[Math] Is this a valid proof of “For all integers m and n, if mn is even, then m is even, or n is even”

Question

Theorem: For all integers $m$ and $n$, if $mn$ is even, then $m$ is even, or $n$ is even.

Proof: Assume for all integers $m$ and $n$, if $mn$ is even, then $m$ is odd and $n$ is odd.

By the definition of odd, $m = 2k + 1$, and $n = 2r + 1$, where $k$ and $r$ are particular but arbitrary integers.

$mn = (2k + 1)(2r + 1)$

$= 4(kr)2 + 2k + 2r + 1$

$= 2((kr)2 + k + r) + 1$

Since $k$ and $r$ are integers, we can replace $(kr)2 + k + r$ with $p$, where $p$ is the integer value of $(kr)2 + k + r$. By the definition of odd, $mn = 2(p) + 1$ is odd, which contradicts $mn$ is even. Hence the supposition is false, therefore the theorem is true.

Answer 1

No, this is not a valid proof. The underlying rule of logic you apply is that the negation of $\forall m\,\forall n\enspace (P\implies Q)$ is $\forall m\,\forall n\enspace (P\implies \neg Q)$, which is wrong: the negation is a counter-example, i.e. $\exists m\,\exists n\enspace(P\wedge( \neg Q))$.

The simplest proof would be by contrapositive, i.e. proving that if none of $m,n$ is even, then $mn$ can't be even.