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I am currently a Philosophy student taking a logic course and I am quite confused about how I can go about tackling this following proof without a premise. I have been able to deal with all the practice proofs so far but this one has me stuck.

$$\exists x\,(P(x) \to \forall y\,P(y))$$

I have tried to solve this problem by reductio ad absurdum, but I have made little progress. Should I instead start by assuming $P(x)$, proving that it leads to $\forall y \, P(y)$, and then introducing the existential quantifier?

Bonnaduck
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NorwegianWood
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2 Answers2

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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.

cinch
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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.

zkutch
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