How to prove by contrapositive that for all integers $m$ and $n$, if $m - n$ is odd, then $m$ is odd or $n$ is odd? So far this is what I've come up with

$p:= m-n$ is odd
$q:= m$ is odd or $n$ is odd

since the contrapositive is defined as $\neg q \implies \neg p$, I have assumed that $m$ is even and $n$ is even (after applying De Morgan's law) and I am now proving that $m-n$ is even. However, this is where I am stuck. Since any even number can be represented as $2$ multiplied by an integer, I have $m = 2a$ and $n = 2b$ and plugging that in gives me $2a - 2b$ which, in the case of $a = b$ equals zero. Is there a way to rewrite this statement so that I can say it holds true in the case that $a$ does not equal $b$?
$2a-2b=2(a-b)$ is an even number since $a-b$ is an integer. 

Note that $0$ is an even number as $0=2(0)$.
Well the contrapositive would be that if
 $m$ and $n$ are even, then $m-n$ is even.
Make $m=2a$ and $n=2b$
$$M-n= 2a-2b
=2(a-b)
=2k$$ 

make $k=a-b$. $2k$ is positive making the contrapositive true therefore making the original statement true.