An adaptation from my prior post: Let's say we have a game that is played with three piles containing $x$, $y$ and $z$ logs, respectively. For practical purposes, $x,y,z\in\mathbb{N}$. The objective of the game is to remove that last log. Player 1 goes first and Player 2, second. During their turn, a player can remove as many logs from only one pile as they want. For example, Player 1 could remove all logs from the $x$ pile in their turn.
Who should win the game dependent on $x$, $y$ and $z$?
My thoughts so far: Case 1) If $x=y$, $y=z$, or $x=z$:
Player 1 should remove the third pile that is not equal to the other two. Then, after Player 2 makes a move, Player 1 need only match the number taken by Player 2 in the opposite pile to force a win.
Case 2) $x=y=z$
Here Player 1's strategy is the same. Simply take any pile. Then, the situation is reduced to the latter part of Case 1) and Player 1 can force a win.
Case 3) $x\neq y \neq z$, $x\neq z$:
My notation is a little sloppy, but here I mean that none of the three piles have equal values of logs. If Player 1 tries to proceed in the same way as Cases 1) and 2), he or she will lose since Player 2 can then implement the same strategy in opposition. Does Player 2 win in this case? If so, how?