# Barber Paradox/Resolution 3

## 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 $S: \mathbb U \to \set {T, F}$ be the propositional function:

- $\forall x \in \mathbb U: \map S x \iff x \text { is shaved by $x$}$

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

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

- $\forall x \in \mathbb U: \map B x \iff x \text { is shaved by $b$}$

The initial premises can be coded:

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

Hence:

- $\map S b \iff \map B b \iff \paren {\neg \map S b}$

and so from Biconditional is Transitive:

- $\map S b \iff \paren {\neg \map S b}$

So from either case there derives a contradiction.

Thus the initial premises are contradictory and cannot both hold.

$\blacksquare$

## Resolution

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

*his task was that he may shave only men in the community who did not shave themselves*.

The initial premises can be coded:

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

Thus it is not the case that:

- $\paren {\neg \map S x} \implies \map B x$

However, from $\map B b \iff \map S b$ this then means that $\neg \map B b$.

So *at least one* person in the community is not shaven by $b$ at all, the barber himself necessarily being one such.

This of course does not preclude the possibility that some other person, who is not the barber, may *also* shave people.

These may or may not include the barber, who may retain his beard.

$\blacksquare$