For instance,
Is $\neg (((p \land q) \lor r) \land s)$ equivalent to $((\neg p \lor \neg q) \land \neg r) \lor \neg s$?
For instance,
Is $\neg (((p \land q) \lor r) \land s)$ equivalent to $((\neg p \lor \neg q) \land \neg r) \lor \neg s$?
Yes. I think you could prove this by induction. You might also note that once your fully parenthesize the intended formula, if you do things step by step, the negation moves to the parentheses first if any exist, and then moves to the letters and switches just one operation from AND to OR or the other way around. You could also note that negation consists of an isomorphism between ({T, F}, AND) and ({T, F}, AND') where " AND' " indicates OR, which implies that as long as you switch all values for the variables accordingly, and switch the first operation with its correlate OR, any given formula will behave exactly the same once the first operation has gotten rewritten as the second. Negation ensures that you've switched all values accordingly, so you can use De Morgan's laws this way.