# To Mock A Mockingbird: Two barbers (logic puzzle)

I suspect an error in the solution given to a logic problem in the book To Mock a Mockingbird (Chapter 3 Problem 2).

Problem

Suppose I told you that the town contains two barbers, Arturo and Roberto, and that Arturo shaves all and only those inhabitants who shave Roberto, and Roberto shaves all and only those inhabitants who don't shave Arturo. Does this lead to a paradox?

Solution

No, this is no paradox. It could be that Roberto shaves himself, Arturo shaves Roberto, Arturo doesn't shave himself, and Roberto doesn't shave Arturo. The other X's in the town don't really matter; indeed, Arturo and Roberto could just as well be the town's only inhabitants.

The problem I have with this answer is that "Arturo doesn't shave himself" and "Roberto doesn't shave Arturo" while the problem states that "Roberto shaves all and only those inhabitants who don't shave Arturo". Therefore shouldn't Roberto also shave Arturo (invalidating the answer)? Or am I missing something?

• Can only one person shave someone else? I mean, is there a uniquely defined person that shaves a determined someone else? Or can there be more than one? – Patrick Da Silva Sep 1 '12 at 4:27
• Since Arturo shaved Roberto, wouldn't he have to shave himself? – Alex Becker Sep 1 '12 at 4:28
• @Alex Good point. The solution given doesn't seem to make much sense at all. – Alex Jasmin Sep 1 '12 at 4:34
• @PatrickDaSilva Having this constraint would make some sense I guess. But as far as I can tell it's not specified. – Alex Jasmin Sep 1 '12 at 5:16

The stated conditions are that for any inhabitant $X$, $AsX = XsR$ and $RsX = \lnot XsA$, where $PsQ$ denotes that $P$ shaves $Q$. Assuming Arturo and Roberto are the town's only inhabitants, there are four variables in the situation, namely $AsA$, $AsR$, $RsA$, and $RsR$. The given conditions then reduce to \begin{align} AsA &= AsR, \\ AsR &= RsR, \\ RsA &= \lnot AsA, \\ RsR &= \lnot RsA, \end{align} or in other words, $$\begin{matrix}AsA & \!\!=\!\! & AsR \\ \lVert & & \lVert \\ \lnot RsA & \!\!=\!\! & RsR\end{matrix}$$ which clearly has the two solutions above.