Having trouble knowing how to approach this proof. I'm not sure why this is unclear to so many, but this is Lemma 2.6 in Section 3.2 of the Robert S. Stoll, Set Theory and Logic textbook.
Define a "difference" to be an ordered pair of natural numbers.
Define a relation $\sim_d$ on differences by letting $(m,n)\sim_d(p,q)$ iff $m+q=p+n$.
We are asked to prove the following:
If $x,y,u$ and $v$ are differences, $x\sim_d u$, and $y\sim_d v$, then $xy\sim_d uv$.
Thank you.
I also need to add that this is not a homework problem. It's something the entire class has been working on for extra practice, and today when we had about $8$ lines written, the professor realised the outcome would not yield the desired result. Past that point, we were all stumped, including him.
You are given $x = (m_1 , n_1)$ $y =(m_2, n_2)$. I would work by simultaneous induction on $(m_1, n_1, m_2, n_2)$: show the result when all four coordinates are zero, then assume it for a general $(m_1, n_1, m_2, n_2)$ and prove it for each of $(m_1 + 1, n_1, m_2, n_2)$, $(m_1, n_1 + 1, m_2, n_2)$, and the other two cases, which follow by symmetry.
Indeed, write $u = (a,b)$, $v= (c,d)$.
Then in the base case $u \sim_d x$ implies $(a + 0) = (b + 0)$, $v \sim_d y$ implies $(c + 0) = (d + 0)$; hence $$uv = (ab +cd, ac+bd) \sim_d (0,0) =xy.$$
And now assume the result for $x = (m_1, n_1)$, $y= (m_2, n_2)$; and suppose that $u \sim_d (m_1+1, n_1)$, while $v\sim_d (m_2, n_2)$. We write $u = (a,b)$ as before.
Then $(a, b+1) \sim_d (m_1,n_1)$, so by inductive hypothesis $(a, b+ 1) \times (c,d) \sim_d xy$, and it follows that $$(a, b) \times (c, d) = (a, b+1) \times (c,d) + (b,0) \sim_d xy + (m_2, n_2) \sim_d (m_1+1, n_1)\times(m_2, n_2)$$ as required.
Now we do the whole thing again for $n_1+1$; and appeal to symmetry for increments to $m_2$ and $n_2$. And by induction we are done.