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Here I've read about the game semantics for logical statements with quantifiers. It gives a good intuition about why $\forall x\exists y\,.\,P(x, y)$ means something different than $\exists y\forall x\,.\,P(x, y)$. The former statement means that one can always win this game: a devil gives you a $x$, and then you have to give a $y$ (that can depend on the $x$ given by the devil) with $P(x, y)$. Now, the latter statement is true if and only if one can win a slightly different game: first, one has to specify the $y$, and then the devil gives you a $x$.

I have a view questions concerning this game semantics for logical statements with quantifiers:

  1. Analogously to the above definition of game semantics for statements of the form $\forall x\exists y\,.\,P(x, y)$ or $\exists y\forall x\,.\,P(x, y)$ I can imagine how to define the game semantics for statements of the form $$Q_1xQ_2yQ_3z\dots\,.\, P(x, y, z, \dots),$$ where $Q_i\in\{\forall, \exists\}$ for all $i$ and $P$ is quantifier-free. But how does one define the game semantics for other statements like $\forall x\,.\,(\forall y \,.\,P(x, y))\lor (\exists z\,.\,Q(x, z))$?

  2. One can one prove that for each quantifier statement $\varphi$ in first-order logic, either $\forall$ has a winning strategy or $\exists$ has a winning strategy?

  • Devils? I rather like the more romantic presentation where the roles are $\forall$belard and $\exists$loise ... ($\forall$belard gets to instantiate universals, $\exists$loise the existentials, of course!) – Peter Smith Mar 29 '18 at 22:50

1 Answers1

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You can treat $\lor$ as an existential quantifier over $2$ options: the left or right argument. So the devil gets to choose the left or right argument of the $\lor$ and the game continues.

Similarly, you can treat $\land$ as a universal quantifier over $2$ options. There you get to choose the left or right argument.

Finally, $\lnot$ switches the roles of the players.

The fact that one of the players has a winning strategy is just a particular instance of the fact that every finite deterministic two-player win/loss/draw game with perfect information has a winning or drawing strategy for one of the players.

Magdiragdag
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  • @user419308 Every formula of classical first-order logic can be equivalently rewritten in terms of only the connectives $\neg$, $\lor$, and $\exists$. For $P_1\lor P_2$, the current player picks a $P_i$ and plays that game. For $\exists x.P(x)$, the current player picks an $a$ and plays $P(a)$. For $\neg P$, the opposite of the current player plays $P$. Using $\forall x.P(x)\equiv\neg\exists x.\neg P(x)$, your original formulas are then: $\neg\exists x.\neg\exists y.P(x,y)$ and $\exists x.\neg\exists y.\neg P(x,y)$ respectively. – Derek Elkins left SE Mar 29 '18 at 20:36