I'm trying to prove:
$\forall x\forall y((x=y)\longrightarrow(x\not<y)$
I tried starting off with
$u=v, u+s(z) = v\vdash u = v$
$u=v, u+s(z) = v\vdash u+s(z) = v$
. . .
$u=v, u+s(z) = v\vdash s(z) = 0$
and try to get a contradiction (since $s(z) \not = 0$, is a theorem), but I'm having a lot of difficulty proving that $s(z) = 0$.
Any help would be greatly appreciated.