Conditional Independence and Complicated Probability

Question

Let $X$, $Y$ and $Z$ be three random variables, which are not degenerated to constants. Assume the measure spaces of $Y$ and $Z$ are the same. Given $$ Y\perp Z\mid X, $$ i.e., $Y$ and $Z$ are independent conditional on $X$, is it possible to impose conditions to guarantee (\ref{equ:tar})? \begin{equation} \mathbb{P} (Y = y\mid X = x, Y \leq Z) = \mathbb{P} (Y = y\mid X = x). \tag{1} \label{equ:tar} \end{equation}
I have made some efforts on derivation. \begin{equation} \begin{split} \mathbb{P} (Y = y\mid Y \leq Z, X = x) &= \frac{\mathbb{P} (Y = y, Z\geq y\mid X = x)}{\mathbb{P} (Y \leq Z\mid X = x)}\\ &= \frac{\mathbb{P} (Y = y\mid X = x) \mathbb{P} (Z\geq y\mid X = x)}{\mathbb{P} (Y \leq Z\mid X = x)}.\\ \end{split} \end{equation} By the similar idea of the comment from @JamieRadcliffe, a naive idea is that $Z$ is a constant and bigger than any value of $Y$, which guarantees the ratio $\frac{\mathbb{P} (Z\geq y\mid X = x) }{\mathbb{P} (Y \leq Z\mid X = x) } = 1$. What if $Z$ does not degenerate to a constant and have the same value ranges as that of $Y$?