# 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?

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.