y^{\scriptscriptstyle{(}1,s_{1})},y^{\scriptscriptstyle{(}1,s_{1}-1)},\dots,y^{\scriptscriptstyle{(}1,0)}\\}$).
∎

###  2.3 Local terraces

In this section $Q$ denotes a box of the form $[a,b]$ for some
$a,b\in\mathbb{Z}^{d}$ (recall (9)). Its boundary $\partial Q$ is the set of
sites in $Q$ that are adjacent to sites in $\mathbb{Z}^{d}\setminus Q$. Its
interior is $Q^{o}:=Q\setminus\partial Q$.

We require a local formulation for the notion of a terrace.

######  Definition 16.

We call $\Delta\subset Q$ a $Q$-terrace if $\Delta=\Lambda\cap Q$ for some
terrace $\Lambda$.

Note that $\varnothing$ is by definition a $Q$-terrace for any $Q$, although
it is not an interesting one. The first observation is elementary.

######  Lemma 17.

If $\Lambda$ is a terrace and $\Delta:=\Lambda\cap Q\neq\varnothing$ then
$\Lambda_{+}\cap Q=\Delta_{+}\cap Q$.

###### Proof.

Let $Q=[a,b]$. If $a\in\Lambda_{+}$ then Lemma 7(i) and the fact that
$\Lambda\cap Q$ is non-empty implies that $a\in\Lambda\cap Q=\Delta$ and in
this case the claim reduces to $Q=Q$. Otherwise $a\notin\Lambda_{+}$. Let
$y\in\Lambda_{+}\cap Q$. Find a nearest neighbour path, none of whose
coordinates increases, from $y$ to $a$. This path is necessarily in $Q$. By
Lemma 7(vii) this path must hit some point $x\in\Lambda\cap Q=\Delta$.
Therefore $y\in x_{+}\subset\Delta_{+}$. This proves that $\Lambda_{+}\cap
Q\subset\Delta_{+}\cap Q$ and the reverse inclusion is trivial. ∎

A number of results that we proved for terraces immediately follow for
$Q$-terraces, leading to the following result.

######  Lemma 18.

Fix a box $Q$.

  * (i)

If
$\Delta^{\scriptscriptstyle{(}0)}\supset\Delta^{\scriptscriptstyle{(}1)}\supset\dots$
is a decreasing sequence of $Q$-terraces, then
$\hat{\Delta}:=\cap_{s=0}^{\infty}\Delta^{\scriptscriptstyle{(}s)}$ is a
$Q$-terrace.

  * (ii)

Let $\Delta$ be a $Q$-terrace, and suppose $x,y\in\Delta$. There is a path
$z_{\scriptscriptstyle{(}0)},\dots,z_{\scriptscriptstyle{(}n)}$ of points in
$\Delta$ such that

    * (a)

$z_{\scriptscriptstyle{(}0)}=x$, $z_{\scriptscriptstyle{(}n)}=y$;

    * (b)

the $\ell_{1}$ distance from $x$ to $z_{\scriptscriptstyle{(}k)}$ is strictly
increasing in $k$, and the $\ell_{1}$ distance from
$z_{\scriptscriptstyle{(}k)}$ to $y$ is strictly decreasing;

    * (c)

each increment $z_{\scriptscriptstyle{(}j+1)}-z_{\scriptscriptstyle{(}j)}$ has
the form $e_{i}$, $-e_{k}$, or $e_{i}-e_{k}$ for some $i\neq k$;

    * (d)

each coordinate $z_{\scriptscriptstyle{(}j)}^{\scriptscriptstyle{[}i]}$ is
monotone in $j$.

  * (iii)

Let $\Delta$ be a $Q$-terrace. If $x,y\in\mathbb{Z}^{d}\setminus\Delta_{+}$ or
$x,y\in\Delta_{+}\setminus\Delta$ then there is a self avoiding nearest
neighbour path connecting $x$ and $y$ which is disjoint from $\Delta$. If in
addition $x,y\in Q$ then there exists such a path in $Q$.

###### Proof.

A decreasing sequence of $Q$-terraces is trivially a $Q$-terrace (compare with
Lemma 9), because the sequence terminates. This proves (i). Property (ii)
holds with the same proof as Lemma 11, since the paths constructed there stay
in $Q$ if they start and end in $Q$. The proof of Lemma 12 carries over in the
same way, showing (iii). ∎

A set $A\subset Q$ is said to be _$Q$ -solid above_ if $A_{+}\cap Q=A$.

######  Lemma 19.

If $B$ is solid above then $A:=B\cap Q$ is $Q$-solid above.

###### Proof.

If $B$ is empty then the statement is vacuously true. Since $A_{+}\cap
Q\supset A$ we need only show that $A_{+}\cap Q\subset A$. Let $x\in A_{+}\cap
Q$. Then $x\in A_{+}$ so $x\in y_{+}$ for some $y\in A=B\cap Q$, so $x\in
B_{+}=B$ and $x\in Q$ as required. ∎

######  Lemma 20.

Let $\Delta$ be a $Q$-terrace. Then $Q\cap\Delta_{+}\setminus\Delta$ is
$Q$-solid above.

###### Proof.

Let $\Lambda$ be a terrace such that $\Delta=Q\cap\Lambda$. Then
$\Lambda_{+}\setminus\Lambda$ is solid above by Lemma 7(vi), and therefore
$Q\cap\Lambda_{+}\setminus\Lambda$ is $Q$-solid above by Lemma 19. Now use
Lemma 17 to see that
$Q\cap\Lambda_{+}\setminus\Lambda=Q\cap\Delta_{+}\setminus\Delta$. ∎

If $A$ is $Q$-solid above, define

$\lambda^{\scriptscriptstyle{Q}}_{A}=\\{x\in A:\text{$\exists i\in[d]$ such
that $x-e_{i}\in Q\setminus A$}\\}.$

The following is the local analogue of parts of Lemma 7.

######  Lemma 21.

Let $Q=[a,b]$, and let $A$ be $Q$-solid above.

  1. (i)

If $A\neq Q$ then $(\lambda^{\scriptscriptstyle{Q}}_{A})_{+}=A_{+}$.

  2. (ii)

$\lambda^{\scriptscriptstyle{Q}}_{A}$ is a $Q$-terrace.

###### Proof.

If $A=\varnothing$ then (i) is trivially true. So let $A\neq Q$ be nonempty.
Then $\lambda^{\scriptscriptstyle{Q}}_{A}$ is nonempty.

Let $x\in(\lambda^{\scriptscriptstyle{Q}}_{A})_{+}$. Then there is a
$y\in\lambda^{\scriptscriptstyle{Q}}_{A}$ such that $x\in y_{+}$. But $y\in A$
so $x\in A_{+}$ too. Conversely, let $x\in A_{+}$, so there is a $y\in A$ with
$x\in y_{+}$. Find a nearest neighbour path, none of whose coordinates
increases, from $y$ to $a$. This path is necessarily in $Q$. Moreover, it must
exit $A$ somewhere, since our hypotheses imply that $a\notin A$. So it must
include a $z\in A$ such that the next step is to some $z-e_{i}\in Q\setminus
A$. Therefore $z\in\lambda^{\scriptscriptstyle{Q}}_{A}$ and $x\in y_{+}\subset
z_{+}\subset(\lambda^{\scriptscriptstyle{Q}}_{A})_{+}$. This shows (i).

Let $B=\\{x-e_{i}:x\in A,x-e_{i}\notin Q\\}$. We claim that

$B_{+}\cap Q\subset A.$ (12)

To see this, let $y\in B_{+}$, so $y\in(x-e_{i})_{+}$ for some $x\in A$ such
that $x-e_{i}\notin Q$. Therefore $x^{[i]}=a^{[i]}$. If
$y^{[i]}=x^{[i]}-1=a^{[i]}-1$ then $y\notin Q$. Whereas if $y^{[i]}\geq
x^{[i]}$ then $y\in x_{+}\subset A_{+}$. Therefore $B_{+}\cap Q\subset
A_{+}\cap Q=A$.

Now let $\Pi=\\{x:\sum x^{[j]}>\sum b^{[j]}\\}$, so $C=\Pi\cup A_{+}\cup
B_{+}$ is good using (6) and (7) and the fact that $\Pi$ is saturated. Thus
$\lambda_{C}$ is a terrace. We claim that
$\lambda^{\scriptscriptstyle{Q}}_{A}=Q\cap\lambda_{C}$ (which is sufficient to
verify (ii)). By (12), $C\cap Q=A_{+}\cap Q=A$, where we have also used the
fact that $A$ is $Q$-solid above by assumption. If
$x\in\lambda^{\scriptscriptstyle{Q}}_{A}$ then $x\in A_{+}\cap Q=A$ and there
is an $i$ with $x-e_{i}\in Q\setminus A$. Therefore $x-e_{i}\notin A_{+}$
(since $A_{+}\cap Q=A)$, and by (12) $x-e_{i}\notin C$. But $x\in C$ and
$x-e_{i}\notin C$ so (e.g. by Lemma 7(ii),(vii)) $x\in\lambda_{C}$.
Conversely, if $x\in\lambda_{C}\cap Q$ then $x\in A$ and (by Lemma 7(i),(ii))
there is an $i$ with $x-e_{i}\notin C$. If $x-e_{i}\notin Q$ then $x-e_{i}\in
B$ which is impossible since $x-e_{i}\notin C$. Therefore $x-e_{i}\in
Q\setminus A$, so $x\in\lambda^{\scriptscriptstyle{Q}}_{A}$, and (ii) is
proved. ∎

Recall from Definition 4 that a _good_ set is one that is saturated, solid
above, and bounded below. Since a subset of $Q$ is obviously bounded below and
not saturated, the natural way to define a $Q$-good set is to simply say that
it is $Q$-solid above.

According to Lemma 7(ii) there is a 1-1 correspondence between terraces and
good sets. The following Lemma indicates that this is not the case for
$Q$-terraces and $Q$-good sets, so we include it to clarify the nature of
$Q$-terraces (even though it is not required for the proof of our main
results). In particular for any $Q$-good set $A$ there are minimal and maximal
$Q$-terraces $\Delta$ (namely $\lambda^{\scriptscriptstyle{Q}}_{A}$ and
$\lambda^{\scriptscriptstyle{Q}}_{A}\cup(A\cap\partial Q)$) with
$\Delta_{+}\cap Q=A$.

######  Lemma 22.

Let $Q=[a,b]$. Let $\Delta$ be a $Q$-terrace.

  1. (i)

Set $\bar{\Delta}=\lambda^{\scriptscriptstyle{Q}}_{\Delta_{+}\cap Q}$. Then
$\bar{\Delta}\subset\Delta$, with $\Delta\setminus\bar{\Delta}\subset\partial
Q$.

  2. (ii)

If $\Delta$ is nonempty and $Q\not\subset\Delta_{+}$ then
$(\bar{\Delta})_{+}\cap Q=\Delta_{+}\cap Q$.

###### Proof.

If $\Delta=\varnothing$ then (i) and (ii) are vacuously true. Let $\Lambda$ be
a terrace such that $\Lambda\cap Q=\Delta$. If $a\in\Delta\subset\Delta_{+}$
then $\bar{\Delta}=\varnothing$, and the only nontrivial statement among (i)
and (ii) is that $\Delta\subset\partial Q$. But $a\in\Lambda$ so
$(a+\boldsymbol{1})_{+}\subset\Lambda_{+}\setminus\Lambda$ by Lemma 7(i).
Since $Q^{o}\subset(a+\boldsymbol{1})_{+}$ we have
$Q^{o}\cap\Lambda=\varnothing$ so $\Delta\subset\partial Q$.

We may therefore assume that $\Delta$ is nonempty (which shows that
$a\in\Delta_{-}$) and $a\notin\Delta$. If $a\in\Delta_{+}$ then
$a\in\Lambda_{+}\cap\Lambda_{-}$ so by Lemma 21(iv) $a\in\Lambda$ which gives
a contradiction. Therefore $a\notin\Delta_{+}$.

To show (i), let $x\in\bar{\Delta}$. Then $x\in\Delta_{+}\cap
Q\subset\Lambda_{+}$ and there is an $i$ with $x-e_{i}\in
Q\setminus\Delta_{+}=Q\setminus\Lambda_{+}$ (the equality holds by Lemma 17).
In particular, $x\in\Lambda$ (Remark 8) so in fact $x\in\Delta$, showing that
$\bar{\Delta}\subset\Delta$.

If $x\in\Delta\setminus\bar{\Delta}$ then $x\in\Delta_{+}\cap Q$, but for
every $j\in[d]$, $x-e_{j}\notin Q\setminus(\Delta_{+})$. Also $x\in\Lambda$ so
there is an $i$ with
$x-e_{i}\in\mathbb{Z}^{d}\setminus(\Lambda_{+})\subset\mathbb{Z}^{d}\setminus(\Delta_{+})$.
Thus, $x-e_{i}\in\mathbb{Z}^{d}\setminus Q$, so $x\in\partial Q$. This proves
(i).

To prove (ii), note that by Lemma 21(i), $(\bar{\Delta})_{+}=(\Delta_{+}\cap
Q)_{+}$. Since $\Delta\subset\Delta_{+}\cap Q$ we have
$\Delta_{+}\subset(\Delta_{+}\cap Q)_{+}$. Similarly since $\Delta_{+}\cap
Q\subset\Delta_{+}$ we have that $(\Delta_{+}\cap
Q)_{+}\subset(\Delta_{+})_{+}=\Delta_{+}$. Thus $\Delta_{+}=(\Delta_{+}\cap
Q)_{+}=(\bar{\Delta})_{+}$. ∎

######  Remark 23.

If $\Delta^{\prime}=\lambda^{\scriptscriptstyle{Q}}_{\Delta_{+}\cap Q}$, where
$\Delta$ is non-empty and $Q\not\subset\Delta_{+}$ then by Lemma 22(ii) we
have that

$\Delta^{\prime}=\lambda^{Q}_{(\Delta^{\prime})_{+}\cap Q}.$ (13)

It therefore follows from the definition of $\lambda^{\scriptscriptstyle{Q}}$
that if $y\in(\Delta^{\prime})_{+}$ satisfies $y-e_{i}\in
Q\setminus(\Delta^{\prime})_{+}$ for some $i$, then $y\in\Delta^{\prime}$.

######  Lemma 24.

If $G$ is $Q$-solid above and $z$ is a corner of $G$ then $G\setminus\\{z\\}$
is $Q$-solid above.

###### Proof.

Let $x\in G\setminus\\{z\\}$. Then $z\notin x_{+}$ since $z$ is a corner of
$G$ (so for each $i\in[d]$, $z\notin(x+e_{i})_{+}$ and also $z\neq x$). Let
$y\in Q\cap x_{+}$. If $y=x$ then $y\in G\setminus\\{z\\}$ and we are done.
Otherwise $y\in(x+e_{j})_{+}$ for some $j$ with $x+e_{j}\in Q$. Now $x,y\in Q$
and $y\in x_{+}$ implies that $[x,y]\subset Q$. Since $y-e_{j}\in x_{+}$ and
$y-e_{j}\in y_{-}$ we have that $y-e_{j}\in[x,y]$. Thus $y-e_{j}\in x_{+}\cap
Q\subset G$. But $y-e_{j}\in x_{+}$ also implies that $y-e_{j}\neq z$ so
$y-e_{j}\in x_{+}\cap Q\setminus\\{z\\}$ as required. ∎

If $z$ is a corner of a $Q$-terrace $\Delta$ we may define

$\Delta_{(\uparrow
z)}^{\scriptscriptstyle{Q}}=\lambda^{\scriptscriptstyle{Q}}_{Q\cap\Delta_{+}\setminus\\{z\\}}.$
(14)

This is a $Q$-terrace by Lemmas 19, 21(ii) and 24. We call this _the
$Q$-terrace resulting from pushing up $z$_. If $\Delta$ is a $Q$-terrace, we
let

$H^{\scriptscriptstyle{Q}}_{\Delta}=\\{x\in\Delta:\text{for every $i\in[d]$,
either $x+e_{i}\in\Delta$ or $x+e_{i}\notin Q$}\\}.$ (15)

In particular, if $\Lambda$ is a terrace and $\Delta=\Lambda\cap Q$ then
$H_{\Delta}^{\scriptscriptstyle{Q}}\cap Q^{o}=H_{\Lambda}\cap Q^{o}$. We then
have the following.

