I am stuck with connecting the integer congruence with the first part of this proof. I've got the outline.
Let $n ∈ Z$, and let $[x]$ denote the equivalence class of $x$ under integer congruence modulo $n$. Let $a, b, x, y$ be integers.
Proof.
Suppose $a ∈ [x]$ and $b ∈ [y]$.
Then $aRx$ and $bRy$.
So $n|(a-x)$ and $n|(b-y)$, by definition of integer congruence modulo $n$.
So there is an integer $g$ such that $a - x = ng$ and there is an integer $f$ such that $b - y = nf.$
So $(a + b) - (x + y) = a - x + b - y = ng + nf = n(g + f).$
Then there is an integer $k$ such that $(a + b) - (x + y) = nk$; namely, $k = g + f$.
Then $n|((a+b) - (x + y))$, by definition of integer congruence modulo $n$.
Therefore $a + b = x + y$ (mod $n$).
Am I on the right path? Can someone help me fill in the '...'? What exactly is contained in [x]? I know what it means to be an equivalence class, but how does it tie to the integer congruence modulo n?
If your notation $a \in [x]$ here indicates $a \equiv x \pmod n$, then by definition of congruence $\pmod n$ this means that $\exists s \in \Bbb Z$ such that $a=sn+x$. Similarly, $b \in [y]$ means that $\exists t \in \Bbb Z$ such that $b=tn+y$. But then $a+b=(s+t)n+(x+y)$ so setting $v=s+t$, $a+b=vn+(x+y)$ which implies $a+b \equiv x+y \pmod n$.