Mendelson logic exercise 2.31.e solution attempt.

Question

This question is from "Introduction to Mathematical Logic" by Elliot Mendelson , forth edition , page 78.I have to prove this following theorem

$\vdash (\forall x)(\forall y)(A(x,y) \to \lnot A(y,x)) \to (\forall x) \lnot A(x,x)$

My attempt:

1. $(\forall x)(\forall y)(A(x,y) \to \lnot A(y,x))$ [hyp]
2. $(\forall y)(A(x,y) \to \lnot A(y,x))$ [Axiom A4]
3. $A(x,x) \to \lnot A(x,x))$ [Axiom A4]
4. $A(x,x)$ [Hyp]
5. $\lnot A(x,x)$ [from 3 & 4 by MP]
6. $A(x,x) \land \lnot A(x,x)$ [Conjunction Introduction]
7. $(\forall x)(\forall y)(A(x,y) \to \lnot A(y,x)) , A(x,x) \vdash A(x,x) \land \lnot A(x,x)$
8. $(\forall x)(\forall y)(A(x,y) \to \lnot A(y,x)) \vdash \lnot A(x,x)$ [contraduction theorem]
9. $(\forall x)(\forall y)(A(x,y) \to \lnot A(y,x)) \vdash (\forall x)\lnot A(x,x)$ [Gen]
10. $\vdash (\forall x)(\forall y)(A(x,y) \to \lnot A(y,x)) \to (\forall x)\lnot A(x,x)$ [Deduction theorem]

I am not sure if my attempt is correct or not.Can someone verify it for me?