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I'm working on some proofs in symbolic logic and I pretty much get it, but I'm having an issue with the last one. Usually the premise sets up the proof and I can walk through it, but this one just starts with a tautology. I may be overthinking it, but I'm not even sure where to start.

⊤ ⊢ ∃x(D(x) → ∀yD(y))

I kinda get the second statement: "There exists an x such that if D(x) then for all y, D(y)." I think it's saying that if D is true for one element, then it is true for all other elements, but how does a tautology prove this? Or I guess a better phrasing is how can I structure this into a proof?

Daniel
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  • To prove $p \implies q$, use its equivalence to $\neg p \vee q$, and then just prove $q$. With that approach, it doesn’t even matter what $p$ is. – mjqxxxx Jun 07 '24 at 18:49
  • In this case, $q$ is a reasonably famous “paradox” (not really a paradox, of course). Either $D$ is always true, or it isn’t. If it isn’t, then there’s some particular $x$ such that $D(x)$ is false, and so $D(x)$ implies anything. – mjqxxxx Jun 07 '24 at 18:52
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    This is the Drinker Paradox. – Rob Arthan Jun 07 '24 at 19:31
  • I'm guessing the author really wanted you to prove this tautology. To do this, you first need to somehow assume the existence of one object $x$. Then consider two cases: Either $\forall x: D(x)$ is true or it is false. In both cases, prove that the above tautology is true. – Dan Christensen Jun 07 '24 at 20:39
  • How are you supposed to prove this? Using a formal derivation? Using formal semantics? Truth tree? Resolution? Etc. – Bram28 Jun 08 '24 at 00:18
  • You can prove it quite easily using a form of natural deduction. I was able to do so in 24 lines as outlined above. (Formal proof available on request.) – Dan Christensen Jun 08 '24 at 02:46
  • Either by contradiction or starting from another tautology, like Excluded Middle. – Mauro ALLEGRANZA Jun 08 '24 at 07:20

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