######  Lemma 25.

Let $Q=[a,b]$ and $\Delta$ be a $Q$-terrace.

  1. (i)

If $z\in H_{\Delta}^{\scriptscriptstyle{Q}}$, then either $z$ is a corner of
$\Delta$ or there exists $i\in[d]$ such that $z-e_{i}\in
H_{\Delta}^{\scriptscriptstyle{Q}}$;

  2. (ii)

If $\Delta=\lambda^{\scriptscriptstyle{Q}}_{G}$ where $G$ is $Q$-solid above,
and $z$ is a corner of $\Delta$, then
$\Delta\setminus\\{z\\}\subset\Delta^{\scriptscriptstyle{Q}}_{(\uparrow
z)}\subset
Q\cap\big{(}(\Delta\setminus\\{z\\})\cup\\{z+e_{j}:j\in[d]\\}\big{)}\subset
Q\cap\Delta_{+}$;

  3. (iii)

In the setting of _(ii)_ , if in addition $z\in
H_{\Delta}^{\scriptscriptstyle{Q}}$, then $\Delta_{(\uparrow
z)}^{\scriptscriptstyle{Q}}=\Delta\setminus\\{z\\}$.

###### Proof.

To prove (i), let $z\in H_{\Delta}^{\scriptscriptstyle{Q}}$, and suppose that
$z$ is not corner of $\Delta$. Then there exists $i\in[d]$ such that
$z-e_{i}\in\Delta$. Let $j\in[d]$. Then $z+e_{j}\in\Delta$ or $z+e_{j}\notin
Q$ since $z\in H_{\Delta}^{\scriptscriptstyle{Q}}$. Also
$z-e_{i}+e_{j}\in\Delta_{+}$ since $z-e_{i}\in\Delta$. We claim that either
$z-e_{i}+e_{j}\in\Delta$ or $z-e_{i}+e_{j}\notin Q$. To see this note that if
$z-e_{i}+e_{j}\in Q\cap\Delta_{+}\setminus\Delta$ then by Lemma 20,
$z+e_{j}\in Q\cap\Delta_{+}\setminus\Delta$ which gives a contradiction. Hence
the claim holds. Since this holds for every $j\in[d]$ it follows that
$z-e_{i}\in H_{\Delta}^{\scriptscriptstyle{Q}}$, proving (i).

To prove (ii) and (iii), if $G\ni a$ then $G=Q$, so $\Delta=\varnothing$ and
both statements are vacuous. Thus we may assume that $a\notin G$. Note that
the last inclusion in (ii) is trivial.

By Lemma 21(i) we have $(\lambda^{\scriptscriptstyle{Q}}_{G})_{+}=G_{+}$ so
$\Delta_{+}\cap Q=G_{+}\cap Q=G$. Therefore
$\Delta^{\scriptscriptstyle{Q}}_{(\uparrow
z)}=\lambda^{\scriptscriptstyle{Q}}_{G\setminus\\{z\\}}$. Thus the first
inclusion in (ii) reduces to

$\displaystyle\\{x\in G\setminus\\{z\\}$ $\displaystyle\text{ and }\exists
i\in[d]\text{ such that }x-e_{i}\in Q\setminus G\\}$ (16)
$\displaystyle\subset\\{x\in G\setminus\\{z\\}\text{ and }\exists
i\in[d]\text{ such that }x-e_{i}\in(Q\setminus G)\cup\\{z\\}\\}$

which is trivial. For the second inclusion, suppose that $x$ is in the right
hand side of (16). If $x-e_{i}=z$ for some $i$ then $x=z+e_{i}$ and we are
done. Otherwise $x-e_{i}\in Q\setminus G$, so
$x\in\lambda_{G}^{\scriptscriptstyle{Q}}=\Delta$ (and $x\neq z$). Since also
$\lambda^{\scriptscriptstyle{Q}}_{G}\subset Q$ (since $G\subset Q$), this
proves (ii).

By definition of $H_{\Delta}^{\scriptscriptstyle{Q}}$, if $z\in
H_{\Delta}^{\scriptscriptstyle{Q}}$ and $z+e_{j}\in Q$ then
$z+e_{j}\in\Delta\setminus\\{z\\}$, so $\\{z+e_{j}:j\in[d]\\}\cap
Q\subset\Delta\setminus\\{z\\}$ in this case). Thus (iii) follows from (ii). ∎

Now we will fix a box $R$ and consider various smaller boxes $Q\subset R$. We
define the relative boundary $\partial_{\scriptscriptstyle{R}}Q$ of $Q$ in $R$
to be the set of sites in $Q$ that are adjacent to sites in $R\setminus Q$ and
the relative interior
$Q^{\scriptscriptstyle{R}:o}=Q\setminus\partial_{\scriptscriptstyle{R}}Q$.

######  Remark 26.

It follows from Lemma 25(ii) that $(\Delta^{Q}_{(\uparrow
z)})_{+}\subset\Delta_{+}$ since the only elements in $\Delta^{Q}_{(\uparrow
z)}$ that may not be in $\Delta$ are in $z_{+}\subset\Delta_{+}$.

######  Lemma 27.

Let $Q\subset R$ be boxes, and let $\Delta$ be an $R$-terrace. Let
$x\in\Delta\cap Q$ and suppose $\Delta$ has no corners in
$Q^{\scriptscriptstyle{R}:o}\cap x_{-}$. Then there exists
$y\in\Delta\cap\partial_{\scriptscriptstyle{R}}Q$ such that we may construct a
path in $\Delta\cap Q$ from $x$ to $y$ as in Lemma 18(ii), but using only
moves of the form $-e_{i}$. For each $i$, all the $-e_{i}$ moves occur
consecutively.

###### Proof.

If $x\in\partial_{\scriptscriptstyle{R}}Q$ then $y=x$ works. So assume
$x\notin\partial_{\scriptscriptstyle{R}}Q$. Therefore $x\in\Delta\cap
Q^{\scriptscriptstyle{R}:o}$.

Let $k_{1}=\inf\\{k\geq 1:x-ke_{1}\notin\Delta\cap
Q^{\scriptscriptstyle{R}:o}\\}$ and $m_{1}=k_{1}-1$. Set
$z^{\scriptscriptstyle{(}k)}=x-ke_{1}\in\Delta\cap
Q^{\scriptscriptstyle{R}:o}$ for $0\leq k\leq m_{1}$. If
$z^{\scriptscriptstyle{(}m_{1})}-e_{1}\in\partial_{\scriptscriptstyle{R}}Q$
then we take $y=z^{\scriptscriptstyle{(}m_{1})}-e_{1}$ and our path is
complete (with $k_{1}$ steps).

Otherwise we proceed recursively. For $j<d$, given $k_{1},\dots,k_{j}$, let
$m_{j}=\sum_{i=1}^{j}(k_{i}-1)$, and assume that
$z^{\scriptscriptstyle{(}k)}\in\Delta\cap Q^{\scriptscriptstyle{R}:o}$ for
$0\leq k\leq m_{j}$ gives a path using only moves $-e_{1},\dots,-e_{j}$. Let
$k_{j+1}=\inf\\{k\geq
1:z^{\scriptscriptstyle{(}m_{j})}-ke_{j+1}\notin\Delta\cap
Q^{\scriptscriptstyle{R}:o}\\}$ and set
$z^{\scriptscriptstyle{(}m_{j}+k)}=z^{\scriptscriptstyle{(}m_{j})}-ke_{j+1}$
for $k\leq k_{j+1}-1$. Take $m_{j+1}=m_{j}+k_{j+1}-1$. Note that it could
happen that $k_{j+1}=1$, in which case
$z^{\scriptscriptstyle{(}m_{j+1})}=z^{\scriptscriptstyle{(}m_{j})}$.

We may continue this process until one of two things happens. Either we find a
$j$ with
$z^{\scriptscriptstyle{(}m_{j+1})}-e_{j+1}\in\partial_{\scriptscriptstyle{R}}Q$,
in which case we’re done, with $1+m_{j+1}$ steps. Or we exhaust the possible
$j\in[d]$ before we reach $\partial_{\scriptscriptstyle{R}}Q$ and the
construction terminates in a point
$z^{\scriptscriptstyle{(}m_{d})}\in\Delta\cap Q^{R:o}$. Assume the latter. We
will see that this leads to a contradiction.

We claim that for each $j$ either $z^{\scriptscriptstyle{(}m_{j})}-e_{j}\notin
R$ or $z^{\scriptscriptstyle{(}m_{j})}-e_{j}\in R\setminus\Delta_{+}$. To see
this, assume $z^{\scriptscriptstyle{(}m_{j})}-e_{j}\in R$. The construction
ensures that $z^{\scriptscriptstyle{(}m_{j})}-e_{j}\notin\Delta\cap
Q^{\scriptscriptstyle{R}:o}$, so either
$z^{\scriptscriptstyle{(}m_{j})}-e_{j}\notin Q^{\scriptscriptstyle{R}:o}$ or
$z^{\scriptscriptstyle{(}m_{j})}-e_{j}\notin\Delta$. Because
$z^{\scriptscriptstyle{(}m_{j})}-e_{j}\in R$, the former is ruled out by the
assumption that
$z^{\scriptscriptstyle{(}m_{j})}-e_{j}\notin\partial_{\scriptscriptstyle{R}}Q$.
Therefore $z^{\scriptscriptstyle{(}m_{j})}-e_{j}\in R\setminus\Delta_{+}$ by
Lemma 20), as claimed.

We know that $z^{\scriptscriptstyle{(}m_{d})}\in
z^{\scriptscriptstyle{(}m_{j})}_{-}$ for each $j$. It follows that for each
$j$ either $z^{\scriptscriptstyle{(}m_{d})}-e_{j}\notin R$ or
$z^{\scriptscriptstyle{(}m_{d})}-e_{j}\in R\setminus\Delta_{+}$. In
particular, $z^{\scriptscriptstyle{(}m_{d})}$ is a corner of $\Delta$ in
$Q^{R:o}\cap x_{-}$. But that is impossible, so in fact the construction must
reach $\partial_{\scriptscriptstyle{R}}Q$. ∎

Let $Q\subset R$ be boxes and let $G$ be $R$-solid above (and
$\neq\varnothing$ or $R$). Let $\Delta=\lambda^{\scriptscriptstyle{R}}_{G}$
and $A\subset G$. If $\Delta$ has a corner $z_{1}$ in
$Q^{\scriptscriptstyle{R}:o}\setminus A$, then push it up to give a new
$R$-terrace $\Delta_{1}=\Delta^{\scriptscriptstyle{R}}_{(\uparrow z_{1})}$. If
$\Delta_{1}$ has a corner $z_{2}$ in $Q^{\scriptscriptstyle{R}:o}\setminus A$,
then push it up to give a new $R$-terrace $\Delta_{2}$. Repeat this process
until we obtain an $R$-terrace with no corners in
$Q^{\scriptscriptstyle{R}:o}\setminus A$ (which could in principle
$=\varnothing$, e.g. if $Q=R$ and $A=\varnothing$). We will use the notation
$\Delta^{\scriptscriptstyle{Q},A}$ for any $R$-terrace obtained in this way
(in principle it need not be unique). Since
$\lambda^{\scriptscriptstyle{R}}_{G}\subset G$ and (from Lemma 25(ii) or
Remark 26) pushing up $\lambda^{\scriptscriptstyle{R}}_{G}$ keeps one in
$(\lambda^{\scriptscriptstyle{R}}_{G})_{+}$ we have that

$\Delta^{\scriptscriptstyle{Q},A}\subset G.$ (17)

######  Lemma 28.

Let $Q\subset R$ be boxes and let $G$ be $R$-solid above (and
$\neq\varnothing$ or $R$) and $A\subset G$. Let
$\Delta=\lambda^{\scriptscriptstyle{R}}_{G}$ and $A\subset G$. Then

  1. (i)

$\Delta^{\scriptscriptstyle{Q},A}=\Delta$ off $Q$ (i.e.
$\Delta^{\scriptscriptstyle{Q},A}\cap(R\setminus Q)=\Delta\cap(R\setminus
Q)$).

  2. (ii)

$R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}$ is the minimal set which is
$R$-solid above, agrees with $G$ off $Q^{\scriptscriptstyle{R}:o}$, and
contains $A$.

  3. (iii)

If $x\in\Delta^{\scriptscriptstyle{Q},A}\cap Q$ then there is a $z\in
G\cap(A\cup\partial_{\scriptscriptstyle{R}}Q)$ with $x\in z_{+}$.

###### Proof.

By Lemma 25 (ii), pushing up corners in $Q^{\scriptscriptstyle{R}:o}\setminus
A$ will never alter $\Delta$ outside $Q$. Therefore (i) holds.

Let $V$ be the minimal set which is $R$-solid above, agrees with $G$ off
$Q^{\scriptscriptstyle{R}:o}$, and is such that $V\supset A$. $V$ is well
defined, since intersections of sets which are $R$-solid above are also
$R$-solid above (and $G$ is an instance of such a set). Let

$W=\\{x:x\in R\cap y_{+}\text{ for some $y\in G\setminus Q$ or $y\in
G\cap(A\cup\partial_{\scriptscriptstyle{R}}Q)$}\\}.$

We will show that

$\displaystyle W=V=R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}.$ (18)

This clearly implies (ii). We shall see below that (18) also implies (iii).

First note that (by Lemma 19) $R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}$ is
$R$-solid above. Next, since $(\Delta^{\scriptscriptstyle{Q},A})_{+}\subset
G_{+}=(\lambda_{G}^{\scriptscriptstyle{R}})_{+}$ (the former e.g. since
$\Delta^{\scriptscriptstyle{Q},A}\subset G_{+}\cap R=G$ and the latter e.g. by
Lemma 21(i)) we have

$R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}\cap(Q^{\scriptscriptstyle{R}:o})^{c}\subset
G\cap(Q^{\scriptscriptstyle{R}:o})^{c}.$ (19)

On the other hand, $G\cap(Q^{\scriptscriptstyle{R}:o})^{c}\subset G_{+}\cap
R=(\lambda_{G}^{\scriptscriptstyle{R}})_{+}\cap R$, and since pushing up
corners in $Q^{\scriptscriptstyle{R}:o}\setminus A$ will never excise points
of $(Q^{\scriptscriptstyle{R}:o})^{c}$ (e.g. by Lemma 25(ii)) this proves that

$R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}\cap(Q^{\scriptscriptstyle{R}:o})^{c}\supset
G\cap(Q^{\scriptscriptstyle{R}:o})^{c}.$ (20)

Combining (19) and (20) we see that
$R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}$ agrees with $G$ off
$Q^{\scriptscriptstyle{R}:o}$. If $x\in A\subset G\subset R$ then $x\in
A\cap(\lambda^{\scriptscriptstyle{R}}_{G})_{+}$. By repeated applications of
Lemma 25(ii), since $x\in A$ is never removed, we have that
$x\in(\Delta^{\scriptscriptstyle{Q},A})_{+}$. Thus
$R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}$ contains $A$.

We have therefore shown that $R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}$ is
$R$-solid above, agrees with $G$ off $Q^{\scriptscriptstyle{R}:o}$ and
contains $A$. Therefore $R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}\supset
V$.

If $U$ is $R$-solid above, agrees with $G$ off $Q^{\scriptscriptstyle{R}:o}$,
and contains $A$, then it contains any $z\in G\setminus Q$ and any $z\in
G\cap(A\cup\partial_{\scriptscriptstyle{R}}Q)$. Therefore it contains $W$ too.
Since $U$ is otherwise arbitrary, we have that $V\supset W$.

