Since there are no premise, the only way to prove must be by way of contradiction, which is to assume ~(∀xFx v ∃x~Fx).
Naturally, one would expand this statement by de Morgan’s law: ~∀xFx v ~∃x~Fx and then use the negation rules of quantifiers to yield ∃x~Fx v ∀xFx which is a contradiction. Sadly, in our course we are not allowed to use the de Morgan’s law and the negation rules of quantifiers.
Another natural approach is of course to observe the law of excluded middle, namely Fx v ~Fx. But again this law is banned from our class.
Under the restriction to use only the introduction and elimination rules of the symbols & v ~ → ↔ ∀ ∃, I am unable to construct any sensible move after supposing for the contrary. May anyone give me some insights? Thank you so much in advance.
The approach by contradiction is correct:
Start with [a1] $\lnot (\forall Fx \lor \exists x \lnot Fx)$ and assume [a2] $\lnot Fx$.
Then $\exists \lnot Fx$ by $(\exists \text I)$ and $\forall Fx \lor \exists x \lnot Fx$.
Now we have a first contradiction and we may derive $Fx$ by Double Negation, discharging [a2].
From it $\forall x Fx$ by $(\forall \text I)$ and $\forall Fx \lor \exists x \lnot Fx$ again.
We have a second contradiction and we may conclude with:
by Double Negation again, discharging [a1].