# Proving that the Nim-sum cannot be zero for two turns in a row

In proving certain results about Nim, I found a lemma that is causing me trouble:

Lemma 1 If the Nim-sum is $$0$$ after a player’s turn, then the next move must change it.

To prove this, let the number of stones in the heaps be $$x_1, x_2, ... x_n$$, and $$s$$ be the nim-sum $$s=x_1⊕x_2⊕x_3⊕ . . . ⊕x_n$$ Let $$t$$ be the sum of the heaps after the move, $$t=y_1⊕y_2⊕y_3⊕ . . . ⊕y_n$$ Then if $$s= 0$$, the next move causes some $$x_k=y_k$$ and the rest of the $$x_i=y_i$$ for $$i \neq k$$, since only one pile of stones is changed.

Then: $$t=0⊕t$$ $$=s⊕s⊕t$$ $$=s⊕(x_1⊕x_2⊕...⊕x_n)⊕(y_1⊕y_2⊕...⊕y_n)$$ $$=s⊕(x_1⊕y_1)⊕(x_2⊕y_2)⊕...⊕(x_k⊕y_k)$$ $$=s⊕x_k⊕y_k$$

If $$s$$ is 0, then $$t$$ must be nonzero, since $$x_k⊕y_k$$ will never be 0. Therefore, if you make the nim-sum $$0$$ on your turn, your opponent must make it nonzero.

I understand why $$x_k ⊕ y_k$$ can't be $$0$$, but why can't it be equal to $$s$$? This would make the nimsum $$0$$ still, and I don't see why this wouldn't be possible.

• $s$ is the state before the move, and is assumed to be zero. So you are effectively asking: I understand $x_k⊕y_k$ can't be zero, but why can't it be equal to zero? – Jaap Scherphuis Jan 30 at 15:26