Proving a Set is a Subset of another (with functions) 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
 A: 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 $m \in B$.
A: Not an answer, but rather advice on writing proofs (too long for a comment).  Here's what you started with:

Proof: Suppose $A = \{m \in \mathbb{Z} \mid m=6r-5, r \in \mathbb{Z}\}, B = \{n \in \mathbb{Z} \mid n = 3s+1, s \in \mathbb{Z}\}$. Suppose P.B.A.C integers $x \in m$ and $y \in n$. By substitution $x=6r-5$ and $y=3s+1$, where $r$ and $s \in \mathbb{Z}$.

Let's break this down one sentence at a time:

Proof: Suppose $A = \{m ∈ \mathbb{Z}\mid m=6r-5,r \in  \mathbb{Z}\}$, $B = \{n ∈  \mathbb{Z}\mid n= 3s+1, s \in \mathbb{Z}\}$

You don't need to suppose this; it is already given as a definition.

Suppose P.B.A.C integers $x \in m$ and $y \in n$.

As I wrote in a comment above, "P.B.A.C." is not a standard mathematical abbreviation, and furthermore doesn't appear helpful. More importantly, $m$ and $n$ are not sets, so it doesn't make sense to have $x \in m$ and $y \in n$. Instead, you're trying to let $x \in A$, so that $x = m$, where $m = 6r-5$. But then notice that $x$ and $y$ aren't actually needed: you can just skip right to $m$ and $n$.

By substitution $x=6r-5$ and $y=3s+1$, where $r$ and $s \in \mathbb{Z}$.

We'll combine this with the previous comment, avoiding the need for "substitution".

Putting it together:

Proof: Let $m \in A$ and $n \in B$, so that $m = 6r - 5$ and $n = 3s + 1$ for some $r, s \in \Bbb Z$.

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