Proposition. An even integer times an odd integer is even.

As an “if-then” statement, we could express this as “If $a$ is an odd integer and $b$ is an even integer, then $ab$ is even.”

Proof. Suppose that $a$ is an odd integer and $b$ is an even integer. Then $a = 2m + 1$ for some integer $m$, by definition of an odd integer, and $b = 2n$ for some integer $n$, by definition of an even integer. Then

$$ ab = (2m + 1)(2n) = 4mn + 2n = 2(2mn + n) \,, $$

and $2mn + n$ is an integer. Therefore $ab$ is an even integer, by definition of an even integer.