# 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

• Solve for s in the following equation. $6r - 5 = 3s+1$ then notice that $s$ will be an integer Commented Apr 28, 2021 at 0:50
• What on earth is P.B.A.C.? Commented Apr 28, 2021 at 1:49
• Particular but arbitrarily chosen. Commented Apr 28, 2021 at 20:11
• @Slowly_Learning Ah, interesting. I've never seen that abbreviation before, and I wouldn't consider it standard. I would go so far as to say that it's not useful either: why write, for example, "let $n$ be a particular but arbitrarily chosen integer" when one can instead just write "let $n$ be an integer"? Commented Apr 28, 2021 at 22:30
• I asked my teacher the same thing (: Commented Apr 29, 2021 at 0:06

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