Let $Y$ be a random variable, $Z$ be another random variable, $X$ be a set of random variables including $Z$ but not $Y$.
Assume $Y$ is conditionally independent of $X$ without $Z$ given $Z$.
Can we state that $Y$ is conditionally independent of $X$ given $Z$? I.e. $X$ now includes $Z$
My attempt:
From the assumption: $P(Y|{X-Z},Z) = P(Y|X) = P(Y|X,Z)$
The last two equations on the right hand side imply that $Y$ is conditionally independent of $X$ given $Z$?
Conditional independence: $$Y \bot (X\backslash Z) \mid Z \iff P(Y \cap (X\backslash Z)\mid Z) = P(Y\mid Z)P((X\backslash Z)\mid Z)$$
So since $\;P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ $$\frac{P(Y \cap (X\backslash Z) \cap Z)}{P(Z)} = \frac{P(Y \cap Z)}{P(Z)}\frac{P((X\backslash Z)\cap Z)}{P(Z)}$$
And obviously $\;(A\backslash B) \cap B = A$ $$\frac{P(Y \cap X)}{P(Z)} = \frac{P(Y \cap Z)}{P(Z)}\frac{P(X)}{P(Z)}$$
By eliminating the common $P(Z)$ $$P(Y \cap X) = \frac{P(Y \cap Z)}{P(Z)} P(X)$$ Which means:$$P(Y\mid X)=P(Y\mid Z)$$
Since $\frac{P(X\cap Z)}{P(Z)} = P(X\mid Z)$
$$\frac{P(Y\cap X)}{P(X)}\cdot \frac{P(X \cap Z)}{P(Z)}=P(Y\mid Z)\cdot P(X\mid Z)$$
However, since $Z\subset X \implies X\cap Z = X$
$$\frac{P(Y\cap (X\cap Z))}{P(X)}\frac{P(X)}{P(Z)}=P(Y\mid Z)P(X\mid Z)$$
So $$P(Y\cap X\mid Z) = P(Y\mid Z)P(X\mid Z)$$ Which means: $$\therefore Y \bot (X\backslash Z) \mid Z \iff Y\bot X\mid Z$$