Formal negation of $((p\rightarrow q) \vee (q \leftrightarrow r)) \rightarrow q$ Can someone give me an outline for how I can negate the following?

$((p\rightarrow q) \vee (q \leftrightarrow r)) \rightarrow q$
You want to work from the inside out. Call the inside of the outer parentheses $s$, so $s=(p\to q)\vee (q\leftrightarrow r)$.

Then, the desired negation is $\lnot (s\to q)$. How do we evaluate that? Well, what actually is $\to$? $s \to q$ means that if $s$ is true, then $q$ is true. If $s$ is false, it doesn't say anything. Thus, $(s\to q)$ is only false if $s$ is true and $q$ is false, or in other words, $\lnot (s\to q)=s \wedge (\lnot q)$

Another way to see this is with truth tables:

$$\begin{array}{c|c|c|c}s&amp; q &amp; s\to q&amp;\lnot(s\to q)\\
T&amp;T&amp;T&amp;F\\
T&amp;F&amp;F&amp;T\\
F&amp;T&amp;T&amp;F\\
F&amp;F&amp;T&amp;F\end{array}$$

Expanding $s$ back out again gives $$((p\to q)\vee (q\leftrightarrow r))\wedge(\lnot q)$$
¬((p → q) ∨ (q ↔ r)) → ¬ q
¬p → ¬q ∧ ¬q ↔ ¬r → ¬ q