⊨∃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?
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.