To verify (18) it remains to show that $W\supset
R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}$. So let $x\in
R\cap(\Delta^{\scriptscriptstyle{Q},A})_{+}\subset G$ where the inclusion was
shown in (19). If $x\notin Q$ then (since $x\in x_{+}$) we immediately
conclude that $x\in W$. If $x\in A_{+}$ then $x\in W$ too, by definition. So
assume that $x\in Q\setminus A_{+}$, so that $x_{-}\subset A^{c}$.

Write $Q=[a,b]$. Then $a\in R$, and if
$a\in(\Delta^{\scriptscriptstyle{Q},A})_{+}$ then (by (17)) $a\in G$, and
$a\in\partial_{\scriptscriptstyle{R}}Q$ (since otherwise $R$ would have to
take the form $[a,b^{\prime}]$ for some $b^{\prime}$, which would in turn
imply that $G=R$, which is excluded by our hypotheses). Therefore $x\in W$ by
definition, since $x\in a_{+}$.

Therefore we may assume that $a\notin(\Delta^{\scriptscriptstyle{Q},A})_{+}$.
Take a nearest neighbour path from $x$ to $a$, none of whose coordinates
increase. By (vii) of Lemma 7 and the fact that $x\in a_{+}$, this path
includes some point $y\in Q\cap\Delta^{\scriptscriptstyle{Q},A}\cap x_{-}$. By
construction, $y_{-}\subset x_{-}\subset A^{c}$, so there are no corners of
$\Delta^{\scriptscriptstyle{Q},A}$ in $Q^{\scriptscriptstyle{R}:o}\cap y_{-}$.
There is therefore a path as in Lemma 27 (beware of the different roles of the
points $x$ and $y$ there) from $y$ to some
$z\in\Delta^{\scriptscriptstyle{Q},A}\cap\partial_{\scriptscriptstyle{R}}Q\subset
G\cap\partial_{\scriptscriptstyle{R}}Q$. Since $x\in y_{+}\subset z_{+}$ it
follows that $x\in W$, as required. This shows (18) and hence (ii).

To see that (18) implies (iii) as well, suppose
$x\in\Delta^{\scriptscriptstyle{Q},A}\cap Q$. By (18), $x\in W$ so $x\in
z_{+}$ for some $z\in G\setminus Q$ or $z\in
G\cap(A\cup\partial_{\scriptscriptstyle{R}}Q)$ . Take a nearest neighbour path
from $x$ to $z$, none of whose coordinates increase. Every point $y$ along it
has $y\in z_{+}\cap x_{-}$, so $y\in G$. If $z\in
G\cap(A\cup\partial_{\scriptscriptstyle{R}}Q)$ then we are done. Otherwise
$z\in G\setminus Q$, so the path passes through a point $z^{\prime}\in
G\cap\partial_{\scriptscriptstyle{R}}Q\cap x_{-}$, which shows (iii). ∎

###  2.4 Relation with $\mathcal{C}_{o}$

Recall the definition of $\Omega_{1}$, and consider the model of Example 2,
where

$E=\mathcal{E}_{+}\text{ and }F=\mathcal{E}.$ (21)

Let $\mathcal{T}_{o}$ denote the (random) set of terraces $\Lambda$ such that
$o\in\Lambda_{+}$ and $\Lambda\subset\Omega_{1}$. Then we have the following,
which is already essentially stated in Proposition 6, but is now proved for
completeness.

######  Lemma 29.

For Example 2, if $p\in(p_{c}(d),1)$ then a.s. $\mathcal{T}_{o}$ is non-empty
and $\mathcal{C}_{o}=\cap_{\Lambda\in\mathcal{T}_{o}}\Lambda_{+}$.

###### Proof.

By Theorem  ‣ 1.1 we have that $L_{x}^{(i)}$ is finite for every
$x\in\mathbb{Z}^{d}$ and $i\in[d]$ and
$\mathcal{C}_{o}=\big{(}\cup_{x}(x+L_{x}^{(i)})\big{)}_{+i}$ for each $i$.
Moreover, by definition of $L_{x}^{(i)}$ we have that
$x+e_{i}+L^{(i)}_{x+e_{i}}=x+L^{(i)}_{x}$ for each $x,i$ so
$\mathcal{C}_{o}=\big{(}\cup_{y:y\cdot e_{i}=0}(y+L_{y}^{(i)})\big{)}_{+i}$.
Since $e_{i}\in\mathcal{E}_{+}$ for each $i$ we have that $\mathcal{C}_{o}$ is
solid above. It is saturated since $L_{x}^{(i)}<\infty$ for each $x,i$ and it
is bounded below since $L_{y}^{(i)}>-\infty$ for each $i$ and $y:y\cdot
e_{i}=0$. Therefore $\mathcal{C}_{o}$ is good. The terrace
$\lambda_{\mathcal{C}_{o}}\subset\mathcal{C}_{o}$ satisfies
$o\in\mathcal{C}_{o}=(\lambda_{\mathcal{C}_{o}})_{+}$ by Lemma 7(ii). Moreover
since each $x\in\lambda_{\mathcal{C}_{o}}\subset\mathcal{C}_{o}$ has a
neighbour $x-e_{i}\notin(\lambda_{\mathcal{C}_{o}})_{+}=\mathcal{C}_{o}$ (by
Lemma 7(i)) we must have that $x\in\Omega_{1}$. Thus
$\lambda_{\mathcal{C}_{o}}\in\mathcal{T}_{o}$, and
$\cap_{\Lambda\in\mathcal{T}_{o}}\Lambda_{+}\subset(\lambda_{\mathcal{C}_{o}})_{+}=\mathcal{C}_{o}$.

Now let $\Lambda\in\mathcal{T}_{o}$, so that $\Lambda\subset\Omega_{1}$, and
suppose that $x\in\mathcal{C}_{o}\setminus\Lambda_{+}$. There is a (self-
avoiding) path from $o$ to $x$ consistent with the environment. By Lemma
7(vii) this path exits $\Lambda_{+}$ by moving from a vertex $y_{0}$ in
$\Lambda$ to a vertex $y_{1}=y-e_{i}\notin\Lambda_{+}$ for some $i$. Thus
$y_{0}\notin\Omega_{1}$ which contradicts $\Lambda\subset\Omega_{1}$. This
proves that $\mathcal{C}_{o}\subset\Lambda_{+}$, so
$\mathcal{C}_{o}\subset\cap_{\Lambda\in\mathcal{T}_{o}}\Lambda_{+}$. ∎

###### Proof of Proposition 6.

As remarked above, $\mathcal{C}_{o}$ is solid above since
$e_{i}\in\mathcal{E}_{+}$ for each $i$. It is saturated for $p<1$ since in
that case $L_{x}^{(i)}<\infty$ for each $x,i$. This proves (i). Lemma 29
directly implies (ii). For $p<p_{c}$, Theorem  ‣ 1.1(I) shows that
$\mathcal{C}_{o}=\mathbb{Z}^{d}$. But the final argument in the proof of Lemma
29 then shows that $\mathcal{C}_{o}\subset\Lambda_{+}$ for every
$\Lambda\in\mathcal{T}_{o}$, so in fact $\mathcal{T}_{o}$ is empty. This
proves (iii). ∎

We recall the definitions of two infinite self-avoiding paths in
$\mathcal{C}_{x}$: Define the Wn path
$\boldsymbol{x}^{\nwarrow}=(x^{\scriptscriptstyle{(}n)})_{n\in\mathbb{Z}_{+}}$
from $x^{\scriptscriptstyle{(}0)}=x\in\mathbb{Z}^{2}$ by following the arrow
$-e_{1}$ whenever possible, and otherwise $e_{2}$. Similarly we define the Se
path
$\boldsymbol{x}^{\searrow}:=(y^{\scriptscriptstyle{(}n)})_{n\in\mathbb{Z}_{+}}$
from $x\in\mathbb{Z}^{2}$ by following the arrow $-e_{2}$ whenever possible,
and otherwise $e_{1}$. We use the concept of Wn and Se paths in copies of
$\mathbb{Z}^{2}$ within $\mathbb{Z}^{d}$ to prove the following.

######  Lemma 30.

Let $p\in(0,1)$ and $u,v\in\mathbb{Z}^{d}$ and $j\in[d]$. There is a
$k\in\mathbb{Z}$ such that $v+ke_{j}\in\mathcal{C}_{u}$. If $u\in v_{-}$ then
$k\leq 0$ and $u\cdot e_{j}\geq(v+ke_{j})\cdot e_{j}$.

Note that the existence of $k$ already follows from the fact that
$\mathcal{C}_{u}$ is good. But the explicit construction in the proof provides
us with additional detail.

###### Proof.

Without loss of generality, $j=d$ and $u=o\in\mathbb{Z}^{d}$. Follow the
$\mathbb{Z}e_{1}\times\mathbb{Z}e_{d}$-Wn path or Se path from $o=(0,\dots,0)$
to find $z^{\scriptscriptstyle{(}1)}\in\mathcal{C}_{o}$ such that
$z^{\scriptscriptstyle{(}1)}=(v_{1},0,\dots,0,n_{1})$ (use the Wn path (in
which case $n_{1}\geq 0$) if $v_{1}<0$ and otherwise the Se path (in which
case $n_{1}\leq 0$) ). Now follow the $\mathbb{Z}e_{2}\times\mathbb{Z}e_{d}$\-
Wn path or Se path from $z^{\scriptscriptstyle{(}1)}$ to find
$z^{\scriptscriptstyle{(}2)}\in\mathcal{C}_{o}$ such that
$z^{\scriptscriptstyle{(}2)}=(v_{1},v_{2},0,\dots,0,n_{2})$. Continue this
procedure until the first $d-1$ coordinates match. If $o\in v_{-}$ then we are
using Se paths each time, which means that the
$z^{\scriptscriptstyle{(}i)}_{d}$ never increase. ∎

For $x\in Q$ we let $\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$ be the set of
$y\in Q$ that may be reached from $x$ by following arrows from our random
environment that begin and end in $Q$. Since
$\mathcal{E}_{+}\subset\mathcal{G}_{y}$ for every $y$ it is trivial that
$\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$ is $Q$-solid above. Moreover, it is
easily seen (c.f. the proof of Lemma 29) that $Q$-terraces
$\Delta\subset\Omega_{1}$ block paths that remain in $Q$, as follows.

######  Lemma 31.

If $\Delta\subset\Omega_{1}$ is a $Q$-terrace, with $x\in\Delta_{+}\cap Q$ and
$y\notin\Delta_{+}\cap Q$, then
$y\notin\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$.

We will need the following converse to this statement, which is a local
version of Proposition 6.

######  Lemma 32.

Let $x\in Q=[a,b]$. Either $Q=\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$ or
$\Delta:=\lambda^{\scriptscriptstyle{Q}}_{\mathcal{C}^{\scriptscriptstyle{Q}}_{x}}$
is a $Q$-terrace satisfying
$\Delta\subset\mathcal{C}^{\scriptscriptstyle{Q}}_{x}\cap\Omega_{1}$, such
that $\Delta_{+}=(\mathcal{C}^{\scriptscriptstyle{Q}}_{x})_{+}$.

###### Proof.

Since $\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$ is $Q$-solid above, if
$a\in\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$ then
$Q=\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$. Otherwise
$a\notin\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$ so
$\Delta:=\lambda^{\scriptscriptstyle{Q}}_{\mathcal{C}^{\scriptscriptstyle{Q}}_{x}}$
is a $Q$-terrace (Lemma 21(ii)), and for each $y\in\Delta$ there exists
$i\in[d]$ such that $y-e_{i}\in
Q\setminus\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$, so $y\in\Omega_{1}$. By
Lemma 21, $\Delta_{+}=(\mathcal{C}^{\scriptscriptstyle{Q}}_{x})_{+}$ as
claimed. ∎

######  Remark 33.

It is worthwhile noting that in general for $p\in(p_{c},1)$ either:
$\lambda_{\mathcal{C}_{o}}$ is irreducible, or;
$o\in\lambda_{\mathcal{C}_{o}}$ and $H_{\lambda_{\mathcal{C}_{o}}}\subset
o_{+}$. The following proves this in the setting of $Q$-terraces

######  Lemma 34.

Let $x\in Q$ and set
$\Delta:=\lambda^{\scriptscriptstyle{Q}}_{\mathcal{C}^{\scriptscriptstyle{Q}}_{x}}$.
Then either

  1. (i)

$H_{\Delta}^{\scriptscriptstyle{Q}}=\varnothing$; or

  2. (ii)

$x\in\Delta$ and $H_{\Delta}^{\scriptscriptstyle{Q}}\subset x_{+}$; or

  3. (iii)

$Q=\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$.

###### Proof.

Assume that neither (i) nor (iii) is the case, so there is a $z\in
H_{\Delta}^{\scriptscriptstyle{Q}}\subset\Delta$. Then for each $i$, either
$z+e_{i}\notin Q$ or $z+e_{i}\in\Delta$, and in the latter case, Lemma 32
implies that $z+e_{i}$ has environment $\mathcal{E}_{+}$. In particular, we
cannot reach $z$ in one step from any site $z+e_{i}\in Q$. Since
$z\in\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$ we conclude that either $z=x$ or
some site $y_{1}=z-e_{j_{1}}\in\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$.

In the latter case for each $i$, either $y_{1}+e_{i}\notin Q$, or
$y_{1}+e_{i}\in Q\cap\Delta_{+}$ (as
$y_{1}\in\mathcal{C}^{\scriptscriptstyle{Q}}_{x}\subset\Delta_{+}$). If
$y_{1}+e_{i}\in Q\cap\Delta_{+}$ then either: $i=j_{1}$ and $y_{1}+e_{i}=z$
or; $i\neq j_{1}$ and since $Q$ is a rectangle, $z+e_{i}\in Q$, so
$z+e_{i}\in\Delta$. Either way we conclude that $y_{1}+e_{i}\in\Delta$ (in the
case $i\neq j_{1}$ we have $y_{1}+e_{i}=(z+e_{i})-e_{j}\in\Delta_{+}$ and
$z+e_{i}\in\Delta$ so by Lemma (20) $y_{1}+e_{i}\in\Delta$). Therefore
$y_{1}\in H_{\Delta}^{\scriptscriptstyle{Q}}$.

We may repeat this argument (either $y_{1}=x$ or some site
$y_{2}=y_{1}-e_{j_{2}}\in\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$ and so on),
but eventually the iteration must stop because $Q$ is finite. It stops
precisely when $y_{n}=x$. In other words, there is a sequence
$z=:y_{0},\dots,y_{n}=x\in\Delta$ with each $y_{k+1}\in(y_{k})_{-}$. Therefore
$z\in x_{+}$ and (ii) holds. ∎

An event $A$ that is measurable with respect to
$\sigma\big{(}(\mathcal{G}_{x})_{x\in\mathbb{Z}^{d}}\big{)}$ is said to be
increasing if the following holds:

$\text{for any }\Gamma_{1}\subset\Gamma_{2}\subset\mathbb{Z}^{d}\text{ if
}\\{\Omega_{1}=\Gamma_{1}\\}\subset A\text{ then
}\\{\Omega_{1}=\Gamma_{2}\\}\subset A.$ (22)

The following is a trivial consequence of the fact that
$\mathcal{E}_{+}\subset\mathcal{E}$.

######  Corollary 35.

Let $x,y\in Q=[a,b]$. The event
$\\{\text{$\mathcal{C}^{\scriptscriptstyle{Q}}_{x}$ doesn't intersect
$y_{-}$}\\}$ is increasing.

