V}(\omega,A)$

where, in going from the first line to the second, we used the fact that
$\mathscr{H}_{V}$ has no causal effect on $A$ in the original causal space,
and in going from the second line to the third, we used $U\cap V=\emptyset$,
which gives us $S\cap U=(S\setminus V)\cap U$ and $(S\cup U)\setminus
V=(S\setminus V)\cup U$. Since $S\in\mathscr{P}(T)$ was arbitrary,
$\mathscr{H}_{V}$ has no causal effect on $A$ in the intervention causal
space.

  3. (iii)

Take any $S\in\mathscr{P}(T)$. Apply Theorem C.3 to see that

$\displaystyle K^{\text{do}(U,\mathbb{Q},\text{hard})}_{S}(\omega,A)$
$\displaystyle=\int\mathbb{Q}(d\omega^{\prime}_{U\setminus S})K_{S\cup
U}((\omega_{S},\omega^{\prime}_{U\setminus S}),A)$
$\displaystyle=\int\mathbb{Q}(d\omega^{\prime}_{U\setminus S})K_{(S\cup
U)\setminus V}((\omega_{S},\omega^{\prime}_{U\setminus S}),A)$ Def. B.1i
$\displaystyle=\int\mathbb{Q}(d\omega^{\prime}_{U\setminus S})K_{((S\setminus
V)\cup U)\setminus V}((\omega_{S},\omega^{\prime}_{U\setminus S}),A)$
$\displaystyle=\int\mathbb{Q}(d\omega^{\prime}_{U\setminus S})K_{(S\setminus
V)\cup U}((\omega_{S},\omega^{\prime}_{U\setminus S}),A)$ Def. B.1i
$\displaystyle=\int\mathbb{Q}(d\omega^{\prime}_{U\setminus(S\setminus
V)})K_{(S\setminus V)\cup U}((\omega_{S\setminus
V},\omega^{\prime}_{U\setminus(S\setminus V)}),A)$
$\displaystyle=K^{\text{do}(U,\mathbb{Q})}_{S\setminus V}(\omega,A),$

where, in going from the second line to the third, we used that $(S\cup
U)\setminus V=((S\setminus V)\cup U)\setminus V$. Since $S\in\mathscr{P}(T)$
was arbitrary, $\mathscr{H}_{V}$ has no causal effect on $A$ in the
intervention causal space.

∎

######  Lemma C.5.

Let $(\Omega,\mathscr{H},\mathbb{P},\mathbb{K})=(\times_{t\in
T}E_{t},\otimes_{t\in T}\mathscr{E}_{t},\mathbb{P},\mathbb{K})$ be a causal
space, and $U\in\mathscr{P}(T)$. For an event $A\in\mathscr{H}$, if
$\mathscr{H}_{U}$ has a dormant causal effect on $A$ in the original causal
space, then there exists a hard intervention and a subset $V\subseteq U$ such
that in the intervention causal space, $\mathscr{H}_{V}$ has an active causal
effect on $A$.

###### Proof.

That $\mathscr{H}_{U}$ has a dormant causal effect on $A$ tells us that
$K_{U}(\omega,A)=\mathbb{P}(A)$ for all $\omega\in\Omega$, but there exists
some $S\in\mathscr{P}(T)$ and some $\omega_{0}\in\Omega$ such that
$K_{S}(\omega_{0},A)\neq K_{S\setminus U}(\omega_{0},A)$. We must have $S\cap
U\neq\emptyset$, since otherwise $S\setminus U=S$ and we cannot possibly have
$K_{S}(\omega_{0},A)\neq K_{S\setminus U}(\omega_{0},A)$. Then we hard-
intervene on $\mathscr{H}_{S\setminus U}$ with the Dirac measure on
$\omega_{0}$. Then apply Theorem C.3 to see that

$\displaystyle K^{\text{do}(S\setminus
U,\delta_{\omega_{0}},\text{hard})}_{S\cap U}((\omega_{0})_{U\cap S},A)$
$\displaystyle=\int\delta_{\omega_{0}}(d\omega^{\prime}_{S\setminus
U})K_{S}(((\omega_{0})_{U\cap S},\omega^{\prime}_{S\setminus U}),A)$
$\displaystyle=K_{S}(\omega_{0},A)$ $\displaystyle\neq K_{S\setminus
U}(\omega_{0},A)$

