I'm having trouble understanding the notation in this proof. The proof begins with 2 lines. I will try to explain the lines, and I would like if you could tell me if I'm correct.
$\forall m\in\mathbb{Z}\ \ p(m) \implies p(3m^5)$
and
$p(m) \equiv \exists y\in\mathbb{Z}\ \ m = 2y $
The 1st line sends all integers $m$ to a condition $p(m)$. Let's say for example $m = 1$. In line 2, we're asserting that there exists an integer $y$ such that $1 = 2y$, which is false. But a false antecedent leads to a true implication, so we can move onto the next integer, $m = 2$, and so on and so forth.
Is this correct?
The second line is the definition (or maybe a repetition of the definition) of the predicate $p$. It says that $p(m)$ means '$m$ is even'.
The first line is the claim you are going to prove: for every integer $m$, if $m$ is even, then $3m^5$ is even as well.
(Note by the way that your title says that 'while proving for every integer $m$ that $3m^5$ is even'. That is wrong.)