# truth table of equivalence relation

I have found the following problem in a book:

Test the validity of the following argument: If 6 is not even, then 5 is not prime But 6 is even Therefore, 5 is prime

I have made a truth table by transforming it into logical propositions like: ~p -> ~q and p then q

and I end up with the values T/F/T/T

according to the answer of the book it is a fallacy. But logically seems that it is not, because 6 is even and 5 is a prime.

Is there any way to prove that this is a tautology?

## 2 Answers

It is true that 5 is prime, of course.

However the argument does not work to show that this is the case. In order for the argument to be logically valid, it has to work no matter what the claims it speaks about. Logically, it is exactly the same as the argument:

If Bob is not sad, then Alice has not broken up with him. But Bob is sad. Therefore Alice has broken up with him.

Logically that's the same argument -- we've just substituted "Bob is sad" for "6 is even" and "Alice broke up with him" for "5 is prime". Such a substitution cannot change the logical validity of the argument.

However, in this case it should be clearer that the argument doesn't hold water. Alice didn't break up with Bob; he's sad for entirely different reasons (his best friend just died).

Let $p$ := "6 is even", $q$ := "5 is prime". Suppose $(\lnot p \rightarrow \lnot q)$. Further suppose that $p$. Does $q$ follow from those two assumptions? The answer is no. Consider the assignment {p := $\top$, q := $\bot$}. Since $p$ is true, $(\lnot p \rightarrow \lnot q)$ is true because it's equivalent to $(p \lor \lnot q)$, and premise $p$ is also true, but $q$ is not. So the argument is invalid.