Answer
By the inductive hypothesis, the second player in that game,
who is also the second player in our actual game, can win, and the proof by strong induction is complete.
Work Step by Step
--Proof by strong induction:
-Basis step:
-When there are n = 2 stones in each pile, if first player takes two stones from a pile,
-then second player takes one stone from the remaining pile and wins. If first player takes one stone from a pile, then second player takes two stones from the other pile and wins.
--Inductive step:
-Assume inductive hypothesis that second player can always win if the game starts with two piles of j stones for all 2 ≤ j ≤ k, where k ≥ 2, and
consider a game with two piles containing k + 1 stones each.
-If first player takes all the stones from one of the piles,
then
-second player takes all but one stone from the remaining pile
and wins.
- If first player takes all but one stone from one of the piles,
- then second player takes all the stones from the other pile and wins. Otherwise first player leaves j stones in one pile,
-where 2 ≤ j ≤ k, and k + 1 stones in the other pile.
-Second player takes the same number of stones from the larger pile,
also leaving j stones there.
-At this point the game consists of two piles of j stones each.
-By the inductive hypothesis, the second player in that game,
who is also the second player in our actual game, can win, and the proof by strong induction is complete.
