How to prove using probability distribution rules that $P(Y_1,Y_2,X_3)*P(Y_3|X_3) = P(Y_1,Y_2,Y_3,X_3)$

Question

How to prove using probability distribution rules that $P(Y_1,Y_2,X_3)*P(Y_3|X_3) = P(Y_1,Y_2,Y_3,X_3)$

I am not considering any other assumptions for this question. Even if I consider expanding $P(Y_1,Y_2,X_3)*P(Y_3|X_3)$ to $\frac{P(Y_1,Y_2,X_3)*P(Y_3,X_3)}{P(X_3)}$, I am stuck with the denominator $P(X_3)$

Currently stuck at trying to show that $\frac{P(Y_1,Y_2,X_3)*P(Y_3,X_3)}{P(X_3)} = P(Y_1,Y_2,Y_3,X_3)$

Answer 1

Well, the factorisation $\mathsf P(Y_1,Y_2,X_3)\,\mathsf P(Y_3\mid X_3)=\mathsf P(Y_1,Y_2,Y_3,X_3)$, is not generally true.

However, consider that because, $\mathsf P(Y_1, Y_2, X_3) = \mathsf P(Y_1, Y_2\mid X_3)\,\mathsf P(X_3)$, you could prove your factorisation if $Y_1\cap Y_2$ and $Y_3$ were conditionally independent given $X_3$.

Are they?