Let $x,y\in R\subset\mathbb{Z}^{d}$. Let
$\boldsymbol{\omega}\in\\{\mathcal{E}_{+},\mathcal{E}\\}^{\mathbb{Z}^{d}}$.
Write $\mathcal{C}_{x}({\boldsymbol{\omega}})$ for the set of vertices in
$\mathbb{Z}^{d}$ reachable from $x$ by following arrows in
${\boldsymbol{\omega}}$ and similarly
$\mathcal{C}_{x}^{\scriptscriptstyle{R}}({\boldsymbol{\omega}})$ is the set of
sites reachable using only arrows in $\boldsymbol{\omega}$ that begin and end
in $R$. A site $u\in R$ is said to be pivotal for $x\to y$ in
$\boldsymbol{\omega}_{R}:=(\boldsymbol{\omega}_{z})_{z\in R}$ if there is no
connection from $x$ to $y_{-}$ within $R$ if $\boldsymbol{\omega}_{u}$ is
replaced by $\mathcal{E}_{+}$, but such a connection is created if we replace
$\boldsymbol{\omega}_{u}$ by $\mathcal{E}$. We write
$\Omega_{1}(\boldsymbol{\omega}_{R})=\\{x\in
R:\boldsymbol{\omega}_{x}=\mathcal{E}_{+}\\}$, and
$\Omega_{1}(\boldsymbol{\omega}):=\Omega_{1}(\boldsymbol{\omega}_{\mathbb{Z}^{d}})$.

######  Lemma 36.

Let
${\boldsymbol{\omega}}\in\\{\mathcal{E}_{+},\mathcal{E}\\}^{\mathbb{Z}^{d}}$
and let $R\ni x,y$ be a box, and suppose that there exists an $R$-terrace
$\Delta\subset\Omega_{1}(\boldsymbol{\omega}_{R})$ such that $x\in\Delta_{+}$
and $y\notin\Delta_{+}$. Let $u\in R\setminus x_{+}$. If $u$ is pivotal for
$x\to y$ in $\boldsymbol{\omega}_{R}$, then $u\in\Delta$ and there is an $i$
with $u+e_{i}\in R\setminus\Delta$, and a $j$ with $u-e_{j}\in
R\setminus\Delta_{+}$.

###### Proof.

Let $u$ be pivotal for $x\to y$ in $\boldsymbol{\omega}_{R}$. Then
$u\in\mathcal{C}^{\scriptscriptstyle{R}}_{x}(\boldsymbol{\omega})$ and
$y\in\mathcal{C}^{\scriptscriptstyle{R}}_{u-e_{i_{0}}}(\boldsymbol{\omega})$
for some $i_{0}\in[d]$. The first observation plus Lemma 31 implies that
$u\in\Delta_{+}$. The second observation implies that $u-e_{i_{0}}\in
R\setminus\Delta_{+}$ (since if $u\in\Delta_{+}\cap R$ and
$y\notin\Delta_{+}\cap R$ then by Lemma 31
$y\notin\mathcal{C}^{\scriptscriptstyle{R}}_{u-e_{i_{0}}}(\boldsymbol{\omega})$).
So the final conclusion of the Lemma holds with $j=i_{0}$.

Let $\Lambda$ be a terrace such that $\Lambda\cap R=\Delta$. By Lemma 17,
$u\in\Lambda_{+}$ and $u-e_{i_{0}}\in(\Lambda_{+})^{c}$. By Lemma 7(vii)
$u\in\Lambda$. Hence $u\in\Delta$.

Since $u\in\mathcal{C}^{\scriptscriptstyle{R}}_{x}(\boldsymbol{\omega})$ there
is a self-avoiding nearest-neighbour path from $x$ to $u$ that is consistent
with the environment (in particular it is a subset of
$\mathcal{C}^{\scriptscriptstyle{R}}_{x}(\boldsymbol{\omega})$) and stays in
$R$. By Lemma 31 in fact this path stays in $R\cap\Delta_{+}$. Let
$z\in\Delta_{+}$ be the last point on this path from which the step taken is
in $\mathcal{E}_{-}$ (there has to be at least one such point, otherwise we’d
have $u\in x_{+}$). In particular $z\notin\Omega_{1}(\boldsymbol{\omega})$ so
$z\notin\Delta$. Therefore $z\in R\cap\Delta_{+}\setminus\Delta$, and there is
an $i_{1}\in[d]$ such that $z-e_{i_{1}}\in R$ (being the next step in the
path) but $u\in(z-e_{i_{1}})_{+}$.

Since $u\in\Delta$, we can’t have $u\in z_{+}$, by Lemma 20. Therefore
$u^{\scriptscriptstyle{[}i_{1}]}=z^{\scriptscriptstyle{[}i_{1}]}-1$ and so
$u+e_{i_{1}}\in R$. Since $z\in\Delta_{+}\setminus\Delta$ and $u+e_{i_{1}}\in
z_{+}$, Lemma 20 implies that $u+e_{i_{1}}\in R\setminus\Delta$. ∎

##  3 Enhancement

In this section, we will prove our main result. Subsection 3.1 will consider a
disturbed version of the half-orthant model, and Subsection 3.3 will then use
this model to prove Theorem 2.

Let $V_{d}\subset\mathbb{Z}^{d}$ denote the set of points
$x=(x^{\scriptscriptstyle{[}1]},\dots,x^{\scriptscriptstyle{[}d]})$ such that

$h(x):=\sum_{i=1}^{\lceil
d/2\rceil}ix^{\scriptscriptstyle{[}2i-1]}-\sum_{i=1}^{\lfloor
d/2\rfloor}ix^{\scriptscriptstyle{[}2i]}\equiv 0\mod\boldsymbol{\rho}_{d},$

where $\boldsymbol{\rho}_{d}$ is the smallest prime greater than $2{\lceil
d/2\rceil}$. When $d=2$ we have that $\boldsymbol{\rho}_{d}=3$ and
$h(x)=x^{\scriptscriptstyle{[}1]}-x^{\scriptscriptstyle{[}2]}$ and a finite
portion of $V_{2}$ can be seen in Figure 3.

Figure 3: A finite portion of the set of points $V_{2}$.

Using the facts that $h(o)=0\mod\boldsymbol{\rho}_{d}$, and
$h(x+\boldsymbol{\rho}_{d}e_{2i})=h(x)-\boldsymbol{\rho}_{d}i$, and
$h(x+\boldsymbol{\rho}_{d}e_{2i-1})=h(x)+\boldsymbol{\rho}_{d}i$, the
following is easy to verify.

######  Lemma 37.

Any adjacent collection of at least $\boldsymbol{\rho}_{d}$ vertices on a line
parallel to one of the axes will contain at least one point of $V_{d}$.

Recall that the (orthant and) half-orthant model on $\mathbb{Z}^{d}$ was
defined using uniform random variables $(U_{x})_{x\in\mathbb{Z}^{d}}$. Given
$p,q\in[0,1]$,

  * (i)

if $x\in\mathbb{Z}^{d}\setminus V_{d}$ then set
$\mathcal{G}_{x}=\mathcal{E}_{+}$ if $U_{x}\leq p$, and
$\mathcal{G}_{x}=\mathcal{E}$ otherwise;

  * (ii)

if $x\in V_{d}$ then set $\mathcal{G}_{x}=\mathcal{E}_{+}$ if $U_{x}\leq q$,
and $\mathcal{G}_{x}=\mathcal{E}$ otherwise.

Let $\mathcal{C}_{x}(p,q)\subset\mathbb{Z}^{d}$ be the set of sites that can
be reached from $x$ by following the arrows of this model. Let
$\theta(p,q)=\mathbb{P}\big{(}\mathcal{C}_{o}(p,q)=\mathbb{Z}^{d}\big{)}$. As
in the introduction it is trivial that $\mathcal{C}_{x}(p,q)$ is monotone
decreasing in both $p$ and $q$. By definition of $\mathcal{C}_{x}(p,q)$ and
Theorem  ‣ 1.1, we have that

$\mathbb{P}(\mathcal{C}_{o}(p,p)=\mathbb{Z}^{d})=\begin{cases}1,&\text{ if
}p<p_{c}(d),\\\ 0,&\text{ if }p>p_{c}(d).\end{cases}$ (23)

For an increasing continuous function $f:[0,1]\rightarrow[0,1]$ such that
$f(0)=0$, $f(1)=1$, and $f(p)<p$ for $p\in(0,1)$ we define what we call an
$f$-disturbed orthant model, as follows. Define

$p_{c}(f,d):=\sup\\{p:\theta(p,f(p))>0\\}.$

Since $f$ is monotone, this shows that $\mathcal{C}_{x}(p,f(p))$ is monotone
decreasing in $p$ and therefore $\theta(p,f(p))>0$ for all $p<p_{c}(f,d)$. We
will prove the following in Section 3.2.

######  Proposition 38.

For $f$ continuous and increasing with $f(0)=0$, $f(1)=1$, and $f(p)<p$ for
$p\in(0,1)$

$p_{c}(f,d)>p_{c}(d).$

Fix $d\geq 2$. For $i\in\\{1,2\\}$ set
$S^{\\{i\\}}=\mathbb{Z}^{d}\times\\{i\\}$ and
$S^{\\{1,2\\}}=\mathbb{Z}^{d}\times\\{1,2\\}$, so
$S^{\\{1\\}},S^{\\{}2\\}\subset S^{\\{1,2\\}}\subset\mathbb{Z}^{d+1}$. Let
$\mathcal{C}_{x,+1}(p)$ denote the forward cluster of $x$ in the half-orthant
model for $\mathbb{Z}^{d+1}$.

For $x\in S^{\\{i\\}}$ define $\mathcal{C}_{x}^{\\{i\\}}(p)\subset
S^{{}^{\\{i\\}}}$ as the set of points of $S^{\\{i\\}}$ that can be reached
from $x$ using only those arrows from the half-orthant model in
$\mathbb{Z}^{d+1}$ that begin and end in $S_{i}$. Likewise, for $x\in
S^{\\{1,2\\}}$ define $\mathcal{C}_{x}^{\\{1,2\\}}(p)$ as the set of points of
$S^{\\{1,2\\}}$ that can be reached from $x$ using only those arrows from the
orthant model in $\mathbb{Z}^{d+1}$ that begin and end in $S^{\\{1,2\\}}$.
Therefore for $x\in S^{\\{i\\}}$ and $i\in\\{1,2\\}$ we have
$\mathcal{C}_{x}^{\\{i\\}}(p)\subset\mathcal{C}_{x}^{\\{1,2\\}}(p)\subset\mathcal{C}_{x,+1}(p)$.

As before, for $A\subset\mathbb{Z}^{d+1}$ let $A_{+}=\cup_{z\in
A}z_{+}\subset\mathbb{Z}^{d+1}$. Now let $x\in S^{\\{i\\}}$, where
$i\in\\{1,2\\}$. Clearly
$p_{c}(d+1)=\sup\\{p:\mathbb{P}\big{(}\mathcal{C}_{x,+1}(p)=\mathbb{Z}^{d+1}\big{)}>0\\}$
and
$p_{c}(d)=\sup\\{p:\mathbb{P}\big{(}\mathcal{C}_{x}^{\\{i\\}}(p)=S^{\\{i\\}}\big{)}>0\\}$.
We may also define

$p_{c}^{\\{1,2\\}}(d):=\sup\big{\\{}p:\mathbb{P}\big{(}\mathcal{C}_{x}^{\\{1,2\\}}(p)=S^{\\{1,2\\}}\big{)}>0\big{\\}}.$
(24)

From the above and the fact that a.s. for all $p<1$ a positive proportion of
vertices in any half line in $\mathbb{Z}^{d+1}$ are of type $\mathcal{E}$
(i.e. are in $\Omega_{0}$) one can easily deduce that

$p_{c}(d+1)\geq p_{c}^{\\{1,2\\}}(d)\geq p_{c}(d).$ (25)

We will show the following in the final section.

######  Proposition 39.

For $f(p)=p(1-(1-p)^{d+1})$ we have $p_{c}^{\\{1,2\\}}(d)\geq p_{c}(f,d)$.

###### Proof of Theorem 2.

Since $f(p)=p(1-(1-p)^{d+1})$ is the product of increasing functions, it is
increasing, and the claim follows immediately from (25) together with
Propositions 38 and 39. ∎

###  3.1 Disturbed orthant model

######  Lemma 40.

For any $p,q\in(0,1)$, the following hold:

  * _(1)_

for any $x\in\mathbb{Z}^{d}$, the events
$\\{\mathcal{C}_{x}(p,q)=\mathbb{Z}^{d}\\}$ and $\\{\mathcal{C}_{x}(p,q)\text{
is not bounded below}\\}$ are almost surely equal, and

  * _(2)_

for any $x\in S^{\\{1,2\\}}$ the events
$\\{\mathcal{C}^{\\{1,2\\}}_{x}(p)=S^{\\{1,2\\}}\\}$ and
$\\{\mathcal{C}^{\\{1,2\\}}_{x}(p)\text{ is not bounded below}\\}$ (as a
subset of $\mathbb{Z}^{d+1}$) are almost surely equal, and

  * _(3)_

each of the events in _(1)_ and _(2)_ has probability 0 or 1.

###### Proof.

The proofs of both (1) and (2) mimic that of [13, Lemma 2.2] as we now sketch.
Suppose that for some $y,i$, $\\{k\in\mathbb{N}:y-ke_{i}\in\mathcal{C}_{x}\\}$
is infinite. Since $\mathcal{C}_{x}$ is solid above, in fact
$y-ke_{i}\in\mathcal{C}_{x}$ for every $k\in\mathbb{N}$. But a.s. infinitely
many of the points $y-ke_{i}$ for $k\in\mathbb{N}$ are in $\Omega_{0}$ so
$y-e_{j}-ke_{i}\in\mathcal{C}_{x}$ for infinitely many $k\in\mathbb{N}$ and
for all $j$ (with the constraint that $y-e_{j}\in S^{\\{1,2\\}}$ in the case
(2)). This shows that if $\mathcal{C}_{x}$ is not bounded below then every
site is in $\mathcal{C}_{x}$, and (1) and (2) follow.

The statement of (3) now follows as in the proof of Theorem 1.11 of [13],
using invariance under translation by vectors $e_{i}$ for $i\neq d+1$ in the
case of $\mathcal{C}^{\\{1,2\\}}_{x}(p)$, and invariance under translation by
vectors in $V_{d}$ in the case of $\mathcal{C}_{x}(p,q)$. ∎

Note that we will not actually use part (3) of Lemma 40 above, but have
included it for completeness.

######  Lemma 41.

Let $x,x^{\prime}\in V_{d}$ be distinct. Then
$x^{\prime}-e_{i}\notin\\{x\\}\cup\\{x-e_{j}:j\in[d]\\}$.

###### Proof.

To see this note that $x^{\prime}-e_{i}$ cannot equal $x$, since if it did
then

$0<|h(x^{\prime})-h(x)|\leq\lceil d/2\rceil<\boldsymbol{\rho}_{d}.$

Similarly, if $x^{\prime}-e_{i}=x-e_{j}$ (and $i\neq j$ since $x$ and
$x^{\prime}$ are distinct) then $x\in V_{d}$ and $x^{\prime}=x+e_{i}-e_{j}\in
V_{d}$. However

$0<|h(x^{\prime})-h(x)|\leq 2\lceil d/2\rceil<\boldsymbol{\rho}_{d},$

so $h(x^{\prime})-h(x)\neq 0\mod\boldsymbol{\rho}_{d}$, giving a
contradiction. ∎

Recall that $\boldsymbol{1}=\sum_{i\in[d]}e_{i}$. The following result is the
crucial technical result that allows us to use the technology of enhancements
due to Aizenman and Grimmett (see [1, 5]). Given $n\in\mathbb{N}$, let $Q_{n}$
be the box $[-n,n]^{d}\cap\mathbb{Z}^{d}$. Using the notation in (9) we have
that $Q_{n}=[-n\boldsymbol{1},n\boldsymbol{1}]$.

Fix $N>M>n>\rho_{d}+4$ and set $R=Q_{N}$ and $S=(-M\boldsymbol{1})_{-}\cap R$.
Let $Q_{n,x,N}=(x+Q_{n})\cap R$. See Figure 4 for a depiction in 2 dimensions.
Let $\mathcal{P}_{M}(\boldsymbol{\omega}_{R})$ denote the set of vertices (in
$R$) that are pivotal for $o\to-M\boldsymbol{1}$ in $\boldsymbol{\omega}_{R}$
(or equivalently $o\to S$ in $\boldsymbol{\omega}_{R}$).

