Statement:
Sum of even and odd integer is odd

$$ \forall(a,b) \in \mathbb{Z} : a \text{ mod } 2 = 0 \wedge b \text{ mod } 2 \neq 0 \implies a + b \text{ mod } 2 \neq 0 $$

Proof:

$$ a \text{ mod } 2 = 0 \implies \exists n \in \mathbb{Z}: a = 2n $$
$$ b \text{ mod } 2 \neq 0 \implies \exists k \in \mathbb{Z}: b = 2k +1 $$

$$ \implies (2n+2k+1) \text{ mod } 2 \neq 0 \implies \exists q \in \mathbb{Z}: 2q+1 = (2n+2k+1)$$

$$ \implies q = n + k $$

Not quite sure if this is correct or not. To me it would seem it's correct but could i have comment on this?