(forgive my poor English)

This can be proved with induction .

Now we calculate the answer for $$$(x,y)$$$

Assume that we already know that the answer is $$$(tx-1)xor(y-1)$$$ for $$$(tx,y),tx<x$$$ and $$$(ty-1)xor (x-1)$$$ for $$$(x,ty),ty<y$$$ .

(make $$$x\leftarrow x-1,y\leftarrow y-1$$$)

What we need to prove is that $$$Mex(tx\ xor\ y,ty\ xor\ x )=x\ xor\ y$$$

We can see that $$$x\ xor\ y$$$ is not in $$${tx\ xor\ y,ty\ xor\ x}$$$ , so we only need to prove that for all $$$k\in[0,x\ xor\ y)$$$ , $$$k$$$ can be represented as $$$tx\ xor\ y$$$ or $$$ty\ xor\ x$$$ which is same as $$$k\ xor\ x<y$$$ or $$$k\ xor\ x<y$$$

We will prove that it is impossible for $$$k\in[0,x\ xor\ y)$$$ to satisfy $$$k\ xor\ x\geq y$$$ and $$$k\ xor\ x\geq y$$$

We can remove the equal sign.

The restriction becomes $$$k\ xor\ x> y$$$ and $$$k\ xor\ y> x$$$

Consider the first bit where $$$x\ xor\ k$$$ and $$$y$$$ differs .

It can be found that for all bits higher than this bit $$$k=x\ xor\ y$$$

And in this bit $$$y=0,x\ xor\ k=1$$$

And it is impossible for $$$x=0,k=1$$$ as it will break the limit of $$$k\in[0,x\ xor\ y)$$$ . So this must be $$$x=1,k=0$$$ , but in this way $$$k\ xor\ y$$$ and $$$x$$$ will differ at this bit and $$$k\ xor y=0,x=1$$$ , which will break the limit $$$k\ xor\ y> x$$$.

So it is impossible for $$$k\in[0,x\ xor\ y)$$$ to satisfy $$$k\ xor\ x\geq y$$$ and $$$k\ xor\ x\geq y$$$.

So $$$Mex(tx\ xor\ y,ty\ xor\ x )=x\ xor\ y$$$