Note that the intervention measure on $A$ is equal to $K_{S\setminus
U}(\omega_{0},A)$:

$\mathbb{P}^{\text{do}(S\setminus
U,\delta_{\omega_{0}})}(A)=\int\delta_{\omega_{0}}(d\omega^{\prime}_{S\setminus
U})K_{S\setminus U}(\omega^{\prime},A)=K_{S\setminus U}(\omega_{0},A).$

Putting these together, we have

$K^{\text{do}(S\setminus U,\delta_{\omega_{0}},\text{hard})}_{S\cap
U}(\omega_{0},A)\neq\mathbb{P}^{\text{do}(S\setminus
U,\delta_{\omega_{0}})}(A),$

i.e. in the intervention causal space
$(\Omega,\mathscr{H},\mathbb{P}^{\text{do}(S\setminus
U,\delta_{\omega_{0}})},K^{\text{do}(S\setminus
U,\delta_{\omega_{0}},\text{hard})}_{S\cap U})$, $\mathscr{H}_{S\cap U}$ has
an active causal effect on $A$. ∎

######  Lemma C.6.

Let $(\Omega,\mathscr{H},\mathbb{P},\mathbb{K})=(\times_{t\in
T}E_{t},\otimes_{t\in T}\mathscr{E}_{t},\mathbb{P},\mathbb{K})$ be a causal
space, and $U,V\in\mathscr{P}(T)$. For an event $A\in\mathscr{H}$, suppose
that $\mathscr{H}_{U}$ has no causal effect on $A$ given $\mathscr{H}_{V}$
(see Definition B.4). Then after an intervention on $\mathscr{H}_{V}$ via any
$(\mathbb{Q},\mathbb{L})$, $\mathscr{H}_{U\setminus V}$ has no causal effect
on $A$.

###### Proof.

Take any probability measure $\mathbb{Q}$ on $(\Omega,\mathscr{H}_{V})$ and
any causal mechanism $\mathbb{L}$ on $(\Omega,\mathscr{H}_{V},\mathbb{Q})$.
Then see that, for any $S\in\mathscr{P}(T)$ and all $\omega\in\Omega$,

$\displaystyle K^{\text{do}(V,\mathbb{Q},\mathbb{L})}_{S}(\omega,A)$
$\displaystyle=\int L_{S\cap V}(\omega_{S\cap V},d\omega^{\prime}_{V})K_{S\cup
V}((\omega_{S\setminus V},\omega^{\prime}_{V}),A)$ $\displaystyle=\int
L_{S\cap V}(\omega_{S\cap V},d\omega^{\prime}_{V})K_{(S\cup
V)\setminus(U\setminus V)}((\omega_{S\setminus(U\cup
V)},\omega^{\prime}_{V}),A)$ $\displaystyle=\int L_{(S\setminus(U\setminus
V))\cap V}(\omega_{(S\setminus(U\setminus V))\cap
V},d\omega^{\prime}_{V})K_{(S\setminus(U\setminus V))\cup
V}((\omega_{S\setminus(U\cup V)},\omega^{\prime}_{V}),A)$
$\displaystyle=K^{\text{do}(V,\mathbb{Q},\mathbb{L})}_{S\setminus(U\setminus
V)}(\omega,A),$

where, in going from the first line to the second, we used the fact that
$\mathscr{H}_{U}$ has no causal effect on $A$ given $\mathscr{H}_{V}$, and in
going from the second line to the third, we used identities $S\cap
V=(S\setminus(U\setminus V))\cap V$ and $(S\cup V)\setminus(U\setminus
V)=(S\setminus(U\setminus V))\cup V$. Since $S\in\mathscr{P}(T)$ was
arbitrary, we have that $\mathscr{H}_{U\setminus V}$ has no causal effect on
$A$ in the intervention causal space. ∎

######  Theorem C.7.

Let $(\Omega,\mathscr{H},\mathbb{P},\mathbb{K})=(\times_{t\in
T}E_{t},\otimes_{t\in T}\mathscr{E}_{t},\mathbb{P},\mathbb{K})$ be a causal
space, where the index set $T$ can be written as $T=W\times\tilde{T}$, with
$W$ representing time and $\mathbb{K}$ respecting time. Take any
$U\in\mathscr{P}(T)$ and any probability measure $\mathbb{Q}$ on
$\mathscr{H}_{U}$. Then the intervention causal mechanism
$\mathbb{K}^{\textnormal{do}(U,\mathbb{Q},\textnormal{hard})}$ also respects
time.

