Which sentence among the three sentences is the lie?

Question

1. The second statement is the lie or the third statement is the lie.
2. This statement is a truth, or the last statement and the second statement cannot both be truths.
3. The first statement is the lie and the second statement is a truth, or this statement is the lie.

There can only be one lie, and the rest of the sentences have to be true. My guess is #3. based on the truth table TTL, TLT, LTT, where 3 contradicts #2. But, I wonder if I have missed anything?

Answer 1

Let $p,q$, and $r$ stand respectively for '(1) is true', '(2) is true', and '(3) is true'. Then the three statements become (1) $\lnot q\lor\lnot r$, (2) $q\lor\lnot(q\land r)$, and (3) $(\lnot p\land q)\lor\lnot r$. Now you can make the full truth table, as shown below:

$$\begin{array}{c} &&&p\equiv&q\equiv&r\equiv\\ p&q&r&\lnot q\lor\lnot r&q\lor\lnot(q\land r)&(\lnot p\land q)\lor\lnot r\\ \hline \text{T}&\text{T}&\text{F}&\color{blue}{\underline{\color{red}{\textbf{F}}}}&\text{T}&\color{blue}{\underline{\color{red}{\textbf{T}}}}\\ \text{T}&\text{F}&\text{T}&\text{T}&\text{F}&\color{blue}{\underline{\color{red}{\textbf{F}}}}\\ \text{F}&\text{T}&\text{T}&\text{F}&\text{T}&\text{T} \end{array}$$

The red entries show the places where the truth values of the statements are inconsistent. As you can see, only the third assignment of truth values is free of inconsistency, so (1) must be the lie.

In fact $q\lor\lnot(q\land r)$ is logically equivalent to $q\lor(\lnot q\lor\lnot r)$ by De Morgan's laws, and this is a tautology, as is easily seen by rewriting it as $(q\lor\lnot q)\lor r$. Thus, (2) must be a true statement. If (1) were also true, (3) would be the lie, but that would imply that (2) was also a lie, which is impossible.

Answer 2

Let the three statements be $A$, $B$ and $C$ respectively. Then we can summarize the statements as:

\begin{align} (A &\Leftrightarrow [\neg B \vee \neg C]) \\ \wedge \ (B &\Leftrightarrow [B \vee \neg (B \wedge C)]) \\ \wedge \ (C &\Leftrightarrow [(\neg A \wedge B) \vee \neg C]). \end{align}

Writing out all implications, we get

\begin{align} (A &\Rightarrow [\neg B \vee \neg C]) \\ \wedge \ (A &\Leftarrow [\neg B \vee \neg C]) \\ \wedge \ (B &\Rightarrow [B \vee \neg (B \wedge C)]) \\ \wedge \ (B &\Leftarrow [B \vee \neg (B \wedge C)]) \\ \wedge \ (C &\Rightarrow [(\neg A \wedge B) \vee \neg C]) \\ \wedge \ (C &\Leftarrow [(\neg A \wedge B) \vee \neg C]). \end{align}

Reducing $P \Rightarrow Q$ to $\neg P \vee Q$, we get

\begin{align} &(\neg A \vee [\neg B \vee \neg C]) \\ \wedge& (A \vee \neg [\neg B \vee \neg C]) \\ \wedge& (\neg B \vee [B \vee \neg (B \wedge C)]) \\ \wedge& (B \vee \neg [B \vee \neg (B \wedge C)]) \\ \wedge& (\neg C \vee [(\neg A \wedge B) \vee \neg C]) \\ \wedge& (C \vee \neg [(\neg A \wedge B) \vee \neg C]). \end{align}

Reducing each term individually, we get

\begin{align} &(\neg A \vee \neg B \vee \neg C) \\ \wedge& (A \vee (B \wedge C)) \\ \wedge&\text{true} \\ \wedge& B \\ \wedge& (\neg C \vee (\neg A \wedge B)) \\ \wedge& C. \end{align}

Or, in short,

$$B \wedge C \wedge (A \vee (B \wedge C)) \wedge (\neg A \vee \neg B \vee \neg C) \wedge (\neg C \vee (\neg A \wedge B)).$$

This statement can only hold if $B$ and $C$ both hold, in which case the third, fourth and fifth term reduce to $\text{true}$, $\neg A$ and $\neg A$. So the statements are consistent if and only if

$$\color{blue}{\neg A \wedge B \wedge C}.$$

So even if multiple statements could be a lie, the only consistent scenario is that statement $1$ is a lie, and $2$ and $3$ are truths.

Answer 3

#3 does not contradict #2, because "This statement is a truth" in #2 and "The first statement is the lie and the second statement is a truth" in #3 can both hold. So #1 must be the only one that is a lie.