If $Z \perp X | V$ does it follow that $Y \perp X | V \implies Y \perp X | V,Z$?

Question

Given random variables $(Y,X,Z,V)$ and letting $\perp$ denote independence I want to show that if $Z \perp X | V$, then $$ Y \perp X | V \implies Y \perp X | V,Z. $$ My current attempt is $$ F_{Y|X,V,Z}(y) = \mathbb{E}(1(Y \leq y) | X,V,Z) = \mathbb{E}(1(Y \leq y) | V,Z) = F_{Y|V,Z}(y), $$ but i cannot mathematically justify the second equality. The intuition is that if knowing $V$, $X$ does not tell me anything about $Y$, then conditioning also on $Z$ which is independent of $X$ given $V$ should not make $X$ anymore relevant... also assuming that a Lebesgue pdf exist, integrating with respect to $f_{Y|X,V,Z}$ is the same as integrating with respect to $f_{Y|V,Z}$ since $$ f_{Y|X,V,Z} = \frac{f_{Y,X,Z|V}}{f_{X,Z|V}} = \frac{f_{Y,Z|V}f_{X|V}}{f_{X|V}f_{Z|V}} = f_{Y|V,Z}. $$ If lebesgue pdfs exist this should be enough to prove it, but i would like to prove it just by using properties of conditional expectations, is this possible? If so, how can I do it?