How come $x \notin N \cup P$ becomes $(x \notin N \text{ and }x \notin P)$ and $x \notin N \cap P$ becomes $(x \notin N \text{ or }x \notin P)$?

Question

The expressions are part of the proofs of the expressions $M\setminus (N \cup P) = (M \setminus N) \cap ( (M\setminus P) \text{ and } M\setminus (N ∩ P) = (M \setminus N) \cup ( (M\setminus P)$. So when I was trying to prove the first one I thought $\{x \mid x \in M \text{ and } x \notin N \cup P\}$ would be expressed as $\{x \mid x ∈ M \text{ and } x \notin N \text{ or } x ∈ M \text{ and } x \notin P\}$ as the "and" from the intersection would distribute over the "or" from the union. Yet to prove the equality, it should be $\{x \mid x \in M \text{ and } x \notin N \text{ and } x \in M \text{ and } x \notin P\}$ so that the equality holds until last step. For the second example just change the middle "and" with "or". Again I thought it should have been "and" as "and"s have associativity. I can simply memorize it and use it this way whenever a similar proof comes. Yet, I would be more than happy if someone tells me the reason behind this expressions, so that I see what's going on.

Edit: My question has been identified as a possible duplicate of Proving DeMorgan's Theorem. First of all, my problem is not about proving the entire theorem. I actually have the proof in front of me. It is about a proof step I couldn't internalize. In the linked question there is limited mention of that specific step except for the fact that it is taken for granted as in the proof in front of me now (so again after reading the question and answers, I am at where I was before). I want an explanation of that step specifically and my question is about that step not a whole proof. Moreover, although I have to admit that my question depicts a step on proving DeMorgan's theorem, in this case M is not fixed as a universe and difference relation is not absolute but relative. As such the equalities can be thought of displaying properties of set difference as well. After I understand these relative difference properties (including a third one) and the specific steps I mentioned in my question, I want to continue with DeMorgan. As such I would appreciate help on a topic that is a step in moving to a more advanced topic and a question on a minor detail rather than the proof of a whole theorem.

Answer 1

As you said, for a point x to be a point in M\ (N U P), it has to satisfy:

• x ∈ M, and
• x ∉ N U P

I believe the first one doesn't give you too much trouble. To put the second condition in words, it is:

x is in neither N nor P

because N U P includes the points in N and P. So the conditions for x becomes:

• x ∈ M, and
• x ∉ N, and
• x ∉ P

In other word:

• x ∈ M\N, and
• x ∈ M\P

which is the desired answer we are looking for.

I think the key thing here is not to simply apply some "rules" that seem true, but to think about what those mathematical symbols actually mean. In this case, I think you might interpret the union symbol "U" as distributive, while as I demonstrated, it is the other way around.

Hope that helps :)