Adjustment formula for counterfactuals: can we get rid of $X=x$?

Pearl et al. "Causal Inference in Statistics: A Primer" (2016) p. 108 contains the following adjustment formula (based on the backdoor criterion) for probabilities of counterfactuals expressed using observed data: \begin{align} P(Y_x|X=x')=\sum_z P(Y=y|X=x,Z=z) P(Z=z|X=x'). \tag{4.21} \end{align} The derivation is as follows: \begin{align} P(Y_x|X=x') &= \sum_z P(Y_x=y|X=x\color{red}{'},Z=z) P(Z=z|X=x') \\ &= \sum_z P(Y_\color{red}{x}=y|X=x,Z=z) P(Z=z|X=x') \\ &= \sum_z P(Y=y|X=x,Z=z) P(Z=z|X=x'). \end{align} The top equality simply partitions the original $$P(Y_x|X=x')$$ w.r.t $$Z$$. Since $$Z$$ satisfies the backdoor criterion, $$(Y_x\perp \!\!\! \perp X)|Z$$, so $$P(Y_x|X,Z)=P(Y_x|Z)$$. This allows replacing $$x\color{red}{'}$$ by $$x$$ and yields the middle equality. The bottom equality follows from the consistency rule: if $$X=x$$ then $$Y_\color{red}{x}=Y$$.

I am interested in taking things one step further. If $$(Y_x\perp \!\!\! \perp X)|Z$$, could we not get rid of conditioning on $$X=x$$ altogether and have \begin{align} P(Y_x|X=x') &= \sum_z P(Y=y|\color{red}{X=x},Z=z) P(Z=z|X=x') \\ &= \sum_z P(Y=y|Z=z) P(Z=z|X=x') \quad ? \end{align}

• Does $(Y_x \perp \!\! \perp X)|Z$ imply $(Y \perp \!\! \perp X)|Z?$ Counterfactuals I find rather tricky and slippery. On the face of it, I wouldn't expect my implication to hold. Isn't $Y_x$ a more specific $Y$ than just $Y?$ Commented Apr 14, 2020 at 18:01
• @AdrianKeister, you might very well be right. $Y_x$ is the counterfactual of $Y$, so they are not the same. Commented Apr 14, 2020 at 18:55

As noted in the comments, this were valid if $$Y_x \perp X|Z$$ would imply $$Y \perp X|Z$$. But it does not. The simplest counterexample is when $$X$$ affects $$Y$$. Then $$Z$$ blocks all back-door paths, but there is an open path $$X \rightarrow Y$$ which is not blocked by $$Z$$, and so $$X$$ is informative about $$Y$$ even conditional on $$Z$$.
If, however, $$X$$ was assumed to not affect $$Y$$, then we would have $$Y_x = Y$$, and the conclusion would follow.