# Is $X\perp Y$ under $\mathbb{P} (\; \cdot\; | \max\{X,Y\} \leq Z)$ if $X,Y,Z$ are independent?

Given $$X,Y,Z$$ are independent random variables, are $$X, Y$$ conditional independent with $$\mathbb{P}(\;\cdot\; |X\vee Y \leq Z)$$?

Suppose $$\mathbb{P} (X\vee Y \leq Z) > 0$$, then by definition, $$\mathbb{P}(\{X\in B\} \cap \{Y\in C\} |X\vee Y \leq Z)\\ =\mathbb{P}(\{X\in B\} \cap \{Y\in C\} \cap \{X \leq Z\}\cap \{Y\leq Z\})/\mathbb{P} (X\vee Y \leq Z)$$ If $$Z$$ were a constant random variable, then this reduces to $$\mathbb{P}(\{X\in B\} |X\vee Y \leq Z) \cdot \mathbb{P}( \{Y\in C\} |X\vee Y \leq Z).$$ If it is true how could I show this?

A sufficient condition is $$\mathbb{P} (X\leq Z, Y\leq Z) = \mathbb{P}(X\leq Z)\mathbb{P}(Y\leq Z)$$, but this is false, take $$X = Y = 1$$.