I know that this is a dumb question in comparison to the problems in the video but when you have a problem such as ~(p∨~q), when simplified, does that problem's quantifier immediately flip because of the "not" at the beginning of the parentheses? If so, why?
There is no quantifier in the statement ~(p v ~q), which would simplify to ~p & q. If there were a quantifier, whether it flips or not depends whether it is "inside" the negation or not. That is, "for all x, ~(p v ~q)" would stay a for all statement, but "~(for all x (p v ~q))" would become a exists statement as you pass the negation through.
There is an easy way to settle this and be sure: truth tables. Write out the truth table for an implication, say A -> B. In the final column you will have 3 T's and 1 F. The F will be in the row in which A is T and B is F. This makes sense: Say someone tells you that if you give them money, then they will give you a car. The way (and only way) they would be lying is if you give them the money, but at the same time, they don't give you a car. Now we want the NEGATION of A -> B. So we want to claim that A -> B is false. This can only happen if A is true and B is false, which is to say A ^ not B. Perhaps you are confusing this with the fact that the implication (not the negation) A -> B is logically equivalent to notA v B?
