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While reading Problem Solving Strategies - Crossing the River with Dogs and Other Mathematical Adventures by Ken Johnson and Ted Herr, I came across a problem I was not sure how to solve.

One of the five members of a company's board of directors was suspected of sleeping during a board meeting. It was known that only one board member had actually slept, but no one (except the five members) knew who it was. The company vice president questioned the members and they made the following statements:

Davis: The snoozer was either Rawls or Charlton

Rawls: Neither Vongy nor I was asleep.

Charlton: Both Rawls and Davis are lying.

Bobbins: Only one of the Rawls or Davis is telling the truth.

Vongy: Bobbins is a liar.

When the board chairperson (she was not questioned) was consulted, she said that three of the board members always tell the truth and two of them always lie. Who slept in the meeting?

Perhaps I am simply glossing over one facet of the information given that I need to solve it. It seems like a simple problem on its hinge, so I am a bit flustered I am not able to solve it!

If you could give me a run-down of the logic used to solve it, that would be lovely.

Note: this is NOT homework - simply for my own benefit.

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This is as much about mathematical logic as rock music is about geology. – Asaf Karagila Mar 27 '12 at 20:24
Are you assuming that the board chairperson is telling the truth? – Graphth Mar 27 '12 at 20:41
@AsafKaragila I removed the logic tag; I knew it wasn't formal mathematical logic. – Joe Mar 27 '12 at 20:52
@Graphth Yes, since she is not part of the five members and the truth or lie tellers only consist of the five members. – Joe Mar 27 '12 at 20:53
@Asaf Karagila: I dont know if I can agree. This kind of problem would probably appeal to a person that likes mathematical logic. Rock musicians, on the other hand are not known to be into geology. – Adam Nov 19 '13 at 20:59
up vote 4 down vote accepted

If Charlton is telling the truth, that means that both Davis and Rawls are the liars; in particular, Vongy is telling the truth, which would mean that Bobbins is also a liar, contradicting the fact that there are only two liars and there three truth-tellers.

So Charlton is lying. That means that at most one of Davis and Rawls are lying.

If Bobbins is telling the truth, then the second liar must be one of Rawls and Davis, which would again mean that Vongy is telling the truth, making Bobbins a third liar; this is impossible, so Bobbins is lying.

So we now know that the two liars are Charlton and Bobbins, and the remaining three are truth-tellers.

Therefore, the snoozer is either Rawls or Charlton (since Davis is telling the truth), but cannot be Rawls (since Rawls is telling the truth).

So Charlton is the one who fell asleep.

Alternatively: Bobbins and Vongy cannot both be telling the truth, so at least one of them is a liar. In particular, Charlton cannot be telling the truth, since that would give at least three liars. So the two liars are among Charlton, Bobbins, and Vongy, hence both Davis and Rawls are telling the truth; this suffices to establish that Charlton is the one who fell asleep.

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  1. Suppose Charlton told the truth, that means Davis & Rawls lied. This means that rest of others told the truth. But this contradicts the statement made by Vongy. Hence Charlton lied.
  2. Suppose Vongy lied, that means Bobbins told the truth, but then either one of Rawls or David must lied. Hence we got 3 persons who lied. Again a contradiction. Hence Vongy told the truth. And hence Bobbin lied.
  3. Hence Davis & Rawls told the truth. This means Charlton slept in meeting.
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In spite of Asaf Karagila's comment on this question, let me try to make this into an exercise in mathematical logic.

Before we start, note that $\;(P \equiv Q) \equiv R\;$ is equivalent to $\;P \equiv (Q \equiv R)\;$: $\;\equiv\;$ is associative, and I will leave out the parentheses. Also, note that $$ (0) \;\;\; \text{exactly 1 of } P,Q,R \text{ is true} \;\equiv\; (P \equiv Q \equiv R) \land \lnot (P \land Q \land R) $$

So we are given distinct $\;d,r,c,b,v\;$ (for Davis, etc.), one of which is equal to $\;s\;$ (for sleeper). Writing $\;T(x)\;$ for "$\;x\;$ always tells the truth", we can formalize "$\;x\;$ says $\;\phi\;$" as $\;T(x) \equiv \phi\;$. Their statements then can be formalized as \begin{align} (1) \;\;\; T(d) & \;\equiv\; s=r \;\lor\; s=c \\ (2) \;\;\; T(r) & \;\equiv\; s \ne v \;\land\; s \ne r \\ (3) \;\;\; T(c) & \;\equiv\; \lnot T(r) \;\land\; \lnot T(d) \\ (4) \;\;\; T(b) & \;\equiv\; (T(r) \;\not\equiv\; T(d)) \\ (5) \;\;\; T(v) & \;\equiv\; \lnot T(b) \\ \end{align} Finally, we are given that exactly 3 of $\;T(d), T(r), T(c), T(b), T(v)\;$ are true.

Since $(5)$ says that $\;T(b)\;$ and $\;T(v)\;$ are each others' opposites, we can use this to go from "exactly 3 of ..." to "exactly 1 of ...": \begin{align} & \text{exactly 3 of } T(d), T(r), T(c), T(b), T(v) \text{ are true} \\ \equiv & \;\;\;\;\;\text{"$\;T(b)\;$ and $\;T(v)\;$ are each others' opposites"} \\ & \text{exactly 2 of } T(d), T(r), T(c) \text{ are true} \\ \equiv & \;\;\;\;\;\text{"negation"} \\ & \text{exactly 1 of } \lnot T(d), \lnot T(r), \lnot T(c) \text{ is true} \\ \Rightarrow & \;\;\;\;\;\text{"use $(0)$"} \\ & \lnot T(d) \;\equiv\; \lnot T(r) \;\equiv\; \lnot T(c) \\ \equiv & \;\;\;\;\;\text{"use $(3)$; DeMorgan"} \\ & \lnot T(d) \;\equiv\; \lnot T(r) \;\equiv\; T(r) \;\lor\; T(d) \\ \equiv & \;\;\;\;\;\text{"cancel two negations"} \\ & T(d) \;\equiv\; T(r) \;\equiv\; T(r) \;\lor\; T(d) \\ \equiv & \;\;\;\;\;\text{"use (what Dijkstra et al. call) the golden rule"} \\ & T(r) \;\land\; T(d) \\ \equiv & \;\;\;\;\;\text{"use $(2)$ and $(1)$"} \\ & s \ne v \;\land\; s \ne r \;\land\; (s=r \;\lor\; s=c) \\ \equiv & \;\;\;\;\;\text{"use $\;s \ne r\;$ in right hand side; simplify"} \\ & s \ne v \;\land\; s \ne r \;\land\; s=c \\ \equiv & \;\;\;\;\;\text{"simplify using $\;c \ne v\;$ and $\;c \ne r\;$"} \\ & s=c \\ \end{align} Therefore Charlton slept.

Note that we did not use Bobbins's statement $(4)$ at all.

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I should probably clarify that pretty much every problem that can be solved mathematically can be posed as an exercise in mathematical logic about deduction rules or whatever. It doesn't mean that every problem that has to be solved mathematically is a problem in mathematical logic. – Asaf Karagila Nov 19 '13 at 21:20



1st scenario. DRC TRUTH BV LIE


2nd. Scenario RCB TRUTH DV LIE


3rd Scenario. CBV TRUTH DR LIE


4th Scenario BVD TRUTH RC LIE


5th Scenario VDR TRUTH CB LIE

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