I've successfully proven $A(x,y)$ for some fixed x and any y with induction but I'm having a hard time proving this for any x and y. I think the next useful step would be proving $A(x,0)= 2[x]3-3 $ but I still having trouble proving it... The definition of $A(x,y)=$ $$\begin{eqnarray*} A(0,y) &=& y+1 \\ A(x+1,0) &=& A(x,1) \\ A(x+1,y+1) &=& A(x,A(x+1,y)) \end{eqnarray*}$$

  • $\begingroup$ Perhaps you should give the definition of $A$, because there are many slightly different definitions... $\endgroup$ – Xoff Dec 20 '15 at 11:31

First, you may see that for all $x$, $2[x]2=4$ for all $x>0$

For $x=0$ $$2[0](y+3 )-3=y+4-3=y+1=A(0,y)$$

Now, by induction on $x$, I suppose this is true for $x$ : $A(x,y)=2[x](y+3 )-3$

For $x+1$


(This is also true when $x=0$, because $A(1,0)=2=5-3=2[1]3-3$)

Now by induction on $y$ :

\begin{align} A(x+1,y+1)&=A(x,A(x+1,y))\\ &=2[x]\bigl(2[x+1](y+3 )-3+3\bigr)-3\\ &=2[x]\bigl(2[x+1](y+3 )\bigr)-3\\ &=2[x+1](y+4 )-3\\ \end{align}


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