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Just for the sake of reminding, the paradox is "The barber shaves all and only those who do not shave themselves". I found a Youtube channel by some professor named Frederic Schuller that contains a few set theory lectures (I recommend them by the way). In the first one he says at some point that this famous statement is a paradox and claims that whether or not it is true is undecidable. I had little background in propositional logic theory but having watched the video up until that point I felt like I could use what I had learnt from him to easily prove that the statement was false assuming every proposition to either be true or false (i.e. assuming from the start that undecidable propositions do not exist). I'm sure Mr Schuller must be right of course, but here's my work.

Let $A$ be the proposition "The barber shaves himself". Then the initial statement merely says $(A \implies \neg A) \wedge (\neg A \implies A)$. If we agree that a proposition is always either true or false, then one of the statements on either side of the wedge is false (whichever one carries $\text{true} \implies \text{false}$, according to the binary truth table). Then the "and" operator represented by $\wedge$ yields false. And here I cannot see how I failed to prove that the Barber's "paradox" is false.

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    $\begingroup$ The paradox is in the truth value of $A$, not of the more complicated initial statement, as you call it. $A$ can be neither true nor false. $\endgroup$ – Arthur Mar 16 '17 at 14:18
  • $\begingroup$ See Barber paradox. $\endgroup$ – Mauro ALLEGRANZA Mar 16 '17 at 14:27
  • $\begingroup$ According to wikipedia it isn't a paradox $\endgroup$ – skyking Mar 16 '17 at 14:35
  • $\begingroup$ @MikhailKatz - I think that the OP is arguing that he proved that it is not the case that "the barber shaves all and only those who do not shave themselves." $\endgroup$ – Ben Blum-Smith Mar 16 '17 at 14:37
  • $\begingroup$ $(A \to \lnot A) \land (\lnot A \to A)$ is $(A \leftrightarrow \lnot A)$. See this post for the formal derivation. $\endgroup$ – Mauro ALLEGRANZA Mar 16 '17 at 14:38
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The statement (premise) is not a paradox, it's a lie - a false statement. If we let $xSy$ mean $x$ shaves $y$ and $b$ is the barber the statement becomes:

$$\forall x(bSx\Leftrightarrow \neg xSx)$$

this means especially $bSb\Leftrightarrow \neg bSb$ which is formally false or expressed mathematically we have that $\neg(\phi\Leftrightarrow\neg\phi)$.

Now the question if the barber shaves himself? Or not? Here comes the power of a false premise - given a false premise any conclusion is true. We can prove both that:

$$\forall x(bSx\Leftrightarrow \neg xSx)\vdash bSb$$ $$\forall x(bSx\Leftrightarrow \neg xSx)\vdash \neg bSb$$

That is the premise leads to a contradiction. But this isn't really a problem, because since the premise is false the consequence does not need to be true anyway - one of them may in fact be true and the other false (which is how we would want things to be).

Also note that forming a contradiction is what RAA is about. When using that we assume a premise and form a contradiction (a statement and it's negation being both conseqences). And from that we conclude that the premise is false (ie it's negation being proven).

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(A) The "Barber paradox" is not really a paradox, properly so called.

What we have here is a perfectly good proof by reductio that there can't exist someone in the village who shaves all and only those in the village who don't shave themselves. For suppose there is such a person, $B$. Then, by hypothesis, for all $y$, $B$ shaves $y$ if and only if $y$ doesn't shave $y$, where the variable $y$ runs over people in the village. So, in particular $B$ shaves $B$ if and only if $B$ doesn't shave $B$ -- contradiction!

Now, we have no antecedent reason to suppose there might be a barber who shaves all and only those who don't shave themselves. Hence it should be no particular surprise to learn that as a matter of simple logic that there can't be one.

