Problem
Definition. Let $V$ be a non-empty set of real numbers. The class VG of $V$-valued two-person deterministic games of perfect information is defined recursively as follows:
Base case. A value $v\in V$ is a VG knows as a payoff.
Constructor case: If $G$ is a non-empty set of VG's, then $G$ is a VG. Each game $M\in G$ is called a possible first move of $G$.
A strategy for a player is a function s from games to games with the property that $s(G)\in G$ for all games $G$. Given which player has the first move, a pair of strategies for the two players determines exactly which moves the players will choose. So, the strategies determine a unique play of the game and a unique payoff.
The max-player wants a strategy that guarantees as high a payoff as possible, and the min-player wants a strategy that guarantees as low a payoff as possible.
The Fundamental Theorem for deterministic games of perfect information says that in any game, each player has an optimal strategy, and these strategies lead to the same payoff. More precisely,
Theorem (Fundamental Theorem for VG's). Let $V$ be a finite set of real numbers and $G$ be a $V$-valued VG. Then there is a value $v\in V$, called a max-value $\text{max}_G$ for $G$, such that if the max-player moves first,
the max-player has a strategy that will finish with a payoff of at least $\text{max}_G$, no matter what strategy the min-player uses, and
the min-player has a strategy that will finish with a payoff of at most $\text{max}_G$, no matter what strategy the max-player uses.
Prove the Fundamental Theorem for VG's.
Can anybody please verify the following solution attempt?
Solution attempt
Proof by structural induction.
Induction hypothesis: $P(G)$ := There is a value $v\in V$, called max-value $\text{max}_G$ for $G$, such that, if the max-player moves first,
the max-player has a strategy that will finish with a payoff of at least $\text{max}_G$, no matter what strategy the min-player uses
the min-player has a strategy that will finish with a payoff of at most $\text{max}_G$, no matter what strategy the max-player uses
Base case ($G = v \in V$): If the max-player is the first player, the strategy is to do nothing and finish with payoff $\text{max}_G = v$.
Constructor case ($G = \left\{M\text{ | }M \in \text{VG}\right\}$): Assume that $P(M)$ is true for all $M\in G$.
By cases:
The max-player has the next move in $G$. For any $M\in G$ that the max-player picks, the min-player will be the first player in $M$. Suppose the max-player picks a move $M$.
1.1. If $M = v \in V$, the max-player's strategy is to do nothing and finish with payoff $\text{max}_G = v$.
1.2. If $M = \left\{N | N \in \text{VG}\right\}$, then suppose the min-player picks some $N \in M$. By the induction hypothesis, since the max-player moves first in $N$, then there is a $\text{max}_N$ such that:
the max-player will have a strategy that ends with payoff at least $\text{max}_N$, and
the min-player will have a strategy that ends with payoff at most $\text{max}_N$.
So, for $G$, the strategy for either max-player or min-player is: first, the max-player makes any available move, then the min-player makes any available move, then both can carry out their strategies for the remaining moves, and the payoff will be $\text{max}_G = \text{max}_N$.
The min-player has the next move in $G$. For any $M\in G$ that the min-player chooses, the max-player will be the first player in $M$.
By the induction hypothesis, for any $M$ that the min-player chooses, there is a $\text{max}_M$ such that:
the max-player will have an optimal strategy that ends with payoff at least $\text{max}_M$, and
the min-player will have an optimal strategy that ends with payoff at most $\text{max}_M$.
So, for $G$, $\text{max}_G$ = $\text{max}_M$.
Therefore, by structural induction, $P(G)$ is true for all $G$ in $\textrm{VG}$.