Proving the balancing score property of propensity score $\newcommand{\P}{\mathbb{P}}$$\newcommand{\E}{\mathbb{E}}$In proving that $X \bot A | ps(x)$ where $X$ denote the baseline covariates, $A$ is binary denoted the Treatment/Exposure and $ps(x)$ denotes the propensity score based on $X$. 
I do not understand a part of the proof. 
The first part was to prove: 
\begin{equation}
\P(A = 1 \vert ps(x), x) = ps(x), \tag{1}
\end{equation}
which I understood. 
The second part was to prove: 
\begin{equation}
\P(A = 1 \vert ps(x))  = ps(x), \tag{2}
\end{equation}
then from the second equation, to prove (1) = (2) and that they are thus, conditionally independent. 
I do not know how: $$\P(A = 1 \vert ps(x) ) = \E(A \vert ps(x)) = \E[\E(A \vert ps(x), x) \vert ps(x)  ] = ps(x). $$ In particular, I do not understand how: $$\E(A \vert ps(x)) = \E[\E(A \vert ps(x), x) \vert ps(x)  ] .$$ I understand there was something to do with Law of Iterated Expectations, but do not see how they equate.
 A: The proof works by showing that conditioning on the propensity score is "sufficient" when you are conditioning on $X=x$ already and you want an expression for the propensity score. For the first part of the proof you refer to, you can immediately see that since $ps(x)$ is a function of $x$, $ps(x)$ is redundant. For the second part we just have to show by conditioning on $x$  that we then condition "too much", and that we just get the propensity score.
First, note that the propensity score is defined as
$$ps(x) = \Pr(A=1\mid x).$$ 
For ease of notation, you can simply see that
$$\mathbb{E}(A)=\mathbb{E}[\mathbb{E}(A\mid x)],$$
by--as you mention--the law of iterated expectations. 
We can make this expression conditional on the propensity score $ps(x)$:
$$\mathbb{E}(A\mid p(x))=\mathbb{E}[\mathbb{E}(A\mid ps(x),x)\mid ps(x)],$$
which is exactly the expression you were looking for. To show--rather elaborately--that this is equal to the propensity score:
\begin{align}
& \mathbb{E}[\mathbb{E}(A\mid ps(x),x)\mid ps(x)]\\
&=\mathbb{E}[\mathbb{E}(A\mid x)\mid ps(x)]\\
&=\mathbb{E}[\Pr(A=1\mid x)\mid ps(x)]\\
&=\mathbb{E}[ps(x)\mid ps(x)]\\
&=ps(x).
\end{align}
A: Could anybody give a clearer proof of P(A=1|ps(x),x)=ps(x)?
Verbally, P(A=1|ps(x),x)=ps(x) is the probability of A=1 given x such that it's propensity score is ps(x)
Then the argument that P(A=1) given x alone is equal to P(A=1) given a particular x that satisfies a particular value of ps, sounds not self evident.
I am very confused.
