Make an probablity relationship easier

Question

If $X$ and $Y$ and $Z$ are independent random variables and $C$ is a fixed value, is the relationship below true? $$\text{Pr} \left\lbrace X+\min \left( Y,Z \right) <C\bigg\vert Z \geq C \right\rbrace = \text{Pr} \left\lbrace X+Y < C \right\rbrace $$ My proof is shown below: $$\text{Pr} \left\lbrace X+\min \left( Y,Z \right) <C\bigg\vert Z \geq C \right\rbrace = \text{Pr} \left\lbrace X+Y<C \, \cup \, X+Z<C \bigg\vert Z \geq C \right\rbrace= \text{Pr} \left\lbrace X+Y<C \bigg\vert Z \geq C \right\rbrace = \text{Pr} \left\lbrace X+Y<C \right\rbrace$$

Answer 1

Let $Y = 1, Z = 0 = C$. Then you ask if $\Pr(X<0) = \Pr(X+1<0)$.