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?
 A: Shorter: $x = (x-y)+y > x-y+0=x-y.\blacksquare$
A: If the statement were only true when $x > 0$ and if $x > y$ then, yes, it would be very worrisome that the proof didn't use the condition that that are necessary and the proof probably has a mistake in it.
But the proof wouldn't be invalid because it didn't use all its premises. The proof would be invalid because it has a mistake. And we know the proof has a mistake, because as the proof didn't use the necessary premise, the proof actually proved the statement for cases when it wouldn't actually be true.  And any proof that proves a false statement must have an error.
For example: If we wanted to prove.  If $n$ is an even integer, then $n^2$ is even and someone gave a proof that $n^2$ is even but it never used the fact that $n$ was even, then we know the proof must be wrong.  If it never assumed $n$ was even, then it would work even if $n$ were odd.    But we know if $n$ is odd, then $n^2$ is odd.  So the proof must be wrong because it reached a wrong conclusion.
But if the statement is true whether our premise is true or not then the premise is unnecessary and we don't need to use it (although we could).
Consider the following ridiculous example:  If $n$ is an even integer and Santa Claus is fat, prove that $n^2$ is even.  Pf: If $n$ is even then there is an integer $k$ so that $n = 2k$.  So $n^2 = (2k)^2 = 2^2k^2 = 2(2k^2)$.  $2k^2$ is an integer and so $n^2$ is two times an integer so $n^2$ is even.
Now imaging if a professor said:  "That proof is invalid because it didn't use Santa Claus is fat".
Well, if $n$ is even then $n^2$ will be even whether or not Santa is fat or thin, or purple, or, gasp, doesn't exist.  We don't need that premise.
And your statement is much the same.  If $y > 0$ then $x > x-y$.  Period.  That will be true if $x > y$.  That will be true if $x = y$ and then will be true if $x < $ negative five hundred gajillion.
The premise $x > y$ and $x>0$ are not necessary as the statement is true without  them and that are completely irrelevent.
....
And the proof you gave is fine.  Valid.  It's just great.
A: Alternative approach
$0 < y \implies  0 = [(-1) \times 0] > (-1) \times y = -y \implies$ 
$x = x + 0 > x + (-y) = x - y.$
A: The proof is indeed valid. And there's a slightly shorter proof which goes like this:
$$y > 0 \implies y+(-y) > 0+(-y) \implies 0 > -y \implies x+0 > x+(-y) \implies x > x-y
$$
