In a village there are two types of people. Type F who always lie and Type T people who always say the truth. X says "according to Y, I always lie". Assume both X and Y belong to same village. Which one of them is possible ?
1)Both X and Y belong to type T
2)Both X and Y belong to type F
I am able to figure out that both cannot belong to type T. But I am not able to figure out the 2nd statement.
For these 'truth-tellers and liars' (often presented as 'Knights and Knaves') puzzles, you can often use some simple algebra.
In general, you use a propositional letter $A$ for some person $A$ being a truth-teller (knights), then you know that they are a truth-teller if and only if what they are saying is true.
Thus, for example, if $A$ says "$B$ is a liar", then what $A$ says can be symbolized as $\neg B$, and that is true if and only if $A$ is a truth-teller, so you would get $A \leftrightarrow \neg B$
For this particular problem, you have $X$ saying something about what $Y$ would say ... which really just means that you first symbolize the latter part. That is, the "$Y$ would say that I ($X$) am a liar" becomes $Y \leftrightarrow \neg X$, and since $X$ is saying all that, you get:
$$X \leftrightarrow (Y \leftrightarrow \neg X)$$
which can be simplified as follows:
$$X \leftrightarrow (Y \leftrightarrow \neg X) \Leftrightarrow \text{ (Commutativity of } \leftrightarrow \text{)}$$
$$X \leftrightarrow (\neg X \leftrightarrow Y) \Leftrightarrow \text{ (Associativity of } \leftrightarrow \text{)}$$
$$(X \leftrightarrow \neg X) \leftrightarrow Y \Leftrightarrow $$
$$\bot \leftrightarrow Y \Leftrightarrow $$
$$\neg Y $$
So, we know that $Y$ is telling lies, while $X$ can be of any type. Hence, 2 is the only viable option out of 1 and 2
P.s. Algebra rules!