Team,
I need a set theoretic definition of a counter model for the following:
$\forall x (Fx \lor \neg Gx) \land (\exists x \neg Fx \land \exists x Gx)$.
This is the equivalent of what I'm trying to show there is no counter model for, namely $\forall x (Fx \lor \neg Gx) \rightarrow (\forall x Fx \lor \neg \exists x Gx).$ In other words, trying to show that there's not counter model for the latter should be finding a counter model for the former.
However, I'm having a hard time finding a counter model where the former comes out true. I've tried a bunch of sets that are subsets of natural numbers, e.g., $\{0\}$, $\{0,1 \}$ $\{0,1,2\}$ ... and nothing seems to be working.
Does anyone have some tips about what possible counter models may work? Have I messed what I take to be equivalent here? Any tips would help.
You say you want to show there is NO counter model to $\forall x (Fx \lor \neg Gx) \rightarrow (\forall x Fx \lor \neg \exists x Gx)$.
But there is one.
Take a two object domain, with objects $a, b$. Suppose $a$ satisfies both $F$ and $G$. Suppose $b$ satisfies neither $F$ not $G$.
Then $\forall x (Fx \lor \neg Gx)$ is true. And $(\forall x Fx \lor \neg \exists x Gx)$ is false.
You should have suspected something is wrong, since if you put $Hx$ for $\neg Gx$, then what you say has NO counter model is equivalent to $\forall x (Fx \lor Hx) \rightarrow (\forall x Fx \lor \forall x Hx)$, which is a well-known fallacy!