Is this contraposition proof complete?

Question

I want to prove the statement If $a>b$ then $X.$

Suppose that I've proved that If $\neg X$ then $a<b.$ This gives $a\leq b,$ which $a<b$ is a subset of.

Can I consider the proof finished, by contraposition?

Answer 1

I've proved that if $\neg X$ then $a<b$.

Taking the contrapositive: $a\ge b\implies X.\tag1$

And we know that $a>b\implies a\ge b.\tag2$

Combining $(1)$ and $(2):$ $a>b\implies X,\tag3$ as required.

P.S. $(1)$ is a stronger statement than $(3)$ because the former has a less stringent condition for attaining the same result as the latter.

P.P.S. In general, if $(A⇒X)$ and $(Y⇒B)$, then the sentence $(X⇒Y)$ is at least as strong as the sentence $(A⇒B).$ This is because $$\big((A→X)∧(Y→B)\big)→\big((X→Y)→(A→B)\big)$$ is logically true and equivalent to $$\big((A→X)∧(Y→B)∧(X→Y)\big)→(A→B).$$