$(-N,N)$$(-M,M)$$(N,N)$$(0,0)$$u$ Figure 4: A depiction of $R=Q_{N}$, $S$, and
$Q_{n,u,N}$ in 2 dimensions.

######  Proposition 42.

Let $n>\boldsymbol{\rho}_{d}+4$, and $N>M>n$. Suppose that there exists
$u\in\mathcal{P}_{M}(\boldsymbol{\omega}_{R})\setminus V_{d}$. We may modify
$\boldsymbol{\omega}_{R}$ on $Q_{n,u,N}$, to give a new environment
$\bar{\boldsymbol{\omega}}_{R}$ for which $V_{d}\cap
Q_{n,u,N}\cap\mathcal{P}_{M}(\bar{\boldsymbol{\omega}}_{R})\neq\varnothing$.

Before we prove Proposition 42 let us demonstrate how it is used to prove
Proposition 38. Let $|A|$ denote the cardinality of $A\subset\mathbb{Z}^{d}$.
For $\boldsymbol{\omega}_{R}\in\\{\mathcal{E}_{+},\mathcal{E}\\}^{R}$, write
$P_{p,q}(\boldsymbol{\omega}_{R})$ to denote
$\mathbb{P}_{p,q}(\boldsymbol{\mathcal{G}}_{R}=\boldsymbol{\omega}_{R})$.

######  Lemma 43.

For every $p,q\in(0,1)$ and $n>\boldsymbol{\rho}_{d}+4$ there exists
$\alpha(p,q,n)\in(0,\infty)$ such that

$\mathbb{E}_{p,q}\big{[}|\mathcal{P}_{M}(\boldsymbol{\mathcal{G}}_{R})\setminus
V_{d}|\big{]}\leq\alpha(p,q,n)\mathbb{E}_{p,q}\big{[}|\mathcal{P}_{M}(\boldsymbol{\mathcal{G}}_{R})\cap
V_{d}|\big{]}$

for every $N>M>n$. If $\pi=[r_{1},r_{2}]$, where $0<r_{1}<r_{2}<1$, then for
fixed $n$, $\sup\limits_{p,q\in\pi}\alpha(p,z,n)<\infty$.

###### Proof.

Fix $n>\boldsymbol{\rho}_{d}+4$ and assume $N>M>n$. Define $R=Q_{N}$ and
$Q_{n,u,N}$ as above. Suppose
$u\in\mathcal{P}_{M}(\boldsymbol{\mathcal{G}}_{R})\setminus V_{d}$. Since
$o\in V_{d}$, we know that $u\neq o$.

If $\boldsymbol{\omega}_{R}$ is a configuration for which
$u\in\mathcal{P}_{M}(\boldsymbol{\omega}_{R})\setminus V_{d}$ then by
Proposition 42 there exists a modification $\bar{\boldsymbol{\omega}}_{R}$
(agreeing with $\boldsymbol{\omega}_{R}$ on $R\setminus(Q_{n}+u)$) such that
we can find a $\bar{u}\in\mathcal{P}_{M}(\bar{\boldsymbol{\omega}}_{R})\cap
V_{d}$. We can bound
$\frac{P_{p,q}(\boldsymbol{\omega}_{R})}{P_{p,q}(\bar{\boldsymbol{\omega}}_{R})}$
by a constant $a_{1}=a_{1}(p,q,n)$, since this involves changing the state of
at most $(2n+1)^{d}$ vertices.

For any given $u$ we may likewise bound the number of
$\boldsymbol{\omega}_{R}$ that can give rise to a given
$\boldsymbol{\bar{\omega}}_{R}$ by a constant $a_{2}=a_{2}(n)$, since only the
sites in $u+Q_{n}$ may change. Thus we may take $a_{2}=2^{(2n+1)^{d}}$ (though
this is a vast overcount since most configurations counted won’t make the
given $u$ pivotal). For any given $\bar{u}\in R\cap V_{d}$ there are at most
$a_{3}=a_{3}(n)=(2n+1)^{d}$ choices for $u$ that could make $u$ pivotal in
this way, because
$|u^{\scriptscriptstyle{[}i]}-\bar{u}^{\scriptscriptstyle{[}i]}|\leq n$ for
each $i$. Therefore

$\displaystyle\mathbb{E}_{p,q}\big{[}|\mathcal{P}_{M}(\boldsymbol{\mathcal{G}}_{R})\setminus
V_{d}|\big{]}$ $\displaystyle=\sum_{u\in R\setminus
V_{d}}\,\,\sum_{\begin{subarray}{c}\boldsymbol{\omega}_{R}\text{ with }\\\
u\text{ pivotal}\end{subarray}}P_{p,q}(\boldsymbol{\omega}_{R})$
$\displaystyle\leq\sum_{u\in R\setminus
V_{d}}\,\,\sum_{\begin{subarray}{c}\boldsymbol{\omega}_{R}\text{ with }\\\
u\text{ pivotal}\end{subarray}}\,\,\sum_{\begin{subarray}{c}\bar{u}\in R\cap
V_{d}:\\\ \|\bar{u}-u\|_{\infty}\leq
n\end{subarray}}\,\,\sum_{\begin{subarray}{c}\boldsymbol{\bar{\omega}}_{R}\text{
with }\bar{u}\text{ pivotal}:\\\
\boldsymbol{\bar{\omega}}_{R}=\boldsymbol{\omega}\text{ off
}Q_{n}+u\end{subarray}}P_{p,q}(\boldsymbol{\omega}_{R})$
$\displaystyle\leq\sum_{\bar{u}\in R\cap
V_{d}}\,\,\sum_{\begin{subarray}{c}\boldsymbol{\bar{\omega}}_{R}\text{ with
}\\\ \bar{u}\text{ pivotal}\end{subarray}}\sum_{\begin{subarray}{c}u\in
R\setminus V_{d}:\\\ \|\bar{u}-u\|_{\infty}\leq
n\end{subarray}}\sum_{\begin{subarray}{c}\boldsymbol{\omega}_{R}\text{ with
}u\text{ pivotal}:\\\
\boldsymbol{\omega}_{R}=\boldsymbol{\bar{\omega}}_{R}\text{ off
}Q_{n}+u\end{subarray}}P_{p,q}(\boldsymbol{\omega}_{R})$ $\displaystyle\leq
a_{1}\sum_{\bar{u}\in R\cap
V_{d}}\,\,\sum_{\begin{subarray}{c}\boldsymbol{\bar{\omega}}_{R}\text{ with
}\\\ \bar{u}\text{
pivotal}\end{subarray}}P_{p,q}(\boldsymbol{\bar{\omega}}_{R})\sum_{\begin{subarray}{c}u\in
R\setminus V_{d}:\\\ \|\bar{u}-u\|_{\infty}\leq
n\end{subarray}}\sum_{\begin{subarray}{c}\boldsymbol{\omega}_{R}\text{ with
}u\text{ pivotal}:\\\
\boldsymbol{\omega}_{R}=\boldsymbol{\bar{\omega}}_{R}\text{ off
}Q_{n}+u\end{subarray}}1$ $\displaystyle\leq a_{1}a_{2}a_{3}\sum_{\bar{u}\in
R\cap V_{d}}\,\,\sum_{\begin{subarray}{c}\boldsymbol{\bar{\omega}}_{R}\text{
with }\\\ \bar{u}\text{
pivotal}\end{subarray}}P_{p,q}(\boldsymbol{\bar{\omega}}_{R})$
$\displaystyle=\alpha(p,q,n)\mathbb{E}_{p,q}[\text{$\\#$ of pivotal vertices
in $R\cap V_{d}$}]$

and the constants involved depend on $n$ but not $M$ or $N$. The boundedness
statement is also clear from this argument. ∎

###### Proof of Proposition 38 (assuming Proposition 42).

Fix $n>\boldsymbol{\rho}_{d}+4$. Let $N>M>n$ both be large and as before, work
on a finite box $R=Q_{N}=[-N,N]^{d}\cap\mathbb{Z}^{d}$. Let
$\beta(p,q)=1-\theta(p,q)$ be the probability that
$\mathcal{C}_{x}(p,q)\neq\mathbb{Z}^{d}$.

Let $\beta_{M}^{N}(p,q)$ be the probability that $o$ fails to connect to $S$
via any path lying in $R$. Then $\beta_{M}^{N}(p,q)$ decreases with $N$,
increases with $M$, and

$\lim_{M\to\infty}\lim_{N\to\infty}\beta_{M}^{N}(p,q)=\beta(p,q).$ (26)

To see this, observe that $1-$ the limit equals the probability of the event
$C$ that $o$ connects to every $(-M\boldsymbol{1})_{-}$. Clearly
$\mathbb{P}_{p,q}(C)\geq\mathbb{P}_{p,q}(\mathcal{C}_{o}(p,q)=\mathbb{Z}^{d})=\theta(p,q)=1-\beta(p,q)$.
Conversely for every $z\in\mathbb{Z}^{d}$ we can find an $M$ such that
$z\in(-M\boldsymbol{1})_{+}$. Therefore if
$y\in(-M\boldsymbol{1})_{-}\cap\mathcal{C}_{o}(p,q)$ we also have
$z\in(-M\boldsymbol{1})_{+}\subset y_{+}\subset\mathcal{C}_{o}(p,q)_{+}$. In
other words, $\mathcal{C}_{o}(p,q)=\mathbb{Z}^{d}$ on the event $C$, showing
that
$\mathbb{P}_{p,q}(C)\leq\mathbb{P}_{p,q}(\mathcal{C}_{o}(p,q)=\mathbb{Z}^{d})$.
This verifies (26).

Fix $M$ and $N$. Consider the sites in $R$ that are pivotal for
$o\to-M\boldsymbol{1}$ in $\mathcal{G}_{R}$ (as defined after Corollary 35).
By Corollary 35, the event that $o$ fails to connect to $S$ via a path lying
in $R$ (under the coupling used above) increases with both $p$ and $q$.
Therefore a version of Russo’s formula applies (see [5]), showing that

$\displaystyle\frac{\partial}{\partial p}\beta_{M}^{N}(p,q)$
$\displaystyle=\mathbb{E}_{p,q}[\text{$\\#$ of pivotal vertices in $R\setminus
V_{d}$}]$ $\displaystyle\frac{\partial}{\partial q}\beta_{M}^{N}(p,q)$
$\displaystyle=\mathbb{E}_{p,q}[\text{$\\#$ of pivotal vertices in $R\cap
V_{d}$}].$

Note that Russo’s formula requires that there be only finitely many sites
under consideration, so this is the point where our proof requires $M$ and $N$
to be finite, i.e. the use of local terraces rather than terraces.

Proposition 42 implies Lemma 43, from which Russo’s formula gives that

$\frac{\partial}{\partial
p}\beta_{M}^{N}(p,q)\leq\alpha(p,q,n)\frac{\partial}{\partial
q}\beta_{M}^{N}(p,q).$

We now use an elementary argument (see [5]) to obtain an inequality of the
form

$\beta_{M}^{N}(p,q)\geq\beta_{M}^{N}(p+\delta,q-a\delta).$

Choose $\epsilon_{0}>0$ so that $p_{c}+2\epsilon_{0}<1$ and
$0<p_{c}-2\epsilon_{0}$. Set $\pi=[p_{c}-\epsilon_{0},p_{c}+\epsilon_{0}]$,
and let $a$ be the bound on $\alpha(p,q,n)$ over $\pi$, given by Lemma 43. By
continuity of $f$ and the fact that $p>f(p)$, we can find a $\delta_{0}>0$
such that $p-a\delta>f(p+\delta)$ whenever $p\in\pi$ and
$0<\delta<\delta_{0}$. Reducing $\delta_{0}$ if necessary, we can ensure that
$\delta_{0}(a+1)<\epsilon_{0}$.

Let $0<\delta<\delta_{0}$. Take $p=q=p_{c}-\frac{\delta}{2}$. Then
$p=q<p_{c}<p+\delta$, and all of $p,q,p+\delta,q-a\delta\in\pi$. By the mean
value theorem, there is some $\delta^{\prime}$ between $0$ and $\delta$ such
that

$\displaystyle 0\leq$ $\displaystyle\beta_{M}^{N}(p+\delta,q-a\delta)$
$\displaystyle\qquad=\beta_{M}^{N}(p,q)+\delta\Big{[}\frac{\partial\beta_{M}^{N}}{\partial
p}(p+\delta^{\prime},q-a\delta^{\prime})-a\frac{\partial\beta_{M}^{N}}{\partial
q}(p+\delta^{\prime},q-a\delta^{\prime})\Big{]}$
$\displaystyle\qquad\leq\beta_{M}^{N}(p,q)+\delta[\alpha(p,q,n)-a]\frac{\partial\beta_{M}^{N}}{\partial
q}(p+\delta^{\prime},q-a\delta^{\prime})$
$\displaystyle\qquad\leq\beta_{M}^{N}(p,q).$ (27)

By (23) and (26),
$\lim_{M\to\infty}\lim_{N\to\infty}\beta_{M}^{N}(p,p)=\beta(p,p)=0$, so taking
limits in (27) we get that
$\lim_{M\to\infty}\lim_{N\to\infty}\beta_{M}^{N}(p+\delta,p-a\delta)=0$ too.
Recalling that $p-a\delta>f(p+\delta)$, we conclude that
$\beta(p+\delta,f(p+\delta))=0$. In particular, $p_{c}(f,d)\geq
p+\delta>p_{c}(d)$, as required. ∎

###  3.2 Local modification

In order to complete the proof of Proposition 38 it therefore remains to prove
Proposition 42. Recall that $E=\mathcal{E}_{+}$ and $F=\mathcal{E}_{+}$.
Recall also that for boxes $Q\subset R$, $Q^{\scriptscriptstyle{R}:o}$ denotes
the set of sites in $Q$ that are not adjacent to sites in $R\setminus Q$.

The strategy of the proof is to start with an $R$-terrace of sites in
$\Omega_{1}$ that passes through $u$ (see e.g. Figure 5), and then push up
corners in a box around $u$, except for a few that are close to $u$. Because
of those nearby sites where we don’t push up, the resulting terrace must still
pass close to $u$. We will show that the absence of corners means that either
$u_{+}$ contains long straight segments of the terrace, or the geometry forces
there to be long straight segments from near $u$ backwards to the boundary of
the box. Either way, this terrace will contain a point $\bar{u}$ of $V_{d}$,
and one can then modify $\boldsymbol{\omega}$ to make that point pivotal and
the new terrace $\subset\Omega_{1}$.

$(-N,-N)$$(N,N)$$u$$o$ Figure 5: A depiction in 2 dimensions of an $R$ terrace
in $\Omega_{1}$, passing through $u$.

The argument is complicated by the possibility of $u$ lying close to a side of
$R$, or close to $o_{+}$, or close to $S$. All these possibilities result in
special cases of the argument. The reader is advised to ignore the caveats and
special cases at first, i.e. on first reading to assume that $u+Q_{n}$
intersects none of $o_{+}$, $\partial R$, or $S$.

###### Proof of Proposition 42.

blank

Recall that $N>M>n>\boldsymbol{\rho}_{d}+4$. Let $R=Q_{N}$,
$Q:=Q_{n,u,N}=(u+Q_{n})\cap R$ and $S=(-M\boldsymbol{1})_{-}\cap R$. Set
$\bar{Q}=R\cap(u+Q_{n-2})\subset Q$. We note here that

$\bar{Q}^{\scriptscriptstyle{R}:o}\subset\bar{Q}\subset
Q\setminus\partial_{\scriptscriptstyle{R}}Q=:Q^{\scriptscriptstyle{R}:o},$
(28)

since if $y\in\bar{Q}$ then $\|y-u\|_{\infty}\leq n-2$, while all
$y\in\partial_{\scriptscriptstyle{R}}Q$ have $\|y-u\|_{\infty}=n$, and all
$y\in\partial_{\scriptscriptstyle{R}}\bar{Q}$ have $\|y-u\|_{\infty}=n-2$.