###### Proof.

Take any $w_{1},w_{@}\in W$ with $w_{1}<w_{2}$. Since $\mathbb{K}$ respects
time, we have that $\mathscr{H}_{w_{2}\times\tilde{T}}$ has no causal effect
on $\mathscr{H}_{w_{1}\times\tilde{T}}$ in the original causal space. To show
that $\mathscr{H}_{w_{2}\times\tilde{T}}$ has no causal effect on
$\mathscr{H}_{w_{1}\times\tilde{T}}$ after a hard intervention on
$\mathscr{H}_{U}$ via $\mathbb{Q}$, take any $S\in\mathscr{P}(T)$ and any
event $A\in\mathscr{H}_{w_{1}\times\tilde{T}}$. Then using Theorem C.3,

$\displaystyle K^{\text{do}(U,\mathbb{Q},\text{hard})}_{S}(\omega,A)$
$\displaystyle=\int\mathbb{Q}(d\omega^{\prime})K_{S\cup
U}((\omega_{S},\omega^{\prime}_{U\setminus S}),A)$
$\displaystyle=\int\mathbb{Q}(d\omega^{\prime})K_{(S\cup
U)\setminus\mathscr{H}_{w_{2}\times\tilde{T}}}((\omega_{S\setminus\mathscr{H}_{w_{2}\times\tilde{T}}},\omega^{\prime}_{U\setminus(S\cup\mathscr{H}_{w_{2}\times\tilde{T}})}),A)$
$\displaystyle=\int\mathbb{Q}(d\omega^{\prime})K_{((S\cup
U)\setminus\mathscr{H}_{w_{2}\times\tilde{T}})\cup(U\cap\mathscr{H}_{w_{2}\times\tilde{T}})}((\omega_{S\setminus\mathscr{H}_{w_{2}\times\tilde{T}}},\omega^{\prime}_{(U\setminus(S\cup\mathscr{H}_{w_{2}\times\tilde{T}}))\cup(U\cap\mathscr{H}_{w_{2}\times\tilde{T}})}),A)$
$\displaystyle=\int\mathbb{Q}(d\omega^{\prime})K_{(S\setminus\mathscr{H}_{w_{2}\times\tilde{T}})\cup
U}((\omega_{S\setminus\mathscr{H}_{w_{2}\times\tilde{T}}},\omega^{\prime}_{U\setminus(S\setminus\mathscr{H}_{w_{2}\times\tilde{T}})}),A)$
$\displaystyle=K^{\text{do}(U,\mathbb{Q},\text{hard})}_{S\setminus\mathscr{H}_{w_{2}\times\tilde{T}}}(\omega,A)$

where, from the second line to the third, we used the fact that
$\mathscr{H}_{w_{2}\times\tilde{T}}$ has no causal effect on $A$, from the
third line to the fourth we used the fact that
$U\cap\mathscr{H}_{w_{2}\times\tilde{T}}$ has no causal effect on $A$ (by
Remark B.2e) and Remark B.2g, and from the fourth line to the fifth, we used
that $((S\cup
U)\setminus\mathscr{H}_{w_{2}\times\tilde{T}})\cup(U\cap\mathscr{H}_{w_{2}\times\tilde{T}})=(S\setminus\mathscr{H}_{w_{2}\times\tilde{T}})\cup
U$ and
$(U\setminus(S\cup\mathscr{H}_{w_{2}\times\tilde{T}}))\cup(U\cap\mathscr{H}_{w_{2}\times\tilde{T}})=U\setminus(S\setminus\mathscr{H}_{w_{2}\times\tilde{T}})$.
Since $S\in\mathscr{P}(T)$ was arbitrary, we have that
$\mathscr{H}_{w_{2}\times\tilde{T}}$ has no causal effect on $A$ (Definition
B.1i). Since $A\in\mathscr{H}_{w_{1}\times\tilde{T}}$ was arbitrary,
$\mathscr{H}_{w_{2}\times\tilde{T}}$ has no causal effect on
$\mathscr{H}_{w_{1}\times\tilde{T}}$, and so
$\mathbb{K}^{\text{do}(U,\mathbb{Q},\text{hard})}$ respects time. ∎