I need to find the negation of a statement on my homework, specifically problem 19 of secton 3.2 in Discrete Mathematics with Applications by Susanna Epp. The problem is as follows:
\begin{align} \neg\left(\forall n\in \mathbb{Z},\:n\:\text{is > prime}\:\longrightarrow\:n\:\text{is odd or}\: n=2.\right)\tag{1} \end{align}
In this case, do I have to negate the if-then, or is it simply as I have written it already in which I have negated either side: \begin{align} \Longleftrightarrow \exists n\in\mathbb{Z},\:n\:\text{is not prime}\longrightarrow\:n\:\text{is not or}\:n\neq 2.\tag{2} \end{align} My gut is telling me that I have it wrong, which is why I would appreciate a second opinion. Once again, I realize that it is not your job to do my homework. Tips and hints to get better along the way would be much appreciated, however.
No, $(2)$ is wrong. Consider the simpler case :
$$ \neg(\forall n\in\mathbb Z)(P(n)\to D(n)). $$
Since $P(n)\to D(n)$ is equivalent to $\neg P(n)\vee D(n)$ we get :
$$ \begin{align} \neg(\forall n\in\mathbb Z)(P(n)\to D(n))&\iff\neg(\forall n\in\mathbb Z)(\neg P(n)\vee D(n))\\ &\iff(\exists n\in\mathbb Z)\neg(\neg P(n)\vee D(n)). \end{align} $$
Can you proceed from there?