I have this equivalence relation (this one is proven already):
$(a,b) R (c,d) ⇔ a + d = b + c$
and now I need to show that for $(a,b),(c,d),(e,f) ∈ ℕ x ℕ$:
$(a,b)R(c,d) ⇒ (a+e,b+f)R(c+e,d+f)$
What I did:
$a+d = b+c$
add f:
$⇒ a+d+f = b+c+f$
add e:
$⇒ a+d+f+e = b+c+f+e$
⇒$a+e+d+f = b+f+c+e$
⇒$(a+e)+(d+f) = (b+f)+(c+e)$
Am I allowed to just add e and f on both sides? If not, how should I get started instead?
Re-write the original equation this way: $$(a,b)R(c,d)\textrm{ iff } a-b=c-d$$
Now you must verify that
$$a+e-b-f=c+e-d-f$$
After doing the obvious simplification, you get $$a-b=c-d$$ which is your assumption $(a,b)R(c,d)$