Question 1101556: Please help me solve this! Anything helps!
One of the six members of a company's board of directors was suspected if sleeping during a board meeting. It was known that only one board member had actually slept, but no one (except the six 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: Both Rawls and Davis are telling the truth
Edwards: The snoozer is lying
When the board chairperson (she was not questioned and known to tell the truth) was consulted, she said that three of the board members always tell the truth and three of them always lie. She also said that Davis and Edwards can be counted on to do the opposite things- when one tells the truth, then the other one lies. Who slept in the meeting?
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! 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.
Another analysis
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.
And a mathematical logical analysis
Before we start, note that (P≡Q)≡R(P≡Q)≡R is equivalent to P≡(Q≡R)P≡(Q≡R): ≡≡ is associative, and I will leave out the parentheses. Also, note that
(0)exactly 1 of P,Q,R is true≡(P≡Q≡R)∧¬(P∧Q∧R)
(0)exactly 1 of P,Q,R is true≡(P≡Q≡R)∧¬(P∧Q∧R)
So we are given distinct d,r,c,b,vd,r,c,b,v (for Davis, etc.), one of which is equal to ss (for sleeper). Writing T(x)T(x) for "xx always tells the truth", we can formalize "xx says ϕϕ" as T(x)≡ϕT(x)≡ϕ. Their statements then can be formalized as
(1)T(d)(2)T(r)(3)T(c)(4)T(b)(5)T(v)≡s=r∨s=c≡s≠v∧s≠r≡¬T(r)∧¬T(d)≡(T(r)≢T(d))≡¬T(b)
(1)T(d)≡s=r∨s=c(2)T(r)≡s≠v∧s≠r(3)T(c)≡¬T(r)∧¬T(d)(4)T(b)≡(T(r)≢T(d))(5)T(v)≡¬T(b)
Finally, we are given that exactly 3 of T(d),T(r),T(c),T(b),T(v)T(d),T(r),T(c),T(b),T(v) are true.
Since (5)(5) says that T(b)T(b) and T(v)T(v) are each others' opposites, we can use this to go from "exactly 3 of ..." to "exactly 1 of ...":
≡≡⇒≡≡≡≡≡≡exactly 3 of T(d),T(r),T(c),T(b),T(v) are true"T(b) and T(v) are each others' opposites"exactly 2 of T(d),T(r),T(c) are true"negation"exactly 1 of ¬T(d),¬T(r),¬T(c) is true"use (0)"¬T(d)≡¬T(r)≡¬T(c)"use (3); DeMorgan"¬T(d)≡¬T(r)≡T(r)∨T(d)"cancel two negations"T(d)≡T(r)≡T(r)∨T(d)"use (what Dijkstra et al. call) the golden rule"T(r)∧T(d)"use (2) and (1)"s≠v∧s≠r∧(s=r∨s=c)"use s≠r in right hand side; simplify"s≠v∧s≠r∧s=c"simplify using c≠v and c≠r"s=c
exactly 3 of T(d),T(r),T(c),T(b),T(v) are true≡"T(b) and T(v) are each others' opposites"exactly 2 of T(d),T(r),T(c) are true≡"negation"exactly 1 of ¬T(d),¬T(r),¬T(c) is true⇒"use (0)"¬T(d)≡¬T(r)≡¬T(c)≡"use (3); DeMorgan"¬T(d)≡¬T(r)≡T(r)∨T(d)≡"cancel two negations"T(d)≡T(r)≡T(r)∨T(d)≡"use (what Dijkstra et al. call) the golden rule"T(r)∧T(d)≡"use (2) and (1)"s≠v∧s≠r∧(s=r∨s=c)≡"use s≠r in right hand side; simplify"s≠v∧s≠r∧s=c≡"simplify using c≠v and c≠r"s=c
Therefore Charlton slept.
All three solutions are from
https://math.stackexchange.com/questions/125197/variation-of-a-who-is-lying-question
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