Claim: The position $(x, y, y)$, where $x \ge y$ is losing (the player in that position will eventually lose the game) Proof: The player in this position will either play $(x + k, y, y - k)$ or $(x, y + k, y - k)$, where $k \in \mathbb{N}$. Notice that in both cases, either all three piles are pairwise unequal, or two are equal and the third is smaller. We will prove that in both cases, the other player can keep two piles equal, and the third pile is bigger or equal. Case 1: $(x, y, y) \to (x + k, y, y - k)$ Then the next player responds with $(x + k, y, y - k) \to (x + 2k, y - k, y - k)$ (since $x + k > y$) and keeps two piles equal, and the third bigger or equal. Case 2: $(x, y, y) \to (x, y + k, y - k)$ Let $x = y + a$. If $y + k \ge x$, then the other player responds with $(y + a, y + k, y - k) \to (y + a - (a + k), y + k + (a + k), y - k) = (y - k, y + 2k + a, y - k)$ and keeps two piles equal, and the third bigger or equal. And if $y + k < x$, then the other player responds with $(x, y + k, y - k) \to (x + 2k, y - k, y - k)$, keeping two piles equal, and the third bigger or equal. Notice that the piles with equal values always decrease in value ($(y, y) \to (y - k, y - k)$) which means that the other player will eventually turn the piles into $(x, 0, 0)$, while the initial player can't as stated above. So the player who initially started with $(x, y, y)$ will have lost. Now if $b = c$, then Player $A$ is in the same position as described above, so Player $B$ will win. If $b \neq c$, let $b = c + n$ $(n \in \mathbb{N})$. Player $A$ can then make the move $(a, c + n, c) \to (a + n, c, c)$ since $a \ge b$, putting Player $B$ in a losing position, so Player $A$ will win. Conclusion: Player $B$ wins if and only if $b = c$.