Prove that $x \gt y \gt 0 \implies x \gt x-y$ Prove that $x \gt y \gt 0 \implies x \gt x-y$
I have approached it as below.
$y \gt 0 \implies y=x+(-x)+y \gt 0 \implies x+((-x)+y)=x+(-(x-y)=x-(x-y)>0 \implies x\gt x-y$
What I observe in this proof is that is does not consider the fact that $x \gt y$ and $x \gt 0$.
Is this proof valid?

The proof is indeed valid. And there's a slightly shorter proof which goes like this:
$$y > 0 \implies y+(-y) > 0+(-y) \implies 0 > -y \implies x+0 > x+(-y) \implies x > x-y
$$