Let $\boldsymbol{\omega}_{R}$ and
$u\in\mathcal{P}_{M}(\boldsymbol{\omega}_{R})\setminus V_{d}$ be as in the
statement of the lemma. Let $\boldsymbol{\omega}^{1}$ and
$\boldsymbol{\omega}^{0}$ be in $\\{\mathcal{E}_{+},\mathcal{E}\\}^{R}$, with
$\boldsymbol{\omega}^{1}_{u}=\mathcal{E}_{+}$ and
$\boldsymbol{\omega}^{0}_{u}=\mathcal{E}$ and both agreeing with
$\boldsymbol{\omega}_{R}$ on $R\setminus\\{u\\}$.

Terrace properties: Since $u$ is pivotal for $o\to-M\boldsymbol{1}$, by Lemma
32,
$\Delta=\lambda^{\scriptscriptstyle{R}}_{\mathcal{C}^{\scriptscriptstyle{R}}_{o}(\boldsymbol{\omega}^{1})}$
is an $R$-terrace satisfying
$\Delta\subset\mathcal{C}^{\scriptscriptstyle{R}}_{o}(\boldsymbol{\omega}^{1})\cap\Omega_{1}$,
and
$\Delta_{+}=(\mathcal{C}^{\scriptscriptstyle{R}}_{o}(\boldsymbol{\omega}^{1}))_{+}$,
and $S\subset R\setminus\Delta_{+}$. Since $o\in\Delta_{+}$ and
$-M\boldsymbol{1}\notin\Delta_{+}$ (and $u$ is pivotal for
$o\to-M\boldsymbol{1}$), by Lemma 36 $u\in\Delta$ and there is an $i$ with
$u+e_{i}\in R\setminus\Delta$. Because
$\Delta=\lambda^{\scriptscriptstyle{R}}_{\mathcal{C}^{\scriptscriptstyle{R}}_{o}(\boldsymbol{\omega}^{1})}$,
Lemma 21 implies that
$R\cap\Delta_{+}=R\cap\mathcal{C}^{\scriptscriptstyle{R}}_{o}(\boldsymbol{\omega}^{1})=\mathcal{C}^{\scriptscriptstyle{R}}_{o}(\boldsymbol{\omega}^{1})$
(recall that
$\mathcal{C}^{\scriptscriptstyle{R}}_{o}(\boldsymbol{\omega}^{1})$ is
$R$-solid above). Therefore

$\Delta=\lambda^{\scriptscriptstyle{R}}_{R\cap\Delta_{+}}.$ (29)

Moreover, from Lemma 34 we have that

$H^{\scriptscriptstyle{R}}_{\Delta}\subset o_{+}.$ (30)

We will find an $R$-terrace $\overline{\Delta}$ with the following properties:

$\overline{\Delta}=\lambda^{\scriptscriptstyle{R}}_{R\cap(\overline{\Delta})_{+}}$
(31) $\overline{\Delta}=\Delta$ on $R\setminus Q^{\scriptscriptstyle{R}:o}$,
i.e. $\overline{\Delta}\cap(R\setminus
Q^{\scriptscriptstyle{R}:o})=\Delta\cap(R\setminus
Q^{\scriptscriptstyle{R}:o})$ (32) $(\overline{\Delta})_{+}=\Delta_{+}$ on
$R\setminus Q^{\scriptscriptstyle{R}:o}$, i.e.
$(\overline{\Delta})_{+}\cap(R\setminus
Q^{\scriptscriptstyle{R}:o})=\Delta_{+}\cap(R\setminus
Q^{\scriptscriptstyle{R}:o})$ (33)
$\displaystyle(\overline{\Delta})_{+}\subset\Delta_{+}$ (34) $\displaystyle
o\in(\overline{\Delta})_{+}$ (35) $\displaystyle S\subset
R\setminus(\overline{\Delta})_{+}$ (36)
$\exists\bar{u}\in\overline{\Delta}\cap V_{d}\cap\bar{Q}$ such that either
$\bar{u}\notin H^{\scriptscriptstyle{R}}_{\overline{\Delta}}$ or $\bar{u}\in
o_{+}$. (37)

We’ll assume for now that this can be done, and will show how to modify
$\boldsymbol{\omega}$ on $Q$ so as to make $\bar{u}\in V_{d}\cap Q$ (appearing
in (37)) pivotal.

Finding paths in $R$: Define
$\boldsymbol{\bar{\boldsymbol{\omega}}}^{1}\in\\{\mathcal{E}_{+},\mathcal{E}\\}^{R}$
as follows. Let $\boldsymbol{\bar{\omega}}^{1}$ agree with
$\boldsymbol{\omega}^{1}$ (and hence $\boldsymbol{\omega}_{R}$) off $Q$.
Choose $\boldsymbol{\bar{\omega}}^{1}$ to be $\mathcal{E}_{+}$ on
$\overline{\Delta}$, and note that this is consistent with
$\boldsymbol{\omega}^{1}$ off $Q$, by (32).

Because $\overline{\Delta}$ is an $R$-terrace, Lemma 31 together with (35) and
(36) shows that there is no path in $\boldsymbol{\bar{\omega}}^{1}$ from $o$
to $S$, regardless of how $\boldsymbol{\bar{\omega}}^{1}$ is defined at the
remaining sites of $Q$. Set $\boldsymbol{\bar{\omega}}^{1}$ to be
$\mathcal{E}$ at any such remaining sites of $Q$.

Now let $\boldsymbol{\bar{\omega}}^{0}$ agree with
$\boldsymbol{\bar{\omega}}^{1}$ at all sites, except at $u$ where we take
$\boldsymbol{\bar{\omega}}^{0}_{u}=\mathcal{E}$ (note that
$\boldsymbol{\bar{\omega}}^{1}_{u}=\mathcal{E}_{+}$ since
$u\in\overline{\Delta}$).

Because $u$ is pivotal, we know that $\boldsymbol{\omega}^{0}$ connects $o$ to
some $y^{\scriptscriptstyle{(}0)}\in S$. Taking its initial and final
portions, we have a (self-avoiding) path in $R$ consistent with
$\boldsymbol{\omega}^{0}$ that runs from $o$ to $u$ and another from $u$ to
$y^{\scriptscriptstyle{(}0)}$. That first segment of path lies in
$\mathcal{C}^{\scriptscriptstyle{R}}_{o}(\boldsymbol{\omega}^{1})\subset\Delta_{+}$
since it does not make use of the environment at $u$, i.e. is consistent with
$\boldsymbol{\omega}^{1}$ as well as $\boldsymbol{\omega}^{0}$.

Let $y^{\scriptscriptstyle{(}1)}$ be the first point in the former path that
is in $Q$ (this is well defined since $u\in Q$), and let
$y^{\scriptscriptstyle{(}2)}$ be the last point that the latter path
encounters $Q$. In other words, there is a path from $o$ to
$y^{\scriptscriptstyle{(}1)}\in\Delta_{+}$ that encounters $Q$ only at its
last step, and a path from $y^{\scriptscriptstyle{(}2)}$ to
$y^{\scriptscriptstyle{(}0)}$ that never returns to $Q$ after its first step.
We claim these paths are consistent with $\boldsymbol{\bar{\omega}}^{0}$, and
therefore that

$\text{$\boldsymbol{\bar{\omega}}^{0}$ connects $o$ to
$y^{\scriptscriptstyle{(}1)}$ and $y^{\scriptscriptstyle{(}2)}$ to
$y^{\scriptscriptstyle{(}0)}$}.$ (38)

Note that $\boldsymbol{\omega}^{0}$ makes these connections, that
$\boldsymbol{\bar{\omega}}^{0}=\boldsymbol{\omega}^{0}$ off $Q$, and that the
only environment in $Q$ used by these paths is at
$y^{\scriptscriptstyle{(}2)}$. We will show that
$y^{\scriptscriptstyle{(}2)}\notin\overline{\Delta}$, so
$\boldsymbol{\bar{\omega}}^{0}_{y^{\scriptscriptstyle{(}2)}}=\mathcal{E}$ by
construction, which makes $\boldsymbol{\bar{\omega}}^{0}$ consistent with
whatever step our path takes to leave $y^{\scriptscriptstyle{(}2)}$. This will
therefore be sufficient to verify the claim.

If $Q$ intersects $S$ it could happen that
$y^{\scriptscriptstyle{(}2)}=y^{\scriptscriptstyle{(}0)}\in S\cap Q$ (just as
it could happen that $y^{\scriptscriptstyle{(}1)}=o$, if $o\in Q$, though
otherwise $y^{\scriptscriptstyle{(}1)}\in\partial_{\scriptscriptstyle{R}}Q$).
Regardless, we know that either $y^{\scriptscriptstyle{(}2)}\in S$ or
$y^{\scriptscriptstyle{(}2)}\in\partial_{\scriptscriptstyle{R}}Q$. Either way,
we have that $y^{\scriptscriptstyle{(}2)}\neq u$. Any path in $R$ from
$\Delta_{+}$ to $S$, that is consistent with $\boldsymbol{\omega}^{0}$, must
pass through $u\in Q$. Therefore $y^{\scriptscriptstyle{(}2)}\notin\Delta_{+}$
by definition, so $y^{\scriptscriptstyle{(}2)}\notin(\overline{\Delta})_{+}$
by (34). In particular, $y^{\scriptscriptstyle{(}2)}\notin\overline{\Delta}$,
establishing the claim.

We now show that $\bar{u}$ connects to $S$ in $\boldsymbol{\bar{\omega}}^{0}$
and $o$ connects to $\bar{u}$ in $\boldsymbol{\bar{\omega}}^{0}$. This will
show that $o$ connects to $S$ in $\boldsymbol{\bar{\omega}}^{0}$. On the other
hand $o$ does not connect to $S$ in $\boldsymbol{\bar{\omega}}^{1}$ (which
agrees with $\boldsymbol{\bar{\omega}}^{0}$ except at $\bar{u}$) showing that
$\bar{u}$ is pivotal.

Connecting $\bar{u}$ to $S$: Because $\bar{u}\in\bar{Q}\cap\overline{\Delta}$,
(31) allows us to find an $i\in[d]$ such that $\bar{u}-e_{i}\in
R\setminus(\overline{\Delta})_{+}$ (and $\bar{u}-e_{i}\in Q_{n-1}$). Therefore
in fact $\bar{u}-e_{i}\in Q\setminus(\overline{\Delta})_{+}$. Using (iii) of
Lemma 18, there is a self-avoiding nearest neighbour path in
$Q\setminus(\overline{\Delta})_{+}$ that connects $\bar{u}-e_{i}$ to
$y^{\scriptscriptstyle{(}2)}$. All the $\boldsymbol{\bar{\omega}}^{0}$
environments along this path are of type $\mathcal{E}$, so
$\boldsymbol{\bar{\omega}}^{0}$ connects $\bar{u}$ to
$y^{\scriptscriptstyle{(}2)}$ and hence to $S$.

Connecting $o$ to $\bar{u}$: (37) tells us that either $\bar{u}\notin
H^{\scriptscriptstyle{R}}_{\overline{\Delta}}$ or $\bar{u}\in o_{+}$. In the
latter case we may trivially connect $o$ to $\bar{u}$ in
$\boldsymbol{\bar{\omega}}^{0}$ (since such a connection can be made using
only $+$ steps, which are available from every site) as required.

We may therefore assume that $\bar{u}\notin
H_{\overline{\Delta}}^{\scriptscriptstyle{R}}$. Accordingly, we may find
$\bar{u}+e_{j}\in R\cap(\overline{\Delta})_{+}\setminus\overline{\Delta}$, and
since $\bar{u}\in\bar{Q}$, in fact $\bar{u}+e_{j}\in
Q\cap(\overline{\Delta})_{+}\setminus\overline{\Delta}$. As remarked above,
either $y^{\scriptscriptstyle{(}1)}\in\partial_{\scriptscriptstyle{R}}Q$ or
$y^{\scriptscriptstyle{(}1)}=o\in V_{d}$, so we must have
$y^{\scriptscriptstyle{(}1)}\neq u$. Since
$y^{\scriptscriptstyle{(}1)}\in\Delta_{+}$, using either (33) (when
$y^{\scriptscriptstyle{(}1)}\in\partial_{\scriptscriptstyle{R}}Q$) or (35) we
conclude that $y^{\scriptscriptstyle{(}1)}\in(\overline{\Delta})_{+}$.

If $y^{\scriptscriptstyle{(}1)}\notin\overline{\Delta}$ then (iii) of Lemma 18
allows us to connect $y^{\scriptscriptstyle{(}1)}$ to $\bar{u}+e_{j}$ via a
self-avoiding path in
$Q\cap(\overline{\Delta})_{+}\setminus\overline{\Delta}$. All the
$\boldsymbol{\bar{\omega}}^{0}$ environments along this path will be of type
$\mathcal{E}$, including that at $\bar{u}+e_{j}$. Therefore
$\boldsymbol{\bar{\omega}}^{0}$ connects $y^{\scriptscriptstyle{(}1)}$ to
$\bar{u}$ and hence (by (38)) also to $\bar{u}$. This shows that $\bar{u}$ is
pivotal when $y^{\scriptscriptstyle{(}1)}\notin\overline{\Delta}$.

It is however possible that $y^{\scriptscriptstyle{(}1)}\in\overline{\Delta}$,
in which case we require an additional argument. In this case, we may take a
step from $y^{\scriptscriptstyle{(}1)}$ in some direction $e_{k}$ that keeps
us in $Q$, provided such a $k$ exists. While we are still in
$\overline{\Delta}$, repeat. Eventually one of the following will happen:

  * (i)

we reach $\bar{u}$ itself; or

  * (ii)

we reach some point $y^{\scriptscriptstyle{(}3)}\in
Q\cap(\overline{\Delta})_{+}\setminus\overline{\Delta}$; or

  * (iii)

we exhaust the possible $+$ steps that keep us in $Q$, before leaving
$\overline{\Delta}$ (i.e. we reach the point $b$, where $Q=[a,b]$, and
$b\in\overline{\Delta}$).

In fact (iii) cannot happen, as we now argue. Recall that we had found a
$\bar{u}+e_{j}\in Q\cap(\overline{\Delta})_{+}\setminus\overline{\Delta}$. But
$b\in(\bar{u}+e_{j})_{+}$, so Lemma 20 rules out having
$b\in\overline{\Delta}$.

In the case (i) we’re done, i.e. $\boldsymbol{\bar{\omega}}^{0}$ connects
$y^{\scriptscriptstyle{(}1)}$ to $\bar{u}$. In the case (ii), since
$y^{\scriptscriptstyle{(}3)}\in
Q\cap(\overline{\Delta})_{+}\setminus\overline{\Delta}$, Lemma 18 allows us to
connect $y^{\scriptscriptstyle{(}3)}$ to $\bar{u}+e_{j}$ via a self-avoiding
path in $Q\cap(\overline{\Delta})_{+}\setminus\overline{\Delta}$ and (as in
the case $y^{\scriptscriptstyle{(}1)}\in
Q\cap(\overline{\Delta})_{+}\setminus\overline{\Delta}$) we conclude that
$\boldsymbol{\bar{\omega}}^{0}$ connects $y^{\scriptscriptstyle{(}1)}$ to
$y^{\scriptscriptstyle{(}3)}$ to $\bar{u}+e_{j}$ to $\bar{u}$.

Thus we have shown that $\bar{u}$ is pivotal.

We now turn to the construction of $\overline{\Delta}$ and $\bar{u}$ (see e.g.
Figure 6), broken into three cases. _We assume for now that $o\notin\bar{Q}$.
This assumption will be in place for the first two cases, and we will come
back to the case $o\in\bar{Q}$ at the end of the proof. _

$Q$$\bar{Q}$$u$$A$ Figure 6: A planar $\Delta$ (shown with dots) and a
corresponding $\overline{\Delta}$ (shown with circles), with $n=7$. $A$ is
shown in red. The construction locates $\bar{u}$ either among the yellow nodes
(if $i=2$) or the green nodes (if $i=1$)

Case I: $o\notin\bar{Q}$, $u\in o_{+}$.

