There are infinitely many monotone games over $L_{5}$

It is easy to see from their definition that the games in the sequence $G_{0},G_{1},G_{2},\ldots$ all have the property that the final score is equal to the number of moves made by Left minus the number of moves made by Right. Such games are automatically monotone. To see why, consider the slightly more general class of games $G$ with the property that the final score is equal to the number of moves made by Left minus the number of moves made by Right plus some constant $C\in\mathbb{Z}$. We write $\mathop{\mathrm{\rm m}}\nolimits(G)=C$ (the mean value of $G$, see [2]). It is then easy to prove by induction that for all such games, $\mathop{\mathrm{\rm m}}\nolimits(G)\leqslant\mathop{\mathrm{\rm m}}\nolimits(H)$ implies $G\vartriangleleft H$ and $\mathop{\mathrm{\rm m}}\nolimits(G)<\mathop{\mathrm{\rm m}}\nolimits(H)$ implies $G\leqslant H$. In particular, since $\mathop{\mathrm{\rm m}}\nolimits(G^{L})=\mathop{\mathrm{\rm m}}\nolimits(G)+1$, we have $G\leqslant G^{L}$, and similarly $G^{R}\leqslant G$, proving that $G$ is monotone. ∎

To show $G_{n}\leqslant G_{n+1}$, we must show three things: First, we must show that all left options $G^{L}_{n}$ satisfy $G^{L}_{n}\vartriangleleft G_{n+1}$. When $n=0$, there is no such left option, and when $n>0$, the unique left option of $G_{n}$ is $1$. But $1$ is also a left option of $G_{n+1}$, so $1\vartriangleleft G_{n+1}$ as claimed. Second, we must show that all right options $G^{R}_{n+1}$ satisfy $G_{n}\vartriangleleft G^{R}_{n+1}$. But the unique right option of $G_{n+1}$ is $P^{n}(G_{n})$, and $G_{n}\vartriangleleft P^{n}(G_{n})$ holds because $G_{n}$ is a left option of $P^{n}(G_{n})$. Third, we must show that if $G_{n}$ or $G_{n+1}$ is atomic, then $G_{n}\vartriangleleft G_{n+1}$. But this only happens when $n=0$. In this case, we must show $0\vartriangleleft G_{1}$. But this holds because $1$ is a left option of $G_{1}$ and $0\leqslant 1$. ∎

From Lemma 3.2 and transitivity, since $0=G_{0}\leqslant G_{1}\leqslant\ldots\leqslant G_{n}$. ∎

The following claims hold for all $n$. We prove them by simultaneous complete induction on $n$. The lemma is claim (8).

1. (1)

   $G_{n}\ntriangleleft-1$.

   This follows from Corollary 3.3. Indeed, if $G_{n}\vartriangleleft-1$ were true, then transitivity would imply $0\vartriangleleft-1$, which is absurd.

2. (2)

   $P^{n}(G_{n})\nleqslant-1$.

   This follows from (1), because $G_{n}$ is a left option of $P^{n}(G_{n})$.

3. (3)

   If $n$ is even, $P^{n}(G_{n})\ntriangleleft-2$.

   Since $-2$ is atomic and $P^{n}(G_{n})$ is not, the only way $P^{n}(G_{n})\vartriangleleft-2$ can hold is if some right option $H$ of $P^{n}(G_{n})$ satisfies $H\leqslant-2$. But this is impossible because $\star$ is the unique right option of $P^{n}(G_{n})$ and $\star\nleqslant-2$.

4. (4)

   If $n$ is odd, $P^{n}(G_{n})\ntriangleleft\star$.

   Since neither $P^{n}(G_{n})$ nor $\star$ is atomic, there are only two ways in which $P^{n}(G_{n})\vartriangleleft\star$ could hold. Either some right option $H$ of $P^{n}(G_{n})$ satisfies $H\leqslant\star$; but this is impossible because $-2$ is the unique right option of $P^{n}(G_{n})$ and $-2\nleqslant\star$. Or else some left option $K$ of $\star$ satisfies $P^{n}(G_{n})\leqslant K$; but this is impossible by (2) because $-1$ is the unique left option of $\star$.

5. (5)

   If $n>0$, then $P^{n}(G_{n})\nleqslant P^{n-1}(G_{n-1})$.

   When $n$ is even, this follows from (3) because $-2$ is a right option of $P^{n-1}(G_{n-1})$.

   When $n$ is odd, this follows from (4) because $\star$ is a right option of $P^{n-1}(G_{n-1})$.

6. (6)

   If $n>0$, then $G_{n+1}\nleqslant G_{n-1}$.

   For the sake of obtaining a contradiction, suppose $G_{n+1}\leqslant G_{n-1}$. By Lemma 3.2, we have $G_{n}\leqslant G_{n+1}$. With transitivity, this implies $G_{n}\leqslant G_{n-1}$. However, this contradicts (8) of the induction hypothesis.

7. (7)

   If $n>0$, then $G_{n+1}\ntriangleleft P^{n-1}(G_{n-1})$.

   Because neither $G_{n+1}$ nor $P^{n-1}(G_{n-1})$ is atomic, there are only two ways in which $G_{n+1}\vartriangleleft P^{n-1}(G_{n-1})$ could hold. Either some right option $H$ of $G_{n+1}$ satisfies $H\leqslant P^{n-1}(G_{n-1})$; but this is impossible by (5) because $P^{n}(G_{n})$ is the unique right option of $G_{n+1}$. Or else some left option $K$ of $P^{n-1}(G_{n-1})$ satisfies $G_{n+1}\leqslant K$; but this is impossible by (6) because $G_{n-1}$ is the unique left option of $P^{n-1}(G_{n-1})$.

8. (8)

   $G_{n+1}\nleqslant G_{n}$.

   When $n=0$, this holds by direct calculation: $G_{1}\nleqslant 0$ because $1$ is a left option of $G_{1}$ and $1\ntriangleleft 0$.

   When $n>0$, this follows from (7) because $P^{n-1}(G_{n-1})$ is a right option of $G_{n}$.∎

By Lemmas 3.2 and 3.4, we have $G_{n}<G_{n+1}$ for all $n$. In particular, the sequence $G_{0},G_{1},\ldots$ consists of infinitely many non-equivalent games. Moreover, they are monotone by Lemma 3.1. ∎

It was shown in [4, Prop. 10.2] that there are infinitely many monotone game values over $A$ when $A$ has two incomparable elements. This takes care of all cases where $A$ is not linearly ordered. It was further shown in [4, Sec. 10.2] that there are only finitely many monotone game values over $A$ when $A$ is a linearly ordered set of size $1$, $2$, $3$, or $4$. (In this case, there are $1$, $3$, $8$, or $31$ such values, respectively). The case of the zero-element poset is trivial, as there are no game values at all. The only remaining cases are linearly ordered sets of $5$ or more elements (including infinite ones). In these cases, there are infinitely many monotone game values by Corollary 3.5. Note that if the atom poset has strictly more than $5$ elements, one may simply disregard the additional atoms. ∎