Is x not in A or B AND not in C?

[kernelpenguin]

I'm trying to prove something small with set theory and since I'm new to it, I've run into a problem. I can't understand what the following means exactly and how to proceed further. Or where the mistake is, if there is one. I think there is, because it seems... freaky.

$$x\notin\left(\left(\left(A\cup B\right)\setminus\left(A\cap B\right)\right)\cap C\right)$$
$$x\notin\left(\left(A\cup B\right)\setminus\left(A\cap B\right)\right)\wedge x\notin C$$
$$\left(x\notin\left(A\cup B\right)\wedge x\in\left(A\cap B\right)\right)\wedge x\notin C$$
$$\left(\left(\left(x\notin A\right)\vee\left(x\notin B\right)\right)\wedge\left(\left(x\in A\right)\wedge\left(x\in B\right)\right)\right)\wedge x\notin C$$

I'd post the entire thing of which this is a small part of, but that's my homework and I don't want to get into the habit of having other people do my homework for me. Plus I want to learn how and why it works, not just do it.

 

[matt grime]

consider this

x is not an element of UnV
x is not an element of U AND x is not an element of V

you've said those two statements are equivalent (I think, since you've not actually said what your deductions are from line to line). find a counter example to show this is false.

negation switches conjunction and disjunction, or union and intersection.

similar observations hold for the other steps in your reasoning.

 

[tacman]

The expression x\notin\left(\left(\left(A\cup B\right)\setminus\left(A\cap B\right)\right)\cap C\right) means "x is not in the intersection of the set (A union B) and the complement of the intersection of A and B, and also not in the set C." This can also be written as x\notin\left(\left(A\cup B\right)\setminus\left(A\cap B\right)\right)\wedge x\notin C, which means "x is not in the set (A union B) minus the intersection of A and B, and also not in the set C."

To further simplify this expression, we can use the distributive property of set operations. This allows us to break up the intersection into two separate parts: (A union B) and the complement of (A and B). This can be written as \left(x\notin\left(A\cup B\right)\wedge x\notin\left(A^c\cap B^c\right)\right)\wedge x\notin C. We can then use De Morgan's laws to simplify the expression even further: \left(\left(x\notin A\right)\wedge\left(x\notin B\right)\right)\wedge\left(\left(x\in A^c\right)\vee\left(x\in B^c\right)\right)\wedge x\notin C.

This final expression can be interpreted as "x is not in A and x is not in B, and either x is not in A's complement or x is not in B's complement, and also not in the set C." This aligns with the original statement of "x not in A or B AND not in C," which means that x is not in either A or B, and also not in the set C.

To prove this statement using set theory, you can use proof by contradiction. Assume that x is in either A or B (or both), and also in the set C. This would contradict the original statement that x is not in A or B AND not in C. Therefore, our assumption must be false and the original statement holds true.

In summary, the expression x\notin\left(\left(\left(A\cup B\right)\setminus\left(A\cap