Proving a Set is a Subset of another (with functions)

Question

we have
$A =\{m ∈ \mathbb{Z}|m=6r-5,r ∈ \mathbb{Z}\} $ and
$B = \{n ∈ \mathbb{Z}| n= 3s+1, s ∈ \mathbb{Z}\}$
prove $A ⊆ B$

I have
Proof: suppose $A = \{m ∈ \mathbb{Z}|m=6r-5,r ∈ \mathbb{Z}\}$,
$B = \{n ∈ \mathbb{Z}| n= 3s+1, s ∈ \mathbb{Z}\}$
suppose P.B.A.C integers x ∈ m and y ∈ n
by substitution $x=6r-5$ and $y=3s+1$, where $r$ and $s ∈ \mathbb{Z}$
and.. I have no idea where to go. should I set these equal to eachother? I have no clue

Answer 1

Let $m \in A$, $r \in \mathbb{Z}$ be such that $m=6r-5$. Then $m=6(r-1)+1=3(2(r-1))+1 $, and so $m \in B$.

Answer 2

$x\in A$ iff $(x-1)/6\in \Bbb Z$.

$x\in B$ iff $(x-1)/3\in \Bbb Z$.

Therefore $x\in A\implies (x-1)/6\in \Bbb Z \implies (x-1)/3=2\cdot (x-1)/6\in \Bbb Z \implies x\in B.$