Who tells the truth?

Three children play a riddle game. Each one either always tells the truth or always lies. The first says: the second is lying. The second says: the third is lying. The third says: the first two are both lying.

Three children $A$, $B$, $C$, each always truthful ($t=1$) or always lying ($t=0$). $A$ says '$B$ lies'; $B$ says '$C$ lies'; $C$ says '$A$ and $B$ both lie'.

$t_A,t_B,t_C\in \{0,1\}$; $t_A=1\iff t_B=0$; $t_B=1\iff t_C=0$; $t_C=1\iff (t_A=0\wedge t_B=0)$.

Suppose the third is truthful. Then the first two both lie. But the first said the second lies, and since the first is a liar, that is false, so the second is truthful. That clashes with the second being a liar. So the third is not truthful.

Try $t_C=1$. Then $t_A=0$ and $t_B=0$. But $t_A=0$ means '$B$ lies' is false, so $t_B=1$, a contradiction. Hence $t_C=0$.

$t_C=1\Rightarrow (t_A,t_B)=(0,0)$, but $t_A=0\Rightarrow t_B=1$: ✗. So $t_C=0$.

So the third lies. The second said the third lies, which is true, so the second is truthful. The first said the second lies, which is false, so the first lies.

With $t_C=0$, the claim of $B$, '$C$ lies', is true, so $t_B=1$. The claim of $A$, '$B$ lies', is false, so $t_A=0$.

$t_C=0\Rightarrow t_B=1\Rightarrow t_A=0$.

The answer: only the second child tells the truth. Check the third: the claim that both lie is false because the second is truthful, which fits a liar. The surprise: the one in the middle was the honest one, and the one with the boldest claim was lying.

The answer is $(t_A,t_B,t_C)=(0,1,0)$: only $B$ tells the truth. The surprise is that the one in the middle was honest.

$(t_A,t_B,t_C)=(0,1,0)$ ✓.