Contrapositve issues I'm currently reading logic and have some issues with the contrapositve of this statement:
The textbook says:

If $n>0$ and $4n−1$ is prime, then $n$ is odd

I understand the contrapositive thinking (I think!) $P\to Q$ becomes with this logic $\neg Q\to\neg P$
Therefore in my mind this statement should be:

If n is even, then $0>n$ and $4n-1$ is not prime.

But this does not even make sense. Are there any rules I'm missing?
 A: You are missing a couple of rules. Let me write your sentences with parentheses:

If (($n>0$) and ($4n-1$ is prime)), then ($n$ is odd)$.

Ok, now we know what $P$ and $Q$ are more clearly. So, we can reverse them and get

If (not ($n$ is odd)), then (not (($n>0$) and ($4n-1$ is prime)))

This should be clear, since I just wrote "not" in front of both sentences. Now let's examine the two parts:


*

*not ($n$ is odd). Well, that one's easy. $n$ is not odd if and only if it is even, so this changes to "$n$ is even" and we are done.

*not (($n>0$) and ($4n-1$ is prime)). Hmm. This is a composite statement, two statements joined with and. You probably know the rule for that: $\neg(P\land Q)$ is the same as $\neg P\lor \neg Q$ (DeMorgan's law), so the original expression becomes

((not $n>0$) OR (not ($4n-1$ is prime))) 

which then simpliy simplifies down to 

($n\leq 0$ OR ($4n-1$ is composite)).

So the total reversal is equivalent to 

If $n$ is even, then $n\leq 0$ or $4n-1$ is composite.

A: Yes. $\lnot$($n>0$ and $4n-1$ is prime) is ($n\leq 0$ or $4n-1$ is composite), not ($n\leq 0$ and $4n-1$ is composite).
Note that both the original statement and the contrapositive statement is false, as evidenced by, for instance, $n=2$ (at least they agree, which is the main point).
