I'll just detail the induction step on $x$.
Let's suppose that $\ \displaystyle A(x,y) < A(x,y+1)\ \forall y \in \mathbb{N}\ $ and notice that this implies that $\ \displaystyle A(x,y) < A(x,w)\ \forall y,w \in \mathbb{N} |w>y$.
Then :
$A(x+1,0)=A(x,1)\ $ and $\ A(x+1,1)=A(x,A(x+1,0))=A(x,A(x,1))$
Notice that the smallest possible result is $0+1=1$ and since $\ 1\le A(x,0) we deduce $1 < A(x,1)$ and infer (by considering $y=1,\ w=A(x,1)$ in the induction hypothesis) that $\ A(x+1,0) < A(x+1,1)$
Let's do a induction on $y$ too and suppose that $A(x+1,z) < A(x+1,z+1)$ for $0\le z\le y\ $ then :
$A(x+1,y+1)=A(x,A(x+1,y))\ $ and $\ A(x+1,y+2)=A(x,A(x+1,y+1))$
but we supposed that $\ A(x+1,y) < A(x+1,y+1)\ $ so that we may use our induction on $x$ again to conclude that $\ A(x+1,y+1) < A(x+1,y+2)\ $ and that : $\ A(x+1,y) < A(x+1,y+1)\ \ \forall\; y \in \mathbb{N}$