# What it means for an object $x$ to be an element in the set?

does this solution is correct for : $A \setminus (B\ \cup\ C)?$. First, I transformed it to: $x \in A\ \land\ \lnot(x\ \in\ (B\ \cup\ C)) - \text{Definition of } \setminus\\ x \in A\ \land\ (\lnot(x\ \in B\ \lor\ x \in C)) - \text{Definition of}\ \cup\\ x \in A\ \land(\lnot(x\ \in B)\ \land\ \lnot(x\ \in C)) - \text{De Morgan}\\ x \in A\ \land (x \notin B\ \land\ x \notin C) - \text{Definition of }\notin\\ x \in A\ \cap\ (x \notin B\ \land x \notin C) - \text{Definition of } \cap\\ \text{ and ended up with}:\ x \in A\ \cap\ (B^c\ \cap\ C^c)$

• Why would $A \cap (x \not\in B \land x\not\in C)$ imply $A \cap ( B \cap C)$ ? – Felipe Dilho Dec 8 '17 at 5:02
• "$(x \not \in B \land x \not \in C)$" is not a set. It is a statement of fact. "$x \in A \cap (x\not \in B \land x \not \in C)$" is a meaningless arrangement of symbols. It's like saying "Harry Potter is a wizard intersected with Harry Potter is not a rabbit and Harry Potter doesn't eat ice cream" – fleablood Dec 8 '17 at 7:37
• I am pretty new to this, so any point of view is welcomed. However, a bit more clarity of how to solve this from your point of view will be really helpful for me to understand it better. Kind regards. – Dimitar Dec 8 '17 at 7:51

No, mistake in the first line$$x \in A\ \land\ \lnot(x\ \in\ (B\ \color{red}\cap\ C))$$

Notice that we have

$$A \setminus D = A \cap D^c$$

Let $D = B \cap C$ and apply De Morgan's Law.

• Edit after the question is edited:

Now mistake at last line, note that $x \notin B$ means $x \in B^\color{red}c$.

• Sorry, I have a typo on the condition line (it's too early here). – Dimitar Dec 8 '17 at 5:01
• answer edited for mistake at last line. – Siong Thye Goh Dec 8 '17 at 5:05
• Sorry for the delay. So in this case if I understand correctly: x belongs of the intersection of A and the intersection of the complements of B and C? – Dimitar Dec 8 '17 at 7:06
• $x$ belongs to the intersection of $A$ and the intersection of the (complements of $B$) and the (complements of $C$). – Siong Thye Goh Dec 8 '17 at 7:09
• Many thanks for the help. – Dimitar Dec 8 '17 at 7:11