Consider there are two tribes living on the Island: Knights and knaves. Knights always tell truth while Knaves always tell lie. Let we encounter two random people A and B, upon asking a question to ‘A’, A says “If B is Knight then I am a Knave”. What can we conclude about person A and B?
(A) A is a knight and B is a knave
(B) A is knave and B is knight
(C) Both A and B are Knight
(D) Both A and B are knave
My Analysis is
The given implication is
if B is a Knight then I am a Knave be denoted as
p: B is a knight
q: I am a knave (Means A is a knave)
$p\rightarrow q$ is our given implication
(A) A is a knight and B is a knave
If A is knight, then we can take the given implication as said by A in it's original form since Knights always tell truth but then if B is a knave, then the antecedent of the implication becomes false,there making the implication true so A can be a knave or knight both.
(B) A is knave and B is Knight
If A is knave, then whatever A said need to be complement to get original result since knaves always lie.
So complement of $p\rightarrow q$ is $p \land \sim q$
which is false as A is knave which makes $\sim q$ false.
So this cannot be our answer.
(C)Both A and B are knight-This will make our implication false. So ruled out.
(D)Both A and B are knave.
So we again need to reverse statement of A and it is $p \land \sim q$
but since B is knight, $p \land \sim q$ becomes false.
So only possible answer is (A).
Is my analysis correct with the answer?
Best Answer
If $A$ is a knave, then the statement $A$ gave is false, which implies that $B$ is a knight and $A$ is a knight is true. This is a contradiction to $A$ being a knave. Therefore, $A$ must be a knight. Then, the statement $A$ gave must be true.
Suppose $B$ were a knight. Then, by $A$'s true statement, it must be that $A$ is a knave, which is a contradiction. Therefore, $B$ must be a knave.
This gives one possible outcome that does not generate contradictions. $A$ is a knight and $B$ is a knave. That is satisfied by (A).