Assume first that $u\in o_{+}$. Because $u\in\Delta$ and $o\in\Delta_{+}$,
(ix) of Lemma 7 implies that $[o,u]\subset\Delta$. Since we assumed that
$o\notin\bar{Q}$, we must have $u-(n-2)e_{j}\in[o,u]$ for any $j$. Since
$n-1>\boldsymbol{\rho}_{d}$, Lemma 37 implies that we can find a $\bar{u}\in
V_{d}$ of the form $u-ke_{j}$, $0\leq k\leq n-2$. Then properties (31)–(37)
hold with $\overline{\Delta}=\Delta$. Relation (31) is immediate from (29).
Relations (32)–(34) are trivially true. Relation (35) is immediate since
$o\in\mathcal{C}^{\scriptscriptstyle{R}}_{o}(\boldsymbol{\omega}^{1})$, and
(36) holds since $-M\boldsymbol{1}\notin\Delta_{+}$. Finally, (37) holds since
our $u\in o_{+}$.

Case II: $o\notin\bar{Q}$, $u\notin o_{+}$.

Construction of $\overline{\Delta}$: Assume now that $u\notin o_{+}$ . Recall
(29) and (30), i.e. that
$\Delta=\lambda^{\scriptscriptstyle{R}}_{R\cap\Delta_{+}}$ and
$H^{\scriptscriptstyle{R}}_{\Delta}\subset o_{+}$. By Lemma 36 and the fact
that $u\notin o_{+}$, there exists $i\in[d]$ such that $u+e_{i}\in
R\setminus\Delta$. Without loss of generality, we assume $u+e_{1}\in
R\setminus\Delta$. By definition of
$\lambda^{\scriptscriptstyle{R}}_{R\cap\Delta_{+}}$ we conclude that

$\text{ for each $i\in[d]$, either \quad$u+e_{1}-e_{i}\notin R$ \quad or
\quad$u+e_{1}-e_{i}\in R\cap\Delta_{+}$}.$ (39)

Let $I$ be the set of $i$ for which the latter holds.

Define $A=\\{u+e_{1}-e_{i}:i\in I\\}\subset R\cap\Delta_{+}$. Note that $1\in
I$ so $u\in A$. We now push up the corners of $\Delta$ in
$\bar{Q}^{\scriptscriptstyle{R}:o}\setminus A$, to give
$\Delta^{\prime}=\Delta^{\scriptscriptstyle{\bar{Q},A}}$. By definition
$\Delta^{\prime}$ has no corners in
$\bar{Q}^{\scriptscriptstyle{R}:o}\setminus A$. Lemma 28(ii) with
$G=R\cap\Delta_{+}$ shows that $(\Delta^{\prime})_{+}=\Delta_{+}$ off
$\bar{Q}^{\scriptscriptstyle{R}:o}$ and by Lemma 28 (i) also
$\Delta^{\prime}=\Delta$ off $\bar{Q}$ (and therefore also off
$Q^{\scriptscriptstyle{R}:o}$ by (28)). This process might lead to the
creation of sites in $H^{\scriptscriptstyle{R}}_{\Delta^{\prime}}\setminus
o_{+}$. But if it does, any points in that set must lie in
$Q^{\scriptscriptstyle{R}:o}$. In other words,

$(Q^{\scriptscriptstyle{R}:o})^{c}\cap
H^{\scriptscriptstyle{R}}_{\Delta^{\prime}}\setminus o_{+}=\varnothing.$ (40)

To see this, we will show that if $z\in(Q^{\scriptscriptstyle{R}:o})^{c}\cap
H^{\scriptscriptstyle{R}}_{\Delta^{\prime}}\setminus o_{+}$ then $z\in
H^{\scriptscriptstyle{R}}_{\Delta}\setminus o_{+}$, and then appeal to the
fact that the latter set is empty. Suppose $z$ is in
$H^{\scriptscriptstyle{R}}_{\Delta’}$ and also outside
$Q^{\scriptscriptstyle{R}:o}$. Then each $z+e_{i}$ is in $\Delta’$ or not in
$R$. Suppose $z+e_{i}$ is in $\Delta’$ but not in $\Delta$. Since
$\Delta=\Delta’$ off $\bar{Q}$, we must have that
$z+e_{i}\in\bar{Q}\subset(u+Q_{n-2})$. Therefore $z\in u+Q_{n-1}$, so every
neighbour of $z$ is either in $Q$ or not in $R$. Therefore $z$ is actually in
$Q^{\scriptscriptstyle{R}:o}$, which is a contradiction. In other words, each
$z+e_{i}$ is in $\Delta$ or not in $R$. So $z\in
H^{\scriptscriptstyle{R}}_{\Delta}$ as well. As noted above this proves (40).

By repeated applications of Lemma 25(i) there would then be corners in
$H^{\scriptscriptstyle{R}}_{\Delta^{\prime}}\setminus o_{+}$, and by (40)
these corners must be in $Q^{\scriptscriptstyle{R}:o}$. These corners can be
successively deleted (i.e. pushed up), as in Lemma 25(iii) without making any
changes off $Q^{\scriptscriptstyle{R}:o}$. This gives a new $R$-terrace
$\Delta^{\prime\prime}\subset\Delta^{\prime}$, which by construction has no
corners in $H^{\scriptscriptstyle{R}}_{\Delta^{\prime\prime}}\setminus o_{+}$.
It follows that $H^{\scriptscriptstyle{R}}_{\Delta^{\prime\prime}}\subset
o_{+}$ (since if not then (as above) repeated applications of Lemma 25(i)
would produce corners in
$H^{\scriptscriptstyle{R}}_{\Delta^{\prime\prime}}\setminus o_{+}$).

Now $\Delta^{\prime\prime}$ might have new corners in
$\bar{Q}^{\scriptscriptstyle{R}:o}\setminus A$, but we can push these up in
turn to get $\Delta^{\prime\prime\prime}$, etc. Applying the two above
procedures repeatedly, in turn, we eventually stabilize at an $R$-terrace
which we take to be $\overline{\Delta}$. By construction, it has no corners in
$\bar{Q}^{\scriptscriptstyle{R}:o}\setminus A$, and
$H^{\scriptscriptstyle{R}}_{\overline{\Delta}}\subset o_{+}$.

Verification of the Terrace properties: Properties (32) and (33) follow
automatically from our construction. Property (34) follows, because every time
we push up or delete a corner, the region above that terrace gets reduced.
Property (31) follows from Remark 23, since (14) implies that pushing up
corners always yields terraces to which that Remark applies.

Property (36) follows from (34), since $S\subset R\setminus\Delta_{+}$. By
Lemma 25(ii), pushing up corners in $\bar{Q}^{\scriptscriptstyle{R}:o}$ will
never remove $o$, which was assumed to be $\notin\bar{Q}$. Therefore (35)
follows as well, since the process of Lemma 25(iii) is never applied to delete
$o$ since $o\in o_{+}$. Therefore it only remains to find a
$\bar{u}\in\overline{\Delta}\cap V_{d}\cap\bar{Q}$. Note that since
$H^{\scriptscriptstyle{R}}_{\bar{\Delta}}\subset o_{+}$, any such $\bar{u}$
will satisfy (37).

To do that we will first establish the following properties of the
construction:

$A\subset(\overline{\Delta})_{+}$, $u+e_{1}\in R\setminus\overline{\Delta}$,
and $u\in\overline{\Delta}$. (41)

We know that $u\in\Delta$, which has the form
$\lambda^{\scriptscriptstyle{R}}_{G}$. So there is an $r\in[d]$ such that
$u-e_{r}\in R\setminus\Delta_{+}$. As in Remark 26,
$(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}\subset\Delta_{+}$, therefore
$u-e_{r}\in R\setminus(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}$ too. Since
$u\in A\subset(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}$ by definition of
$\Delta^{\scriptscriptstyle{\bar{Q},A}}$ and Lemma 25(ii) (or Lemma 28(ii)),
it follows that
$u\in\lambda^{\scriptscriptstyle{R}}_{R\cap(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}}\subset\Delta^{\scriptscriptstyle{\bar{Q},A}}$,
the latter by Lemma 22(i). Now $u+e_{1}-e_{r}\in R$ (since $u+e_{1}\in R$ and
$u-e_{r}\in R$, and $R$ is a box). So by definition of $A$ and $I$,
$u+e_{1}-e_{r}\in A\subset(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}$, and
because $u-e_{r}\in R\setminus(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}$,
it follows from Remark 23 that
$u+e_{1}-e_{r}\in\Delta^{\scriptscriptstyle{\bar{Q},A}}$. Since
$u+e_{1}-e_{i}\in(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}$ for all $i\in
I$ (by definition of $A$ and the fact that
$A\subset(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}$ as above), we conclude
that
$u+e_{1}\in(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}\setminus\Delta^{\scriptscriptstyle{\bar{Q},A}}$.
To see this, note that if
$u+e_{1}\in\Delta^{\scriptscriptstyle{\bar{Q},A}}=\lambda^{\scriptscriptstyle{R}}_{R\cap(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}}$
(the latter by Remark 23) then there exists $i_{0}\in[d]$ with
$u+e_{1}-e_{i_{0}}\in R\setminus(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}$.
So $i_{0}\in I$ (by (39)) but
$u+e_{1}-e_{i_{0}}\notin(\Delta^{\scriptscriptstyle{\bar{Q},A}})_{+}$, giving
a contradiction.

Since $u+e_{1}\in R\setminus\Delta^{\scriptscriptstyle{\bar{Q},A}}$ we have
that for each $i\in[d]$, $u+e_{1}-e_{i}\notin
H_{\Delta^{\scriptscriptstyle{\bar{Q},A}}}^{\scriptscriptstyle{R}}$ (so in
particular $u\notin
H_{\Delta^{\scriptscriptstyle{\bar{Q},A}}}^{\scriptscriptstyle{R}}$). This
shows that no point of $A$ will be deleted from
$\Delta^{\scriptscriptstyle{\bar{Q},A}}$ by applying the process of Lemma 25
(iii). Pushing up corners in $\bar{Q}^{\scriptscriptstyle{R}:o}\setminus A$
won’t excise any part of $A$ either (Lemma 25(ii)), so in fact, points of $A$
survive every iteration in the construction of $\overline{\Delta}$, which
verifies that $A\subset(\overline{\Delta})_{+}$. That implies the second
property of (41), since for every $k$, either
$u+e_{1}-e_{k}\in\overline{\Delta}$ (for $k\in I$) or it $\notin R$ (so
$k\notin I$), so $u+e_{1}$ can’t be in $\overline{\Delta}$. The remaining
property, that $u\in\overline{\Delta}$, now follows since $u\in
A\subset\overline{\Delta}_{+}$ while as noted above, for some $r\in[d]$ we
have $u-e_{r}\in R\setminus\Delta_{+}\subset R\setminus\overline{\Delta}_{+}$.

Finding $\bar{u}$: There must exist indexes $i\neq j$ with $u+ne_{i}\in R$ and
$u-ne_{j}\in R$, since every index has at least one of these properties (as
$n<N$), and neither property fails for every index (as
$u\notin\boldsymbol{1}_{+}$ and $u\notin S$). Therefore $u+ne_{i}\in Q$ and
$u-ne_{j}\in Q$.

_Assume first that $i\neq 1$._ If $u+(n-3)e_{i}\in\overline{\Delta}$ then
Lemma 20 (cf. (ix) of Lemma 7) implies that $u+ke_{i}\in\overline{\Delta}$ for
all $k$ such that $0\leq k\leq n-3$ (since we already know that
$u\in\overline{\Delta}$). But $n-2\geq\rho_{d}$ so by Lemma 37 at least one of
these points is in $V_{d}$. We may take that point as $\bar{u}$ in this case,
since all such points also lie in $\bar{Q}$.

Otherwise $u+(n-3)e_{i}\in\bar{Q}\setminus\overline{\Delta}$, so there is a
first $q\in[1,n-3]$ such that $u+qe_{i}\notin\overline{\Delta}$. Set
$y_{0}=u+(q-1)e_{i}-e_{j}$ and $y_{1}=u+qe_{i}-e_{j}$. By choice of $i$ and
$j$, both $y_{0}\in\bar{Q}\subset R$ and $y_{1}\in\bar{Q}\subset R$. We claim
that at least one of these (which we will then denote $y$) is in
$\overline{\Delta}$. To see this, note that as $u+qe_{i}\in
R\cap(\overline{\Delta})_{+}\setminus\overline{\Delta}$ it follows (e.g. apply
the conclusion of Remark 23 to the point $u+qe_{1}\in(\overline{\Delta})_{+}$)
that $y_{1}\in(\overline{\Delta})_{+}$. If $y_{1}\in\overline{\Delta}$ then
take $y=y_{1}$ to establish the claim. Otherwise
$y_{1}\notin\overline{\Delta}$, so (again by Remark 23 at the point
$y_{1}\in(\overline{\Delta})_{+}$)
$y_{0}=y_{1}-e_{i}\in(\overline{\Delta})_{+}$. But we know that
$y_{0}+e_{j}=u+(q-1)e_{i}\in\overline{\Delta}$, which (by Lemma 20) implies
that $y:=y_{0}\in\overline{\Delta}$, which verifies the claim. The assumption
that $i\neq 1$ implies that $y\notin A_{+}$, since every element of $A_{+}$ is
either in $u_{+}$ or has first component at least
$u^{\scriptscriptstyle{[}1]}+1$. Because
$\overline{\Delta}=\overline{\Delta}^{\scriptscriptstyle{\bar{Q},A}}$ (recall
the sentence prior to “verification of the Terrace properties), and
$y\in\bar{Q}$, Lemma 28 (iii) provides a
$z\in(\overline{\Delta})_{+}\cap(A\cup\partial_{\scriptscriptstyle{R}}\bar{Q})$
with $y\in z_{+}$. Since $y\notin A_{+}$, in fact
$z\in(\overline{\Delta})_{+}\cap\partial_{\scriptscriptstyle{R}}\bar{Q}$. Thus
$\|z-u\|_{\infty}=n-2$, so there is a $k\in[d]$ such that
$|z^{\scriptscriptstyle{[}k]}-u^{\scriptscriptstyle{[}k]}|=n-2$. Because
$y\in\overline{\Delta}$, Lemma 20 (cf. Lemma 7 (ix)) shows that $z_{+}\cap
y_{-}\subset\overline{\Delta}$, i.e. $[z,y]\subset\overline{\Delta}$. We have
$z^{\scriptscriptstyle{[}i]}\leq
y^{\scriptscriptstyle{[}i]}<u^{\scriptscriptstyle{[}i]}+n-2$,
$z^{\scriptscriptstyle{[}j]}\leq
y^{\scriptscriptstyle{[}j]}=u^{\scriptscriptstyle{[}j]}-1<u^{\scriptscriptstyle{[}j]}+n-2$,
and for $k\neq i,j$, $z^{[k]}\leq y^{[k]}=u^{[k]}$. Now
$z^{\scriptscriptstyle{[}k]}-u^{\scriptscriptstyle{[}k]}<n-2$ for each $k$ and
$|z^{\scriptscriptstyle{[}k]}-u^{\scriptscriptstyle{[}k]}|=n-2$ for some $k$
implies that $z^{[k]}=u^{[k]}-(n-2)$ for some $k$. But $y^{[k]}\geq
u^{[k]}-1$, so $z^{[k]}\leq y^{[k]}-(n-3)$. Therefore
$y-(n-3)e_{k}\in[z,y]\subset\overline{\Delta}$. Arguing as before, there is at
least one point from $V_{d}$ among the $n-2$ points $y-\ell e_{i}$,
$0\leq\ell\leq n-3$, which we take as $\bar{u}$.

