Smullyan's Drinking Principle

Paradox
Suppose that there is at least one person in the pub.

Then there is a person $x$ in the pub such that if $x$ is drinking, then everyone in the pub is drinking.

Theorem
Let $P$ be a set interpreted as the pub, with its members interpreted as the people in the pub.

Let $\map D x$ be interpreted as the statement "$x$ is drinking".

Then:
 * $P \ne \O \implies \exists x \in P : \paren {\map D x \implies \forall y \in P : \map D y}$

Proof
We have two choices:
 * $\forall y \in P : \map D y$

and
 * $\neg \forall y \in P : \map D y$

Suppose $\forall y \in P : \map D y$.

By True Statement is implied by Every Statement, we have:
 * $\map D x \implies \forall y \in P : \map D y$

By the assumption that $P$ is non-empty, we have:
 * $x \in P$

for some $x$.

By Existential Generalisation, we have our theorem:
 * $\exists x \in P : \paren {\map D x \implies \forall y \in P : \map D y}$

Now suppose:
 * $\neg \forall y \in P : \map D y$

This gives:
 * $\exists y \in P: \neg \paren {y \in P \implies \map D y}$

Switch the variable $y$ with $x$.

Thus, for some $x$:
 * $x \in P \land \neg \map D x$

By False Statement implies Every Statement, we have:
 * $\map D x \implies \forall y \in P : \map D y$

By Existential Generalisation, we have our theorem:
 * $\exists x \in P : \paren {\map D x \implies \forall y \in P : \map D y}$

Thus, $\exists x \in P : \paren {\map D x \implies \forall y \in P : \map D y}$ holds both when:
 * $\forall y \in P : \map D y$

and when:
 * $\neg \forall y \in P : \map D y$

concluding the proof.

Resolution
This is a counter-intuitive result, but it is a veridical paradox resolved by analyzing the possible scenarios.

Either everyone in the pub is drinking or someone in the pub is not drinking.

Suppose that everyone in the pub is drinking.

By True Statement is implied by Every Statement, the statement "everyone in the pub is drinking" is implied by the statement "$x$ is drinking" for any $x$.

Since the pub is by assumption non-empty, there is some $x$ in the pub.

Thus, there is some $x$ in the pub such that, if $x$ is drinking, then everyone in the pub is drinking.

Now suppose that there is some $x$ in the pub who is not drinking.

By False Statement implies Every Statement, if $x$ is drinking then everyone in the pub is drinking.

Thus, there is some $x$ in the pub such that, if $x$ is drinking, then everyone in the pub is drinking.

Thus, a careful analysis of the possible scenarios, as well as certain Paradoxes of Material Implication, show this to be a veridical paradox.

Note that as this is a theorem for arbitrary non-empty sets $P$ and arbitrary propositional functions $\map D x$, this theorem is valid for all interpretations of $P$ and $\map D x$.

For example, interpreting $P$ as the set of things in Philadelphia and $\map D x$ as "$x$ is a dog," we have the following:
 * If there are things in Philadelphia, then there is something in Philadelphia such that, if it is a dog, then everything in Philadelphia is a dog.