Causality: Models, Reasoning and Inference, by Judea Pearl: Causal Bayesian Networks and the Truncated Factorization Background:
$\newcommand{\doop}{\operatorname{do}}\newcommand{\op}[1]{\operatorname{#1}}$
Definition 1.2.2 (Markov Compatibility) If a probability function $P$ admits the factorization of 
$$P(x_1,\dots,x_n)=\prod_i P(x_i|\operatorname{pa}_i)$$
relative to the Directed Acyclic Graph (DAG) $G,$ we say that $G$ represents $P,$ that $G$ and $P$ are compatible, or that $P$ is Markov relative to $G.$
Here the $\operatorname{pa}_i$ are the parents of $x_i.$ 
Definition 1.3.1 (Causal Bayesian Network) Let $P(v)$ be a probability distribution on a set $V$ of variables, and let $P(v|\doop(x))$ denote the distribution resulting from the intervention $\doop(X=x)$ that sets a subset $X$ of variables to constants $x.$ Denote by $P_*$ the set of all interventional distributions $P(v|\doop(x)), X\subseteq V,$ including $P(v),$ which represents no intervention (i.e., $X=\varnothing$). A DAG $G$ is said to be a causal Bayesian network compatible with $P_*$ if and only if the following three conditions hold for every interventional $P\in P_*:$


*

*$P(v|\doop(x))$ is Markov relative to $G;$

*$P(v_i|\doop(x))=1$ for all $V_i\in X$ whenever $v_i$ is consistent with $X=x;$

*$P(v_i|\operatorname{pa}_i,\doop(x))=P(v_i|\operatorname{pa}_i)$ for all $V_i\not\in X$ whenever $\operatorname{pa}_i$ is consistent with $X=x,$ i.e., each $P(v_i|\operatorname{pa}_i)$ remains invariant to interventions not involving $V_i.$
Truncated Factorization:
$$P(v|\doop(x))=\prod_{i|V_i\not\in X}P(v_i|\operatorname{pa}_i)\qquad\text{for all } v \text{ consistent with }x.$$
Problem Statement: Prove that the three conditions of Definition 1.3.1 (Causal Bayesian Network) are necessary and sufficient for the truncated factorization. 
My Answer So Far:
$(\to)$ Assume the three conditions of Definition 1.3.1 hold. We know by (i) that we can write
$$P(v|\doop(x))=\prod_iP(v_i|\op{pa}_i,\doop(x)).$$
Then we factor into two products, depending on where the $v_i$ are:
\begin{align*}
P(v|\doop(x))
&=\prod_iP(v_i|\op{pa}_i,\doop(x))\\
&=\left[\prod_{i, v_i\in X}P(v_i|\op{pa}_i,\doop(x))\right]\left[\prod_{i, v_i\not\in X}P(v_i|\op{pa}_i,\doop(x))\right].
\end{align*}
According to (ii), the first product is $1,$ yielding
$$P(v|\doop(x))=\prod_{i, v_i\not\in X}P(v_i|\op{pa}_i,\doop(x)).$$
Finally, we argue that 
\begin{align*}
P(v|\doop(x))
&=\prod_{i, v_i\not\in X}P(v_i|\op{pa}_i,\doop(x))\\
&=\prod_{i, v_i\not\in X}P(v_i|\op{pa}_i)
\end{align*}
by invoking (iii), since we assumed that $P(v_i|\op{pa}_i)$ are invariant to interventions 
involving $X.$ Am I correct so far?
$(\leftarrow)$ Other than the obvious first step of assuming we can write the truncated factorization, I don't have any ideas on this one. How can I proceed? Would the steps in the $(\to)$ direction all be reversible?
Many thanks for your time!
 A: $\newcommand{\doop}{\operatorname{do}}\newcommand{\op}[1]{\operatorname{#1}}$
I think your proof of the forward implication is correct. For the backward implication I may have something.
Suppose the Truncated Factorization: for all $v$ consistent with $x$,
$$P(v\mid \mathrm{do}(x))=\prod_{i\mid Vi\notin X}P(v_i \mid \mathrm{pa}_i)$$  for a non cyclic oriented graph $G$.
Proof that condition 3 is verified
Let be $i$, $v_i$, and an intervention $X = x$ be such that $V_i \notin X$ and a realization of $\mathrm{pa}_i$ is compatible with $X = x$. We need to prove that: $$P(v_i|\mathrm{pa}_i, \mathrm{do}(x)) = P(v_i |\mathrm{pa}_i).$$ 
To do so, let's take an intervention $X' = x'$ such that:


*

*$X \subset X'.$

*$X = x$ and $X' = x'$ are compatible.

