# Barber Paradox

## 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 1

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.

$\blacksquare$

## Analysis 2

Let $\map M x$ be defined as:

- $x$ is a man in the community.

Let $\map S {x, y}$ be defined as:

- $x$ shaves $y$.

Let $b$ be the barber.

Suppose $\map M b$.

Suppose that:

- $\forall x, y: \paren {\map S {x, y} \implies \map M x, \map M y}$

Suppose to the contrary that:

- $\forall x: \paren {\map M x \implies \paren {\map S {b, x} \iff \neg \map S {x, x} } }$

For $x = b$ we obtain the contradiction:

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

Therefore, it must be false that:

- $\forall x: \paren {\map M x \implies \paren {\map S {b, x} \iff \neg \map S {x, x} } }$

## Resolution

This paradox has been demonstrated to be an antinomy, and can be resolved by several means, for example:

### Resolution 1

Let $b$ be defined so that $b \notin \mathbb U$.

That is, suppose $b$ is *not* one of the men of the community.

This could be the case by, for example:

- $(1): \quad$ The barber is a woman
- $(2): \quad$ The barber is a boy too young to shave.

Then as $b \notin \mathbb U$, it is not necessarily the case that:

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

Thus $b$ is allowed not to be shaved, by himself or anyone else.

$\blacksquare$

### Resolution 2

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:

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

Thus it is not the case that:

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

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

$\blacksquare$

### Resolution 3

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.

$\blacksquare$

## Historical Note

Bertrand Russell himself used the **Barber Paradox** as an illustration of his paradox, claiming it was invented by an unnamed acquaintance of his.

## Sources

- 1918: Bertrand Russell:
*The Philosophy of Logical Atomism*: $7.$ The Theory of Types and Symbolism: Classes - 1964: Donald Kalish and Richard Montague:
*Logic: Techniques of Formal Reasoning*... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $(2)$ - 2008: David Joyner:
*Adventures in Group Theory*(2nd ed.) ... (previous) ... (next): Chapter $1$: Elementary, my dear Watson: $\S 1.2$: Elements, my dear Watson: Example $1.2.1$ - 2010: Raymond M. Smullyan and Melvin Fitting:
*Set Theory and the Continuum Problem*(revised ed.) ... (previous) ... (next): Chapter $1$: General Background: $\S 8$ Russell's paradox