And it is indeed a matter of simple logic. Generalizing, take any binary relation $R$ defined over the domain $U$ of widgets. Then this is a straightforward theorem of first-order logic: $$\neg\exists x\forall y(Rxy \leftrightarrow \neg Ryy)$$ with the quantifiers running over the domain $U$. Hence, whatever relation $R$ you take, there can't be a widget which is $R$-related to all and only those widgets which are not $R$-related to themselves. (Exercise, prove that formal wff in your favourite system of first-order logic!)

So far, there is nothing paradoxical going on. We can only speak really of a paradox or antinomy when a (seemingly) compelling proof clashes with other (seemingly) compelling ideas. And that isn't yet the case here.

(B) Suppose though -- to make the famous historical connection -- that we are thinking not about the shaving relation but about set-membership. Then we have this particular instance of our first-order theorem: $$\neg\exists x\forall y(x \in y \leftrightarrow y \notin y)$$ with the quantifier running over all sets. Hence there is no set containing all and only the normal sets (where a set is normal if and only if it doesn't contain itself.

But now note that unlike the Barber case, this result does clash with some assumptions that we might rather naturally have had about sets. For suppose we start off wedded to the ideas that (i) we can collect together all the $X$s into a set of $X$s, whatever $X$s might me -- so in particular, there is a super-collection $U$ that collects together all the sets (so $U$ is a set of all the sets), and (ii) for any given set $X$ and property $P$, there will be a subset of it containing just those members of $X$ with property $P$, and (iii) there is a perfectly good property of being a normal set, i.e. one which doesn't have itself as a member.

Then (i), (ii) and (iii) commit us to the existence of a set $Russell$, the subset of the super-collection $U$ which contains all and only the normal sets -- so $Russell$ is the set of all sets which are not members of themselves.

But as we've seen the $Russell$ set can't exist because of our simple logical theorem. Which shows that we can't hold (i), (ii) and (iii) together. And this traditionally is called a paradox -- though it is, strictly speaking, only paradoxical to the extent to which we previously might have been firmly wedded to (i), (ii) and (iii).

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The Barber Paradox is a straight-forward contradiction, and your logical analysis shows exactly that. Good job!

OK, but then what about the professor's claim?

The professor used the Barber's paradox to illustrate the self-reference or diagonalization method that is behind Godel's proof of the incompleteness of arithmetic. That is, at some point in Godel's proof, he introduces a sentence (the Godel sentence) $G$ that roughly says "I cannot be derived from the axioms". This $G$ feels very much like the statement about the barber, except where the barber sentence is a genuine contradiction, this $G$ is not a contradiction ... It is in fact true: If it were false, then the statement $G$ can be proven from the axioms, so assuming the axioms reflect basic truths, sentence $G$ would have to be true as well. So: If $G$ is false, then $G$ is true. So: $G$ is true! And so, $G$ is some truth that cannot be proven from the axioms. Which axioms? Any (recursive) set of axioms for arithmetic (each axiom set will have its own Godel sentence!) Meaning: there is no (recursive ... think: finite) set of axioms for arithmetic such that all arithmetical truths can be derived from it. And that is Godel's incompleteness result. Relatedly: there is no decision procedure for arithmetic: for any procedure that tries to decide whether some claim about arithmetic is true or false: if it is sound (i.e. It never getsthe wrong answer), then for some clais it cannot figure out whether it is true or false.

So ... yeah ... fascinating stuff, but we're far beyond any basic Barber's paradox/contradiction here!

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  • $\begingroup$ Here is the video : m.youtube.com/…. --- If the whole playlist comes up it's the 1st vid. I tried hard to find the spot in question but gave up I think it's like 1/3rd near the end $\endgroup$ – James Well Mar 17 '17 at 2:24
  • $\begingroup$ @JamesWell Got it, and found it! It's near the end, at about 1:33:00. OK, my intuition was basically correct; it was about the Godel results ... I'll add something to my answer. $\endgroup$ – Bram28 Mar 17 '17 at 2:49

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