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This problem involves two people. Person A and person B. They can either always tell the truth, or always lie. When asked, person A replies that: "At least one of us is a liar." Does person A and B tell the truth or do they lie?

What I have so far is that person A can either tell the truth or lie. If he tells the truth, then his statement can be expressed by "A$\lor$B", if he lies then it must be given that they are both speaking the truth, which contradicts that he would be a liar. So person A must be speaking the truth.

But if person A is speaking the truth, then person B could be either lying or speaking the truth. Both would still make person A tell the truth.

So how can I figure out whether person B is speaking the truth or not?

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  • $\begingroup$ Discrete mathematics ??? $\endgroup$ – creative Sep 19 '14 at 8:59
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    $\begingroup$ If at least one of $A,B$ is a liar and $A$ is not a liar, then ... $\endgroup$ – Hagen von Eitzen Sep 19 '14 at 9:05
  • $\begingroup$ Jesus.. You are right Hagen... Been thinking too much about this I guess.. Thank you. $\endgroup$ – Attaque Sep 19 '14 at 9:12
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$$\begin{array}{cccc} &&&\text{A says}\\ \text{A lies} &\text{B lies}&\text{A or B lies}&\text{A or B lies}\\ a&b&a\lor b&(a\lor b)\oplus a\\ F& F& F& \color{red}F\\ T& F& T& \color{red}F\\ F& T& T& \color{green}T\\ T& T& T& \color{red}F\\ \end{array}$$

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  • $\begingroup$ A or B lies is not $ a \lor b $ but $ \lnot a \lor \lnot b $ $\endgroup$ – Willemien Sep 19 '14 at 13:31
  • $\begingroup$ @Willemien: no, $a$ (resp. $b$) denotes "A (B) lies"; the $\lor$ is non-exclusive. "A or B lies" is $a\lor b$. There is no doubt when you look at the truth values. The last column is the third one xored with "A lies", i.e. the truth value is changed when A lies. $\endgroup$ – Yves Daoust Sep 19 '14 at 13:48
  • $\begingroup$ you are right sorry, I mis interpreted the $\oplus $ (Xor) (I do prefer the translation as the expressing $(\lnot a \lor \lnot b ) \equiv a $ so with using equivalences) $\endgroup$ – Willemien Sep 19 '14 at 13:49
  • $\begingroup$ To my great surprise, this way of computing with logical expression makes it very straightforward. For instance "A says he lies" is $a\oplus a$, indeed a false statement. $\endgroup$ – Yves Daoust Sep 19 '14 at 13:57
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If B would speak the truth then both of them would speak the truth - a contradiction to A's statement that at least one of them is a liar. So B will obviously not speak the truth.

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I find it easier to always suppose that A, B tell the truth in any of these riddles with two types of people (that either always tells truth, or always lies). This is my approach:

$a$: A tells the truth

$b$: B tells the truth

Proposition "at least one of us is a liar" is $\lnot a \lor \lnot b$. Person A tells us this proposition, so $a$ and $\lnot a \lor \lnot b$ must have the same truth value (if A tells the truth, his proposition is true and if he lies, his proposition is false). Therefore, the answer to the problem is if there is one or more $T$ values of the proposition $a\leftrightarrow (\lnot a \lor \lnot b$). Let's see:

\begin{array}{cccccc} a& b& \color{grey}{\lnot a}& \color{grey}{\lnot b}& \lnot a \lor \lnot b& a\leftrightarrow (\lnot a \lor \lnot b)\\ F& F& \color{grey}T& \color{grey}T& T& \color{red}F\\ T& F& \color{grey}F& \color{grey}T& T& \color{green}T\\ F& T& \color{grey}T& \color{grey}F& T& \color{red}F\\ T& T& \color{grey}F& \color{grey}F& F& \color{red}F\\ \end{array}

So, we see that $a\leftrightarrow (\lnot a \lor \lnot b)$ is true only when A tells the truth and B tells lies.

The same approach can be applied to any of this kind of problems.

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