*$V_i \notin X'$ and $\forall j\neq i, V_j\in X'.$

*The realization of $\mathrm{pa_i}$ considered is compatible with $X'.$
Intuitively, we are fixing everything but $V_i$ by an intervention, without contradicting the intervention $X = x$ nor the considered realization of $\mathrm{pa}_i$.
Then, using the factorization, $$P(v|\mathrm{do}(x')) = P(v_i| \mathrm{pa}_i)$$ since only index $i$ is left in the product, and thus $$P(v|\mathrm{do}(x'), \mathrm{do}(x)) = P(v_i|\mathrm{pa}_i, \mathrm{do}(x)).$$ 
But as $X = x$ is included in $X' = x'$, $P(v|\mathrm{do}(x'), \mathrm{do}(x)) = P(v|\mathrm{do}(x'))$. So we have that: 
$$P(v_i|\mathrm{pa}_i, \mathrm{do}(x)) = P(v_i|\mathrm{pa}_i),$$
which is what we wanted.
Proof that condition 1 is verified
If we use the truncated factorization on a null intervention, we obtain that $G$ and $P$ are Markov compatible:
$$P(v) = \prod_i P(v_i|\mathrm{pa}_i).$$ 
Conditioning the last equation on an intervention $X = x$, we get that
$$P(v|\mathrm{do}(x)) = \prod_i P(v_i|\mathrm{pa}_i, \mathrm{do}(x)),$$ which is that $P(v|\mathrm{do}(x))$ and $G$ are Markov compatible.
Proof that condition 2 is verified
Let's consider an intervention $X =x$. Using condition 1, we have: 
\begin{align*}
P(v|\doop(x))
&=\prod_i P(v_i|\op{pa}_i, \doop(x))\\
&=\prod_{i|V_i\in X}P(v_i|\op{pa}_i,\doop(x))\cdot\prod_{i|V_i\not\in X}P(v_i|\op{pa}_i,\doop(x))\\
&=\prod_{i|V_i\in X}P(v_i|\op{pa}_i,\doop(x))\cdot\prod_{i|V_i\not\in X}P(v_i|\op{pa}_i),
\end{align*}
using condition 3. As $P(v|\mathrm{do}(x))$ can also be expressed with the truncated factorization, we get that:
$$\prod_{i|V_i \notin X}P(v_i|\mathrm{pa}_i) = \prod_{i|V_i \in X}P(v_i|\mathrm{pa}_i, \mathrm{do}(x))\prod_{i|V_i \notin X}P(v_i| \mathrm{pa}_i)$$ 
and so, simplifying by dividing out $P(v_i|\mathrm{pa}_i)$:
$$ \prod_{i|V_i \in X}P(v_i|\mathrm{pa}_i)  = 1 .$$ 
(To be allowed to make this simplification, we need to suppose that $P(v_i|\mathrm{pa}_i) \neq 0$, which is necessarily the case if we suppose $P(v|\mathrm{do}(x)) \neq 0$ for instance.)
In the end we have that $P(v_i|\mathrm{pa}_i, \mathrm{do}(x)) = 1$ for all $i$ such that $i \in X$ (since their product is $1$). To get to condition 2, write 
$$P(v_i|\mathrm{do}(x)) = \mathbb{E}_{\mathrm{pa}_i}\left[P(v_i| \mathrm{pa}_i,  \mathrm{do}(x))\right] = \mathbb{E}_{\mathrm{pa}_i}\left[1\right] = 1.$$
I hope this is understandable, correct and helping..
