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A very special island is inhabited only by knights and knaves. Knights always tell the truth, and knaves always lie. You meet four inhabitants: Bozo, Marge, Bart and Zed.

  • Bozo says," Bart and Zed are both knights".
  • Marge tells you that both Bart is a knight and Zed is a knave
  • Bart tells you," Neither Marge nor Zed are knaves".
  • Zed says that neither Bozo nor Marge are knaves.

Can you determine who is a knight and who is a knave?

I am having extreme difficulty with this can anyone help me? I assume is starts like this.

So

$Bo\equiv(Ba\land Ze)$

$Ma≡(Ba\land \lnot Ze)$

$Ba\equiv(Ma\lor Ze)$

$Ze≡(Bo\lor Ma)$

Where

$Bo$= Bozo is a knight

$Ma$= Marge is a knight

$Ba$= Bart is a knight

$Ze$= Zed is a knight

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    @Don: What I meant is, you suggested a duplicate, and it appears at least superficially that it poses a different question, so I was asking you to explain how you're mapping the problems onto each other in order to conclude that they're in fact the same problem even though it doesn't look like that at first sight.2012-12-04

4 Answers 4

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First, your symbolic translations of Bart’s and Zed’s statements are incorrect. Bart actually said $\text{Ma}\land\text{Ze}\;,$ and Zed said $\text{Bo}\land\text{Ma}\;.$

A quick way to solve it is to suppose that Bart is a knight. Then he’s telling the truth, so Marge and Zed are also knights. But that’s impossible, because Marge said that Zed is a knave: if she’s a knight, she’s telling the truth, and Zed isn’t knight. Thus, Bart cannot be a knight and must therefore be a knave. Can you finish it from there?

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    I agree with your correction. If you make a of truth table, you will find that only one combination is true. (For a nice truth table generator, see http://mathdl.maa.org/images/upload_library/47/mcclung/index.html )2012-12-04
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Bozo says, "Bart and Zed are both knights."

Marge tells you that both Bart is a knight and Zed is a knave.

Bart tells you, "Neither Marge nor Zed are knaves."

Zed says that neither Bozo nor Marge are knaves.

Marge and Bart contradict each other. If Marge is telling the truth, Zed lies, and Bart tells the truth. Bart, however, Zed and Marge tell the truth. Thus, Marge and Bart are Knaves. Therefore, Zed is a knave (Because Marge said he was a knight). Bozo says Bart and Zed are knights, which has been established false. Thus all are knaves.

The way I typically solve these problems is to look for contradictions.

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If Bozo is a knight, that means that Bart and Zed are knights, and if Zed is a knight, then Bozo and Marge are knights. If Marge is a knight, then Zed is a knave, which contradicts what we have, so Bozo is a knave.

If Marge is a knight, then Bart is a knight and Zed is a knave. This means that (from Bart) that Zed is not a knave, so Marge is a knave.

If Bart is a knight then that means Marge is a knight, which we already established to be false, so Bart is a knave.

If Zed is a knight, then that means Bozo is a knight, which we already establish to be false, so Zed is a knave.

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You have chosen a good formalization here: with the corrections from Brian M. Scott's answer, you are given that \begin{align} (0) \;\;\; & Bo \equiv Ba \land Ze \\ (1) \;\;\; & Ma \equiv Ba \land \lnot Ze \\ (2) \;\;\; & Ba \equiv Ma \land Ze \\ (3) \;\;\; & Ze \equiv Bo \land Ma \\ \end{align} Now, looking at the shape of these formulae, we note that $\;Bo\;$'s $(0)$ and $\;Ma\;$'s $(1)$ have a similar structure where we may expect to get a contradiction, and these $\;Bo, Ma\;$ are used symmetrically in $(3)$. Therefore we calculate \begin{align} & Ze \\ \equiv & \;\;\;\;\;\text{"by (3)"} \\ & Bo \land Ma \\ \equiv & \;\;\;\;\;\text{"by (0); by (1)"} \\ & Ba \land Ze \;\land\; Ba \land \lnot Ze \\ \equiv & \;\;\;\;\;\text{"logic: contradiction; simplify"} \\ & \text{false} \\ \end{align} Using this in $(0)$ and $(2)$ immediately leads to $\;Bo \equiv \text{false}\;$ and $\;Ba \equiv \text{false}\;$, respectively. And plugging that last conclusion into $(1)$ gives us $\;Ma \equiv \text{false}\;$.

Therefore all are knaves.