Solving Each Other

Logic Level 2

$\begin{array}{|l|} \hline {\text{ The other statement is true. }} \\ {\text{ The other statement is false. }} \\ \hline \end{array}$

How many of the sentences above is/are true?

0 1 2 This is an impossible scenario

2 solutions

Kishore S. Shenoy
Oct 8, 2016

Relevant wiki: Truth-Tellers and Liars

Let's see, there are only two possibilities for the credibility of the statement - true or false.

Case 1 - Statement 1 is TRUE

So Statement 2 is TRUE
So Statement 1 is FALSE

Contradiction...

Case 2 - Statement 1 is FALSE

So Statement 2 is FALSE
So Statement 1 is TRUE

Contradiction...

Since only two possibilities are there and both are not possible, this is an IMPOSSIBLE SCENARIO.

But the question doesn't ask whether the scenario is true or false. It asks how many of the statements are true. That is, one of them is true, but we don't know which one!

David Lewis - 4 years, 8 months ago

Since Statement 1 can neither be true nor false, it is an impossible scenario.

Kishore S. Shenoy - 4 years, 8 months ago

Incorrect.

If statement one is true then statement two is true implying that statement one is false. This tells you that statement two is false. However, it's impossible that both statements are false because then that contradicts both statements.

If statement two is true, than statement one is false implying that statement two is false. Again, it's impossible that both statements are false because then that contradicts both statements.

Digital Gaming 227 - 2 years, 3 months ago

I'm sorry, I don't understand.

Kishore S. Shenoy - 2 years ago

Agnishom Chattopadhyay
Oct 7, 2016

Relevant wiki: Propositional Logic

Let's call the statements $p$ and $q$ for clarity. Now, we can express them as follows:

$p := Tq$
$q := T \neg p$

where $T$ is the truth predicate, defined as $Tx \equiv x$

Now, we are going to show that it is not possible to assign consistent classical truth values to $p$ and $q$

Suppose, $p$ is True . Then by $p$ , $q$ is True . But $q$ says, the negation of $p$ is True . Which means $p$ is False . This is a contradiction, since $p$ cannot be both True and False

On the other hand, if $p$ is False . Then, the negation of $p$ is True . So, the negation of $q \equiv \neg p$ is True as well. Which means $\neg \neg p$ is True or $p$ is True . This is a contradiction, since $p$ cannot be both True and False

Hence, there is no way to assign a classical truth value to $p$ and hence, no way for the system in general.

This is a well known paradox called the Liar paradox. One way to resolve this paradox is to introduce a third truth value, thus allowing a statement to be neither true, nor false.

Solving Each Other

2 solutions

0 pending reports