Questions about Quantifier Proofs with No Premise in Fitch

Solution 1:

There're already several posts in this site about the proof of this seeming paradox in natural deduction systems, see here and here, for instance.

In case you're interested in a proof for Hilbert system you can refer here, it's much more cumbersome.

Solution 2:

Let me suggest the following considerations more as a topic for some discussion, then as an answer, but I hope it works as simple mathematics, not as philosophy. It's too long for a comment, but nowhere else to write. I am using formal statement from Drinker_paradox $$\exists x\in P.\,(D(x) \to \forall y\in P.\,D(y))$$ $$\exists x\in P.\,D(x)\lor \exists x\in P.\neg(\forall y\in P.\,D(y))$$ $$\exists x\in P.\,D(x)\lor \exists x\in P.(\exists y\in P.\,\neg D(y))$$ Now assuming $D$ is not dependent on $x$, then second member in disjunction is only $\exists y\in P.\,\neg D(y)$ i.e. $\exists x\in P.\,\neg D(x)$, so we have $$\exists x\in P.\,D(x)\lor \exists x\in P.\neg D(x)$$ $$\exists x\in P.\,(D(x)\lor \neg D(x))$$ Of course I used some properties of quantifiers and wait, that somebody brings more clearness, if needed.