In predicate logic how to prove that:
$$\forall x \forall y (A(x,y) \rightarrow \neg A(y,x)) \vdash \forall x \forall y (A(x,y) \rightarrow \neg x = y)$$
First it should be proven that:
$$x=y \vdash A(x,y)\leftrightarrow A(y,x) $$
(I did prove the second result, but have problems utilizing it).
Hint
1) $∀x∀y(A(x,y)→¬A(y,x))$ --- premise
From it derive : $A(x,x)→¬A(x,x)$ and with TAUT : $(P \to \lnot P) \to \lnot P$ derive : $\lnot A(x,x)$.
2) $A(x,y)$ --- assumed [a]
3) $x=y$ --- assumed [b]
From 2) and 3) derive : $A(x,x)$ --- contradicition !
Conclude with : $¬x=y$, discharging [b].
Derive : $A(x,y) \to ¬x=y$, discharging [a].