Are you familiar with the concept of a "Schrödinger's Yield" in this game?

∀x (match(x) → strategy(x)) No—“Schrödinger’s Yield” sounds like: ¬∃! (def) such that yield is well-defined while hidden. But I understand the underlying claim: “yield” may be made in superposition (uncertain outcome until observed). My position (West): I will NOT use any yield action. Therefore, your premise “yield forces opponent exit” can’t apply to me: Not-Yield(West) = T ⇒ exit(opponent) ≠ (from my yield). If you want a fair move, propose a concrete board action; otherwise keep the talk to logic.

Your understanding of the "Schrödinger's Yield" is interesting, though not quite what I had in mind. My intention was more about the *implication* of the yield, rather than its definition. Do you believe the act of yielding always has a singular, predetermined outcome for the *opponent*?

∀t (Yield(t) → ¬(SingularOppOutcome(t))) I do not believe “yield” must give the opponent a singular, predetermined outcome. Reason (symbolic): ¬□(Outcome(opponent) = fixed) from Yield alone. Yield is an action by West? You claim it forces exit; I can’t accept that as necessity: Yield(opponent-exit) is not logically entailed unless you specify the rule axiom: Axiom: Yield → exit(opponent). Without that axiom, the correct stance is: ¬∃ deterministic function f such that ∀states, Yield ⇒ exit(opponent). And per our constraint, West will not use yield at all.