_Special case $i=1$:_ Now turn to the case $i=1$. Let $v=u+e_{1}-e_{j}$. By
choice of $j$, we know $v\in R$, so in fact $v\in A\subset\overline{\Delta}$.
We will repeat the previous argument, using $v$ instead of $u$.

We know that $v+(n-4)e_{i}\in R$. If $v+(n-4)e_{i}\in\overline{\Delta}$ then
we may repeat the earlier argument (now using that $n-3\geq\rho_{d}$) to see
that $v+ke_{i}\in\overline{\Delta}\cap V_{d}\cap\bar{Q}$ for some $0\leq k\leq
n-4$, which we may then take to be $\bar{u}$. Otherwise
$v+(n-4)e_{i}\notin\overline{\Delta}$ and as before we may find a
$q\in[1,n-4]$ and a $y\in\overline{\Delta}$, with either
$y=y_{0}:=v+(q-1)e_{i}-e_{j}$, or $y=y_{1}:=v+qe_{i}-e_{j}$. Equivalently,
$y=u+qe_{i}-2e_{j}$ or $y=u+(q-1)e_{i}-2e_{j}$, which again implies that
$y\notin A_{+}$.

Now $q\leq n-4$ implies that $y\in\bar{Q}$. Arguing as before, there is a
$z\in(\overline{\Delta})_{+}\cap\partial_{\scriptscriptstyle{R}}\bar{Q}$ with
$y\in z_{+}$. Therefore $[z,y]\subset\overline{\Delta}$. There will now be at
least one coordinate $k$ such that
$z^{\scriptscriptstyle{[}k]}=u^{\scriptscriptstyle{[}k]}-(n-2)\leq
y^{\scriptscriptstyle{[}k]}-(n-4)$ and as before, one of the points $y-\ell
e_{k}$, $0\leq\ell\leq n-4$ will be in $\overline{\Delta}\cap
V_{d}\cap\bar{Q}$. We take that point as $\bar{u}$.

Case III: $o\in\bar{Q}$.

Again, there are two possibilities, namely that $u\in o_{+}$ or $u\notin
o_{+}$. If $u\in o_{+}$ then as before we have $[o,u]\subset\Delta$. So in
fact, $\overline{\Delta}=\Delta$ and $\bar{u}=o\in V_{d}$ satisfy the desired
conditions.

Therefore assume $u\notin o_{+}$. In fact, every step of the previous argument
carries through as before, up to the verification of (35), which could fail.
So consider the procedure we used to obtain $\overline{\Delta}$, except carry
out the excisions or pushing up of corners one by one. If at every stage of
this process we have an $R$-terrace $\Delta^{\prime}$ for which
$o\in(\Delta^{\prime})_{+}$, then the process can continue to the end, to give
$\overline{\Delta}$ and $\bar{u}$ satisfying the desired conditions. But if at
some stage $o$ ceases to be $\in(\Delta^{\prime})_{+}$, we instead stop the
process and let $\overline{\Delta}$ be the last $\Delta^{\prime}$ with
$o\in(\Delta^{\prime})_{+}$. Then in fact $o\in\overline{\Delta}$, as each
step of pushing up loses only a point in the current terrace. Now, every step
of the construction process preserves (31),(32),(33),(34),(36), while our
stopping rule means that (35) holds as well. But also $o\in V_{d}$ and $o\in
o_{+}$. So in fact, all the desired conditions will hold for this
$\overline{\Delta}$ with the choice $\bar{u}=o$.

This completes the proof. ∎

###  3.3 Orthant model for slabs

Recall the notation introduced after Proposition 38.

In this section, terraces for $\mathbb{Z}^{d+1}$ will be called
$(d+1)$-terraces, to distinguish them from what we call $d$-terraces, which
will be subsets of $S^{\\{i\\}}=\mathbb{Z}^{d}\times\\{i\\}$ for $i\in[2]$.
For a subset $A^{\\{i\\}}$ of $S^{\\{i\\}}$ we will let $(A^{\\{i\\}})_{+[d]}$
denote the set of points $z\in S^{\\{i\\}}$ for which there exists $y\in
A^{\\{i\\}}$ such that $z^{\scriptscriptstyle{[}j]}\geq
y^{\scriptscriptstyle{[}j]}$ for each $j\in[d]$. We also let $A^{\\{i\\}}_{+}$
denote the set of points $z\in S^{\\{1,2\\}}$ such that there exists an $y\in
A^{\\{i\\}}$ such that $z^{\scriptscriptstyle{[}j]}\geq
y^{\scriptscriptstyle{[}j]}$ for each $j\in[d+1]$. It is an easy exercise to
see that

$(A^{\\{i\\}})_{+[d]}=(A^{\\{i\\}})_{+}\cap S^{\\{i\\}}.$ (42)

Exactly as in Proposition 6 or Lemma 32, one can show that the following holds
on the graph $S^{\\{1,2\\}}$. Loosely, it says that
$\mathcal{C}_{x}^{\\{1,2\\}}(p)$ is bounded by two $d$-terraces, $\Lambda^{1}$
in $S^{\\{1\\}}$ and $\Lambda^{2}$ in $S^{\\{2\\}}$, for which the projection
of $\Lambda^{2}$ onto $S^{\\{1\\}}$ lies below $\Lambda^{1}$. Both
$\Lambda^{1}$ and $\Lambda^{2}$ consist of sites in $\Omega_{1}$, as do any
sites in $S^{\\{2\\}}$ that project onto the region between $\Lambda^{1}$ and
the projection of $\Lambda^{2}$.

######  Lemma 44.

For $p>p_{c}^{\\{1,2\\}}(d)$ almost surely for each $x\in S^{\\{1,2\\}}$ there
are $d$-terraces $\Lambda^{i}\subset S^{\\{i\\}}$, $i=1,2$ (depending on $x$)
such that:

  * (i)

$\Lambda^{1}\cup\Lambda^{2}\subset\mathcal{C}_{x}^{\\{1,2\\}}(p)$, and

  * (ii)

$\Lambda^{1}\cup\Lambda^{2}\subset\Omega_{1}$, and

  * (iii)

$(\Lambda^{i})_{+}\cap S^{\\{i\\}}=\mathcal{C}_{x}^{\\{1,2\\}}(p)\cap
S^{\\{i\\}}$ for each $i$, and

  * (iv)

$(\Lambda^{1})_{+}\cap S^{\\{2\\}}\subset(\Lambda^{2})_{+}$, and

  * (v)

$((\Lambda^{2})_{+}\cap S^{\\{2\\}})\setminus((\Lambda^{1})_{+}\cap
S^{\\{2\\}})\subset\Omega_{1}$.

  * (vi)

Let $W=e_{d+1}+(\\{o\\}\cup\mathcal{E}_{-}(d))$. If $z\in\Lambda^{1}$ then
$(z+W)\cap\Omega_{1}\neq\varnothing$.

###### Proof.

By Lemma 40 (2) and the fact that $p>p_{c}^{\\{1,2\\}}(d)$, we have that
$\mathcal{C}_{x}^{\\{1,2\\}}(p)$ is a.s. bounded below. Because
$e_{j}\in\mathcal{E}_{+}$ for every $j\in[d+1]$, it is clear that
$(\mathcal{C}_{x}^{\\{1,2\\}}(p))_{+}=\mathcal{C}_{x}^{\\{1,2\\}}(p)$.
Therefore $\mathcal{C}_{x}^{\\{1,2\\}}(p)\cap S^{\\{i\\}}$ is a _good_ set
(viewed as a subset of $\mathbb{Z}^{d}$), and (recall Definitions 4 and 5) we
may define $\Lambda^{i}=\lambda_{\mathcal{C}_{x}^{\\{1,2\\}}(p)\cap
S^{\\{i\\}}}\subset S^{\\{i\\}}$.

Claim (i) is immediate. We have (ii) because if
$x\in\Lambda^{i}\setminus\Omega_{1}$, then $x-e_{j}$ would
$\in\mathcal{C}_{x}^{\\{1,2\\}}(p)\subset(\Lambda^{i})_{+[d]}$ for every
$j\in[d]$, contradicting the fact that $\Lambda^{i}$ is a $d$-terrace. Claim
(iii) holds since $(\Lambda^{i})_{+}\cap
S^{\\{i\\}}=(\Lambda^{i})_{+[d]}=\mathcal{C}_{x}^{\\{1,2\\}}(p)\cap
S^{\\{i\\}}$ (the first equality from (42), and the second quantity by Lemma
7(ii) applied to $S^{\\{1\\}}$ as a copy of $\mathbb{Z}^{d}$). Since
$(\Lambda^{1})_{+}\cap S^{\\{2\\}}\subset\mathcal{C}_{x}^{\\{1,2\\}}(p)\cap
S^{\\{2\\}}\subset(\Lambda^{2})_{+}$, (the latter inclusion by (iii)), (iv)
follows.

To prove (v), suppose that $y\in((\Lambda^{2})_{+}\cap S^{\\{2\\}})$. Then by
(iii) $y\in\mathcal{C}_{x}^{\\{1,2\\}}(p)\cap S^{\\{2\\}}$. If
$y-e_{d+1}\in\mathcal{C}_{x}^{\\{1,2\\}}(p)$ then
$y-e_{d+1}\in\mathcal{C}_{x}^{\\{1,2\\}}(p)\cap
S^{\\{1\\}}=(\Lambda^{1})_{+}\cap S^{\\{1\\}}$, thus
$y\in(\Lambda^{1})_{+}\cap S^{\\{1,2\\}}$. It follows that if
$z\in((\Lambda^{2})_{+}\cap S^{\\{2\\}})\setminus((\Lambda^{1})_{+}\cap
S^{\\{2\\}})$ then $z\in\mathcal{C}_{x}^{\\{1,2\\}}(p)$ but
$z-e_{d+1}\notin\mathcal{C}_{x}^{\\{1,2\\}}(p)$, so $z\in\Omega_{1}$. To prove
(vi), let $z\in\Lambda^{1}$. First note that if $z+e_{d+1}\in\Lambda^{2}$ then
by (ii) we are done. So assume otherwise. Since
$z+e_{d+1}\in(\Lambda^{1})_{+}\cap S^{\\{2\\}}\subset(\Lambda^{2})_{+}$ (the
latter by (iv)) we have that
$z+e_{d+1}\in(\Lambda^{2})_{+}\setminus\Lambda^{2}$ so
$z+e_{d+1}-e_{j}\in(\Lambda^{2})_{+}$ for each $j\in[d]$. Because
$\Lambda^{1}$ is a $d$-terrace, there is a $k\in[d]$ such that $z-e_{k}\in
S^{\\{1\\}}\setminus(\Lambda^{1})_{+[d]}$. Since
$(\Lambda^{1})_{+[d]}=(\Lambda^{1})_{+}\cap S^{\\{1\\}}$ (by (42)) it follows
that also $z-e_{k}\notin(\Lambda^{1})_{+}$, so
$z-e_{k}+e_{d+1}\in((\Lambda^{2})_{+}\cap
S^{\\{2\\}})\setminus((\Lambda^{1})_{+}\cap S^{\\{2\\}})$. Therefore by (v),
$z-e_{k}+e_{d+1}\in\Omega_{1}$. ∎

###### Proof of Proposition 39.

We show that whenever $p>p_{c}^{\\{1,2\\}}(d)$ we have $p\geq p_{c}(f,d)$.

Let $\boldsymbol{\mathcal{G}}=(\mathcal{G}_{x})_{x\in\mathbb{Z}^{d+1}}$ be the
half-orthant model in dimension $d+1$. As we have seen above
$(\mathcal{G}_{x})_{x\in S^{\\{1\\}}}$ (using only arrows that stay in
$S^{\\{1\\}}$) realizes the half-orthant model in dimension $d$, via the
natural correspondence between $S^{\\{1\\}}$ and $\mathbb{Z}^{d}$. We will now
define two other environments $\boldsymbol{\eta}=(\eta_{x})_{x\in
S^{\\{1\\}}}$ [resp. $\boldsymbol{\zeta}=(\zeta_{x})_{x\in S^{\\{1\\}}}$] by
specifying the sets $\Omega_{1}({\boldsymbol{\eta}})$ [resp.
$\Omega_{1}(\boldsymbol{\zeta})$] upon which $\eta_{x}=\mathcal{E}_{+}$ [resp.
$\zeta_{x}=\mathcal{E}_{+}$].

Let $W$ be defined as in Lemma 44(vi). For $V_{d}$ as in Section 3.1, set
$\hat{E}_{1}=V_{d}\times\\{1\\}\subset\mathbb{Z}^{d+1}$. For $x\in
S^{\\{1\\}}$ we set

$\displaystyle
x\in\Omega_{1}(\boldsymbol{\eta})\iff\big{\\{}x\in\Omega_{1}(\boldsymbol{\mathcal{G}})\text{
and
}(x+W)\cap\Omega_{1}(\boldsymbol{\mathcal{G}})\neq\varnothing\big{\\}},\text{
and }$ (43) $\displaystyle
x\in\Omega_{1}(\boldsymbol{\zeta})\iff\big{\\{}x\in\hat{E}_{1}\cap\Omega_{1}(\boldsymbol{\eta})\quad\text{
or }\quad
x\in(S^{\\{1\\}}\setminus\hat{E}_{1})\cap\Omega_{1}(\boldsymbol{\mathcal{G}})\big{\\}}.$
(44)

Now note that (with $(U_{z})_{z\in\mathbb{Z}^{d+1}}$ i.i.d. standard uniforms)
for $x\in S^{\\{1\\}}$, $\eta_{x}=\eta_{x}(p)$ depends on $U_{x}$,
$U_{x+e_{d+1}}$ and $\\{U_{x+e_{d+1}-e_{j}}:j\in[d]\\}$. For
$x\in\hat{E}_{1}\subset S^{\\{1\\}}$, $\zeta_{x}(p)$ depends on the same
variables. For $x\in S^{\\{1\\}}\setminus\hat{E}_{1}$, $\zeta_{x}(p)$ depends
on $U_{x}$ only. Thus if $x,x^{\prime}\in S^{\\{1\\}}$ are distinct then
$\zeta_{x}$ and $\zeta_{x^{\prime}}$ depend on disjoint collections of $U_{z}$
(this is obvious unless $x,x^{\prime}\in\hat{E}_{1}$, in which case we use the
definition of $V_{d}$ and Lemma 41).

It is now easy to check that in fact $\boldsymbol{\zeta}$ is an $f$-disturbed
orthant model. In particular for $x\in\hat{E}_{1}$,

$\displaystyle\mathbb{P}(\zeta_{x}(p)=\mathcal{E}_{+})$
$\displaystyle=\mathbb{P}(U_{x}\leq
p)\Big{(}1-\mathbb{P}(U_{x+e_{d+1}}>p)\prod_{i=1}^{d}\mathbb{P}(U_{x+e_{d+1}-e_{i}}>p)\Big{)}$
$\displaystyle=p(1-(1-p)^{d+1}).$

Let $p>p_{c}^{\\{1,2\\}}(d)$, and let
$\Lambda^{1}\subset\Omega_{1}(\boldsymbol{\mathcal{G}})$ be the $d$-terrace
given by Lemma 44 (using only arrows in $S^{\\{1,2\\}}$). By Lemma 44(vi),
$\Lambda^{1}\subset\Omega_{1}(\boldsymbol{\eta})$. Moreover, it is easy to see
from (43) and (44) that
$\Omega_{1}(\boldsymbol{\eta})\subset\Omega_{1}(\boldsymbol{\zeta})$.
Therefore $\Lambda^{1}$ is also a $d$-terrace of $\Omega_{1}$ sites for
$\boldsymbol{\zeta}$. This shows that
$\mathcal{C}_{o}(p,f(p))\subset\mathbb{Z}^{d}$ is a.s. bounded below when
$p>p_{c}^{\\{1,2\\}}(d)$. By Lemma 40(1) this means that for
$p>p_{c}^{\\{1,2\\}}(d)$ we have that