Can this set-theoretic resolution of the Barber Paradox be proven in FOL?

Question

Using set theory, I can prove that, in that proverbial village with a resident barber, that barber can shave those and only those men in the village who do not shave themselves if and only if that barber is not a man. Can something similar to this be proven in FOL (edit: using predicate logic without set theory)?

In set theory, I have proven:

$\forall V, b, M: [b\in V \land M\subset V \implies [\exists S: \forall x\in M: [(b,x)\in S \iff (x,x)\notin S] \iff b\notin M]]$

where

$V=$ the set of villagers

$M=$ the set of men in the village

$b=$ the barber

$S=$ the shave relation, $(b,x)\in S$ means $b$ shaves $x$

See full text of formal proof

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EDIT

Is it possible, for example, to prove:

$\forall b: [V(b) \land \forall x:[M(x) \implies V(x)] $

$\implies [\forall x: [M(x)\implies [S(b,x)\iff \neg S(x,x)]] \iff \neg M(b)]] $

It is easy to prove (changing the biconditional to a conditional):

$\forall b: [V(b) \land \forall x:[M(x) \implies V(x)] $

$\implies [\forall x: [M(x)\implies [S(b,x)\iff \neg S(x,x)]] \implies \neg M(b)]] $

The problem seems to be in proving the converse:

$\forall b: [V(b) \land \forall x:[M(x) \implies V(x)]$

$ \implies [\neg M(b)\implies\forall x: [M(x)\implies [S(b,x)\iff \neg S(x,x)]] ]] $

I don't think this is true in general, e.g. if one man in the village is never shaved.

Answer 1

As usual, things hinge on exactly what it is you are trying to do.

One possibility is to - as you do initially? - convert the Barber Theorem to a statement in the first-order language of set theory. It is then easily provable, from the ZFC axioms or indeed vastly less. Indeed, this is the natural thing to do: the Barber Theorem is a statement about sets, so we should expect to formalize and prove it in first-order set theory.

Alright, now what about those pesky axioms? We have a first-order sentence $\beta$ in the language of set theory, which is provable from ZFC; is it a validity? It turns out the answer (unsurprisingly) is no. For example, there are $\{\in\}$-structures which satisfy "there are no nontrivial binary relations at all"! In such a structure, obviously $\beta$ is not true.

Interestingly, it's not hard to see that the "paradox" half of the Barber Theorem - that the barber can't be a man - is a validity. So it's easy to see the "bad" side, but hard to see the "safe" side. Basically, this amounts to the fact that set existence principles tend to be nontrivial (that is, not validities), whereas the classical paradoxes - which are really set nonexistence principles - use nothing more than pure logic.

The right way to express $\beta$ as a first-order sentence which is provable from no axioms at all, is to bring the relevant axioms in as hypotheses: namely, we look at $$T\implies \beta,$$ where $T$ is a conjunction of a few of the ZFC axioms. This will then be a validity.

When you "prove the Barber Theorem with set theory," it may feel like you're proving $\beta$ from no axioms at all. But this isn't really accurate: implicitly in your proof, you're using various properties about sets. You can view them as built into the underlying logic, but that's a cheap way out: what's really going on is that you're working in a first-order "background" theory, which proves some statements about sets. You should either prove $\beta$ from the axioms of that theory, or instead bring the relevant axioms in as hypotheses.

Answer 2

With $M(x)$ meaning $x$ is a man, and $S(x,y)$ meaning $y$ shaves $x$, we have: $$\forall y \,. [\forall x \,. M(x) \rightarrow (S(x,y) \leftrightarrow \neg S(x,x))] \rightarrow \neg M(y) \enspace, $$ whose validity is easily proved (regardless of the intended interpretation).

On to the proof. To prove validity, we want to show that the negation of our sentence is unsatisfiable. That is, that

$$\exists y \,. [\forall x \,. M(x) \rightarrow (S(x,y) \leftrightarrow \neg S(x,x))] \wedge M(y) $$

is unsatisfiable. We instantiate the existential quantifier with a fresh constant symbol $b$ (as in barber):

$$[\forall x \,. M(x) \rightarrow (S(x,b) \leftrightarrow \neg S(x,x))] \wedge M(b) \enspace. $$

Now we instantiate the universal quantifier with $b$:

$$[M(b) \rightarrow (S(b,b) \leftrightarrow \neg S(b,b))] \wedge M(b) \enspace. $$

This is clearly contradictory.

