Smullyan's Drinking Principle

Theorem

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.

Formal Exposition

Let the universe of discourse be the non-empty set of people in the pub.

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

Then:

$\exists x : \paren {\map D x \implies \forall y : \map D y}$

Informal Proof

There are two cases:

$(1): \quad$ Everyone in the pub is drinking.
$(2): \quad$ Someone in the pub is not drinking.

Suppose first that everyone in the pub is drinking.

Then $x$ can be chosen to be any person in the pub.

Suppose instead that someone in the pub is not drinking.

Then $x$ can be chosen to be any person in the pub who is not drinking.

$\blacksquare$

Semi-Formal Proof

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

Suppose that everyone in the pub is drinking.

everyone in the pub is drinking

is implied by the statement:

$x$ is drinking

for any $x$ in the pub.

Since the pub is by assumption non-empty, there exists 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.

$\blacksquare$

Formal Proof

We have two choices:

$\forall y : \map D y$

and

$\neg \forall y : \map D y$

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

$\map D x \implies \forall y : \map D y$
$\exists x : \paren {\map D x \implies \forall y : \map D y}$

Now suppose:

$\neg \forall y : \map D y$
$\exists y : \neg \map D y$

Switch the variable $y$ with $x$.

Thus, for some $x$:

$\neg \map D x$

By False Statement implies Every Statement, we have:

$\map D x \implies \forall y : \map D y$
$\exists x : \paren {\map D x \implies \forall y : \map D y}$

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

$\forall y : \map D y$

and when:

$\neg \forall y : \map D y$

concluding the proof.

$\blacksquare$

Smullyan's Drinking Principle is often considered to be a counter-intuitive result.

Hence it is often classified as a veridical paradox.

However, it can be resolved by analyzing the possible scenarios.

The semi-formal proof attempts to do just that.

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 this is a theorem for arbitrary non-empty universes of discourse and arbitrary propositional functions $\map D x$.

So this theorem is valid for all interpretations of $\map D x$.

For example:

let the universe of discourse be the objects in Philadelphia
let $\map D x$ mean:
$x$ is a dog
Then there exists something in Philadelphia such that, if it is a dog, then everything in Philadelphia is a dog.

Source of Name

This entry was named for Raymond Merrill Smullyan.