# Island of knights and knaves

This question is about an island of knights and knaves, where knights always speak the truth and knaves always lie. You encounter two people A and B. Determine, if possible, what each of them are if they address you as follows: A says "B is a knave" and B says "A is a knight".

Here's my solution:

Let p be the proposition 'A is a knight' and let q be the proposition 'B is a knight.'

     p  q
F   F      x A cannot speak the truth
F   T      x B cannot lie
T   F      x B cannot speak the truth
T   T      x A cannot lie


Therefore, all possibilities are eliminated, so this means that the identity of A and B cannot be deduced.

-

You have that $$A\implies \neg B\\\neg A\implies B\\B\implies A$$ but we can stop right there because $A\implies \neg B$ means that the contrapositive is also true, so that $B\implies\neg A$, which contradicts the last statement. This world cannot exist, which is a little bit different than saying that the identies cannot be determined, because there is no possible set of identies for both A and B.

-

EDIT: Sorry I did this wrong :(

A says "B is a knave" and B says "A is a knight".

Let $a$ be the statement "A is knight" and $b$ be the statement "B is knight". Then, based on what they said, we have either both knights, a knight and b knave, ... etc. $$abb'a+ab'b'a'+a'bba+a'b'ba'=0+0+0+0=0$$ There is no solution, as you claimed.

-

If B is a knave, then A is a knight but then B cannot say A is a knight. If B is a knight, then A is a knave but then B cannot say A is a knight. This is an impossible situation by analyzing the two possible scenarios. No matter what, B cannot say that A is a knight given what A says.

-

As you can see in other answers of mine, I would formalize this using $\newcommand{\says}[2]{#1\text{ says }\unicode{x201C}#2\unicode{x201D}} \newcommand{\cansay}[2]{#1\text{ can say }\unicode{x201C}#2\unicode{x201D}}$\begin{align} \tag{0} & \says{x}{P} \;\Rightarrow\; (T(x) \equiv P) \\ \end{align} where $\;T(x)\;$ stands for "$\;x\;$ speaks the truth" or "$\;x\;$ is a knight".

For this specific puzzle, using $(0)$ twice, the statements of $\;A\;$ and $\;B\;$ immediately imply \begin{align} \tag{1} T(A) & \equiv \lnot T(B) \\ \tag{2} T(B) & \equiv T(A) \\ \end{align} From these it immediately follows that $T(B) \equiv \lnot T(B)$, which is a contradiction. So this specific world cannot exist.

(This is different from your incorrect conclusion that we cannot know anything about $\;A\;$ and $\;B\;$. That would be the case if, e.g., they both said, "I am a knight".)

-