Answer 3

Assume $V$ as the universe of the interpretation and let $M \subseteq V$.

Thus, we will not use $V$ in the formulae and we'll use a predicate $M(x)$ to match the set $M$, i.e. $M(a)$ iff $a \in M$.

Let :

$S(a,b) \leftrightarrow a= \text {barber} \land M(b)$;

this is the translation of line 23): the definition of: $S= \{ (a,b) \mid a= \text {barber} \text { and } b \in M \}$.

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Part 1 : from line 6) : $\text {barber} \in M$.

6) $M(\text {barber})$ --- assumed

8) $\forall a \ (M(a) \to [S(\text {barber},a) \leftrightarrow \lnot S(a,a)])$ --- assumed

9) $M(\text {barber}) \to [S(\text {barber},\text {barber}) \leftrightarrow \lnot S(\text {barber},\text {barber})])$ --- instantiating 8)

10) $S(\text {barber},\text {barber}) \leftrightarrow \lnot S(\text {barber},\text {barber})$ --- contradiction; thus, by $(\lnot \text I)$, discharging assumption 8), followed by $(\to \text I)$, discharging assumption 6):

12) $M(\text {barber}) \to \lnot \forall a \ (M(a) \to [S(\text {barber},a) \leftrightarrow \lnot S(a,a)])$

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Part 2 : from line 13) : $\text {barber} \notin M$.

13) $\lnot M(\text {barber})$ --- assumed

23) $S(a,b) \leftrightarrow a= \text {barber} \land M(b)$

43) $\forall a \ [ M(a) \to \lnot S(a,a)]$ -- intuitively, if $S(a,b)$ holds only when $a= \text {barber}$ and we have assumed that $\lnot M (\text {barber})$, then when $M(a)$ we are forced to conclude that $a$ cannot be equal to $\text {barber}$ (simple derivation by contradiction).

63) $\forall a \ [ M(a) \to S(\text {barber},a)]$ -- by definition of $S$.

Using the tautological implication : $p \to q \vDash p \to (r \to q)$ we may derive respectively :

• from 43) : 68) $M(x) \to [S(\text {barber},x) \to \lnot S(x,x)]$

• from 63) : 72) $M(x) \to [\lnot S(x,x) \to S(\text {barber},x)]$.

Now, using the tautological implication : $(p \to q), (p \to r) \vDash p \to (r \leftrightarrow q)$, we conclude with :

75) $\forall a \ (M(a) \to [S(\text {barber},a) \leftrightarrow \lnot S(a,a)])$.

Finally, by $(\to \text I)$ from 13) and 65), discharging assumption 13):

76) $\lnot M(\text {barber}) \to \forall a \ (M(a) \to [S(\text {barber},a) \leftrightarrow \lnot S(a,a)])$.

Finally, from 12) and 76) by $(\leftrightarrow \text I)$ :

80) $\forall a \ (M(a) \to [S(\text {barber},a) \leftrightarrow \lnot S(a,a)]) \leftrightarrow \lnot M(\text {barber})$.

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See Barber paradox :

A modified version of the barber paradox is phrased nearly identically to the standard paradox, but omitting a detail that allows an answer to escape the paradox entirely. For example, the puzzle can be stated as occurring in a small town whose barber claims: I shave all and only the men in our town who do not shave themselves. This version identifies the sex of the clients, but omits the sex of the barber, so a simple solution is that the barber is a woman. The barber's claim applies to only "men in our town," so there is no paradox if the barber is a woman. Such a variation is not considered to be a paradox at all: the true barber paradox requires the contradiction arising from the situation where the barber's claim applies to himself.

The "original" version of the paradox is expressed in FOL as :

$(\exists x) \ [{\text{man}}(x)\wedge (\forall y)({\text{man}}(y)\rightarrow ({\text{shaves}}(x,y)\leftrightarrow \neg {\text{shaves}}(y,y)))]$.

In this form, the predicate $\text{man}$ must be read as "human" and not as "male", and thus we cannot assume that $(\exists x) \ \lnot {\text{man}}(x)$.

In the modified (non-paradoxical) version, we assume that the "universe" $U$ is made of "men" and "women"; i.e. $W=U \setminus M$ and thus, assuming $x \notin M$ is equivalent to say : $x \in W$.