I'm struggling with proof by induction for the following game
Two players called P2 and P2 are playing a game with a starting number of stones. Player always plays first, and the two players move in alternating turns. The game's rules are as follows: In a single move, a player can remove either 2, 3, or 5 stones from the game board. If a player is unable to make a move, that player loses the game.
A player wins where the number of stones $n$ is $n≥1$, and $n mod 7 = 0$ or $n mod 7 = 1$.
For a proof by induction, I pick $n = 0$ as by base case (here P1 will win)
Do I need to have a second base case for $n = 1$, since if I choose that as the inductive step surely I'm just proving that $2 mod 7 = 1$ which is self-evident.
How do I express these base cases?
My attempt at the proof is as follows:
Base case.: $n = 0$ (is a multiple of 7) first player loses
Inductive case: Assume the theorem is true for $n = k$. Prove it is true for $n = k$
Show it is true for $n = 1$ $1 mod 7 = 1$ so p1 loses
Assume it is true for $n = k$ $n mod 7 == 0$ OR $n mod 7 == 1$ so p1 loses.
Clearly this is not correct!