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⊨∃x(Px⇒∀xPx) I was trying to solve the above drinker's paradox by contradiction and came up with the below expression by negating the above expression -

⊨ ∀x ( Px ∧ ¬ ∀x Px )

This is where I got stuck because to distribute the universal quantifier I need all the variables inside the parenthesis to be bound variables but as x is a free variable I was not able to go further. Is the approach correct considering that the universal set is not empty and what should I do next?

Mauro ALLEGRANZA
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Random
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1 Answers1

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In your approach, you have not given an equivalent expression, but rather the negation of the expression.

You can however equivalenty write it as $\models \exists x (\neg Px \lor \forall y Py)$

It should be quite clear why this is true. Just read it out. "There is an $x$ such that either $Px$ is false, or $Px$ is true for all $x$. This is evident by just considering the two cases of when $Px$ is true for all $x$ and when it is not.

Saad Haider
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  • I have updated my question a little just for the language. Can i solve it through contradiction? And if yes then how? – Random May 19 '20 at 17:59
  • @Random Yup. Since the x in the second term is not bound by the x in the first quantifier, you can pull it out and you get $\forall x Px \land \neg(\forall x Px)$ which is a contradiction – Saad Haider May 20 '20 at 18:16