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From "Discrete mathematics and its applications", a book by Kenneth H. Rosen, chapter 1.1 exercise 57, goes as:

  1. A says "I am a knave or B is a knight" and B says nothing.

Knight always tell the truth and knaves always lie. We are to determine of which type are A and B.

Assuming that,

p: A is a knight

q: B is a knight

Can I arrive to the answer (provided by the book), which is "A is a knight and B is a knight" using a truth table? And if not, then how?

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    $\begingroup$ I think you are supposed to say knights always tell the truth and knaves never do. $\endgroup$
    – Henry
    Oct 1, 2016 at 15:23
  • $\begingroup$ Yes, a truth table might be helpful. $\endgroup$
    – Mark Viola
    Oct 1, 2016 at 15:24
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    $\begingroup$ What happens if you try a truth table? Clearly (Knight, Knight) works as A's statement would be true and B has not said anything. What about the other three possibilities? $\endgroup$
    – Henry
    Oct 1, 2016 at 15:25

3 Answers 3

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You don't even have to use a truth table. If A says that he is a knave or B is a knight, he cannot be a knave because if he was, then his statement would be true, even though knaves always tell lies. Now let's assume A is a knight. Then, since he isn't a knave, the second part of the statement, that B is a knight, must be true. So A and B are both knights.

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Yes, you can do it with a four line truth table. For each line assess the truth of $A$'s statement based on whether each is a knight. Then see if the truth of the statement matches whether $A$ lies or tells the truth.

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  • $\begingroup$ Can you provide the table? My problem is mostly with notation. I don't know how to properly devise the table. $\endgroup$ Oct 1, 2016 at 15:32
  • $\begingroup$ You just make the four lines for the four choices of true and false for $p$ and $q$. $A$'s statement is $\lnot p \wedge q$, so that is the third column. The statement needs to be true when $p$ is true and false when $p$ is false, because of the definition that knights always tell the truth and knaves always lie. $\endgroup$ Oct 1, 2016 at 15:36
  • $\begingroup$ That should be $\vee$, not $\wedge$ in the last comment $\endgroup$ Oct 1, 2016 at 15:57
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A truth table would help.

In that table, there are four possible truths; (i) A and B are knights, (ii) A is a knight and B is a knave, (iii) A is a Knave and B is a knight, and (iv) A and B are knaves.

Let's proceed with testing whether (i) is true or false. If both A and B are knights, then the statement by A that "I am either a knave or B is a knight" cannot be refuted.

Next, let's test whether (ii) is true or false. If A is a knight and B is a knave, then the statement by A that "I am either a knave or B is a knight" cannot be true. By hypothesis, A is a knight and is telling the truth. So, A is not a knave and the statement by A must mean that B is a knight. Inasmuch as B is a knave by hypothesis, we have a contradiction. Therefore, the hypothesis that A is a knight and B is a knave is false.

Can you continue from here?

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