# Barber Paradox/Resolution 2

## Paradox

There exists a community, one of whose members is a barber.

This barber operated under an unusual rule: his task was to shave every man in the community who did not shave himself, and *only* those men.

Who shaves the barber?

If he does not shave himself, then he must shave himself.

But if he shaves himself, he must not shave himself.

## Analysis

This is an application of Russell's Paradox.

Let $\mathbb U$ be the set of all the men of the community.

Thus $\mathbb U$ is considered to be the universe.

Let $S: \mathbb U \to \left\{{T, F}\right\}$ be the propositional function:

- $\forall x \in \mathbb U: S \left({x}\right) \iff x \text { is shaved by $x$}$

Let $b \in \mathbb U$ be the barber.

Let $B: \mathbb U \to \left\{{T, F}\right\}$ be the propositional function:

- $\forall x \in \mathbb U: B \left({x}\right) \iff x \text { is shaved by $b$}$

The initial premises can be coded:

- $(1): \quad \forall x \in \mathbb U: \left({\neg S \left({x}\right)}\right) \iff B \left({x}\right)$
- $(2): \quad B \left({b}\right) \iff S \left({b}\right)$

Hence:

- $S \left({b}\right) \iff B \left({b}\right) \iff \left({\neg S \left({b}\right)}\right)$

and so from Biconditional is Transitive:

- $S \left({b}\right) \iff \left({\neg S \left({b}\right)}\right)$

So from either case there derives a contradiction.

Thus the initial premises are contradictory and cannot both hold.

$\blacksquare$

## Resolution

Let the *only* condition above be relaxed, and rewrite it as

*his task was to shave every man in the community who did not shave himself*.

The initial premises would be coded:

- $(1): \quad \forall x \in \mathbb U: \left({\neg S \left({x}\right)}\right) \implies B \left({x}\right)$
- $(2): \quad B \left({b}\right) \iff S \left({b}\right)$

Thus it is not the case that:

- $B \left({x}\right) \implies \left({\neg S \left({x}\right)}\right)$

and so the barber is allowed to shave *at least one* man who *does* shave himself, the barber himself necessarily being one such.

$\blacksquare$