Conditional probability proof involving inequality

Question

We have a discrete random variable $X$. I have a problem showing algebraically (if that is even possible) that $P(X\gt a\mid X \gt a+b) = 1$. Given that $a$ and $b$ are both positive integers.

When I rewrite it as $P(X\gt a\mid X-b\gt a)$ I can see that it makes sense that if we know that $X$ minus anything is more than $a$ then $X$ is definitely more than $a$. But is there a way to show this algebraically somehow?

Answer 1

$$\mathbb{P}(X>a\mid X>a+b)=\frac{\mathbb{P}(X>a;X>a+b)}{\mathbb{P}(X>a+b)}=\frac{\mathbb{P}(X>a+b)}{\mathbb{P}(X>a+b)}=1$$

Answer 2

The event $\Big[ X>a\ \&\ X>a+b\Big]$ is the same as the event $\Big[X>a+b\Big],$ i.e. either occurs if and only if they other occurs.

Therefore they both have the same probability.