# De Morgan's Laws Question

1. Sep 21, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
Prove $A\setminus(B\cap C)=(A\setminus B)\cup(A\setminus C)$

2. Relevant equations
3. The attempt at a solution

We will show that every element in $A\setminus(B\cap C)$ is contained in $(A\setminus B)$ or $(A\setminus C)$. If $x\in A\setminus (B\cup C)$ then $x\in A$, $x\notin B$ and $x\notin C$, thus $x\in(A\setminus B)$ and $x\in(A\setminus C)$ now I'm assuming the next step is to say that this shows that $x\in(A\setminus B)\cup(A\setminus C)$ but I'm confused as to why we this is a union instead of an intersection. If $x$ is not in B and C then why do we use and/or instead of just and? I am obviously interpreting this incorrectly so hopefully someone can clarify.

2. Sep 21, 2015

### Potatochip911

Okay so since it's possible that x is in either B or C we have $x\in A$, $x\in (B\cup C)$?

3. Sep 21, 2015

### verty

It might be helpful to "de morgan" each side. For example, the left side becomes $\overline{A} \cup (B \cap C)$.

4. Sep 21, 2015

### Potatochip911

So I should also try to show that $(A\setminus B)\cup(A\setminus C)=A\setminus(B\cap C)$? Also I haven't yet learnt what $\overline{A}$ represents.

5. Sep 21, 2015

### verty

Oh right, in that case do it the way you were doing it but be careful not to get the symbols wrong. I thought you had to apply De Morgan's laws somehow, so the obvious thing was to take the complement of each side.

6. Sep 21, 2015

### Ray Vickson

If $E^c$ denotes the complement ("not-$E$) of a set $E$, then $D \backslash E = D \cap E^c$, essentially by definition. Do you know how one of the DeMorgan laws relates $(B \cap C)^c$ to $B^c$ and $C^c$?

Last edited: Sep 22, 2015
7. Sep 21, 2015

### Potatochip911

So for the other side: If $x\in (A\setminus B)\cup (A\setminus C)$ then $x\in A$ and $x\in (B\cup C)$
In my statistics class we had $P(B\cap C)^c=P(B^c)+P(C^c)-P(B\cup C)$

8. Sep 22, 2015

### Ray Vickson

This is irrelevant: the question has nothing to do with statistics or probability. It is just a question about set manipulation.

9. Sep 22, 2015

### Potatochip911

Hmm I think I have solved this question, if $x\in A\setminus(B\cap C)$ then $x\in A$ and $x\notin (B\cap C)$, it follows that $x\notin B$ or $x\notin C$ which leads to $x\in (A\setminus B)\cup (A\setminus C)$

10. Sep 22, 2015

### Dick

I think you have solved it as well.

11. Sep 22, 2015

### Potatochip911

I would really appreciate it if someone could look over this one as well, I'm supposed to prove that $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$, In order to prove this I must show that that x can go from an element of the LHS to the RHS (Not really sure how to word this suggestions would be appreciated) $x\in (A\cup B)\cap (A\cup C)$, Starting from the LHS I have that if $x\in A\cup (B\cap C)$ then $x\in A$ and $x\in (B\cap C)$, this leads to two cases, the first case is when $x\in A$, this implies that $x\in (A\cup B)\cap (A\cup C)$ since if $x\in A$ this statement holds true. The second case is when $x\in (B\cap C)$, this also implies the RHS since if $x\in (B\cap C)$ then the statement $(A\cup B)\cap (A\cup C)$ also holds, therefore $A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$. I'm curious as to whether or not this is sufficient or if I have to manipulate the expressions more.

12. Sep 23, 2015

### Ray Vickson

You have gone one way (proving that $A\setminus(B \cap C) \subset (A\setminus B) \cup (A\setminus C)$. You also need to establish the reverse.

13. Sep 23, 2015

### Potatochip911

Is it possible that the proof changes when going the other way around or is it always the same steps just proving them in reverse?

14. Sep 23, 2015

You tell me.