# Prove $(A \cap B) \cup (A \cap B')= A$ using Set Identities

I recently started a Discrete Mathematics course in college and I am having some difficulties with one of the homework questions. I need to learn this, so please guide me through at least two steps to get the ball rolling.

The question reads: Show that if $A$ and $B$ are sets, then: $(A \cap B) \cup (A \cap B')=A$

We are supposed to use set identities. I had a question prior, but it was simple: $(A \cap B \cap C)' = A'\cup B' \cup C'$ - Which would be one of De Morgan's laws.

I am at a loss. I have been reading the textbook and tried looking up some videos, but I am not sure exactly where to start. Any help you can provide, will be greatly appreciated!

Thanks, Kei

• I'm not sure what you're asking. Can you clarify what specifically you're confused with? Also I highly reccomend taking a look at "How to Prove It" as a supplementary text. It's a fantastic introduction to this subject. – lordoftheshadows Jan 23 '17 at 18:58
• Use distributivity! LHS is equal to $A \cap (B \cup B')$ – Crostul Jan 23 '17 at 18:58
• @lordoftheshadows: I wasn't sure on the first two steps to prove that the left side is indeed equal to the right side? If I am explaining it right. Our professor showed us an example in class such as this: (A ∪ (BnC)) = (C ∪ B) ∩ A = A ∩ (B ∩ C) - De Morgan Law = A ∩ (B u C) De Morgan Law = (B u C) ∩ A = Commutative Law = (C u B) ∩ A = Commutative Law So I am looking for the first two steps, I see that @Crostul had posted the first step. Thank you! I am going to see if I can take it from there. I do really appreciate the help! – Kei U. Jan 23 '17 at 19:00
• I have edited the question for formatting and some grammatical things for you, @KeiU. We will all see it when it is approved. Good luck! – The Count Jan 23 '17 at 19:03
• Thank you @TheCount! I appreciate that, I really should have looked over the question better, sorry! @Crostul: Do I distribute (A ∩ B) U (A ∩ B`) together? – Kei U. Jan 23 '17 at 19:05

The idea is to achieve get close $B$ and $B^c$. Then we use distributive property: $(A\cap B)\cup(A\cap B^c)=A\cap (B\cup B^c)=A\cap X=A$, with $X$ the universe

• Hello, I really appreciate your answer you provided! I think where I was confused was, being able to reverse an identity. I was using the set identity, though not using it in reverse if that makes sense. This is what I did: I used Distributive law, Absorption law, and then Identity law. – Kei U. Jan 23 '17 at 19:24
• It would not let me edit the above comment. The last sentence should be a question. I wanted to ensure that I did use the right laws above, as you shown in your post Julio Maldonado Henriquez. Thank you! – Kei U. Jan 23 '17 at 19:33

Consider an element of $A$ - either it is in $B$, or it isn't, and thus is in the complement of $B$. Thus $A \subset (A \cap B)$ $\cup$ $(A \cap B')$. Now, try to argue on your own that the reverse "inclusion" holds: that we have $A \supset (A \cap B)$ $\cup$ $(A \cap B')$.

You can use identities such as $(A\cap B) \cup (A \cap C) = A \cap (B \cup C)$ to get $(A \cap B) \cup (A \cap B') = A \cap (B \cup B') = A \cap U = A$.

But I prefer to think of what it is saying. $A \cap B$ means "everything in A and in B" and $A \cap B'$ means "everything that is in A that is not in B" and $(A\cap B) \cup (A \cap B')$ means "every thing that is in A and B combined with everything that is not in B". Is there a logical reason that "everything in A and B combined with everything in A and not in B" would be "A"?

Well, I hope it should be obvious. Everything in A is either in B or not in B so combining the items of A that are not in B with those that are should give you all the items in A.

So the best way to express that idea directly would be:

$A = A \cap U = A \cap (B \cup B') = (A \cap B) \cup (A\cap B')$. Or if that's a little too abstract, I rather like to do an element by element proof:

Let $x \in A$ either $x \in B$ or $x \in B'$. If $x \in B$ then $x \in A \cap B$. If $x \in B'$ then $x \in A \cap B'$. Either way $x \in A \cap B$ or $x \in A\cap B'$ so $x \in (A \cap B) \cup (A\cap B)$. So $A \subseteq (A\cap B) \cup (A \cap B)$. Likewise if $y \in (A \cap B) \cup (A\cap B)$ then either $y \in (A \cap B) \subset A$ or $y \in (A\cap B') \subset A$. Either way, $y \in A$ so $(A\cap B)\cup (A\cap B') \subseteq A$.

$A \subseteq (A\cap B) \cup (A \cap B)$ and $(A\cap B)\cup (A\cap B') \subseteq A$, so (A\cap B)\cup (A\cap B') = A$. A fourth way is the big guns. Let$x \in U$now one of four things might happen: 1)$x \in A$and$x \in B$. Then$x \in A$and$x \in A\cap B$and$x \in (A \cap B) \cup (A \cap B')$. 2)$x \in A$and$x \not \in B$. Then$x \in A$and$x \in B'$and$x \in A \cap B'$and$x \in (A \cap B) \cup (A \cap B')$3)$x \not \in A$and$x \in B$. Then$x \not \in A$and$x \not \in A \cap B$and$x \not \in A \cap B'$so$x \not \in (A \cap B) \cup (A \cap B')$. 4)$x \not \in A$and$x \not \in B$. Then$x \not \in A$and$x \not \in A \cap B$and$x \not \in A \cap B'$so$x \not \in (A \cap B) \cup (A \cap B')$. Looking at the four cases we see$x \in A \iff x \in (A \cap B) \cup (A \cap B')$. Thus$A$and$(A \cap B) \cup (A \cap B')$have precisely the same elements and neither has any element the other doesn't. In other words,$A = (A \cap B) \cup (A \cap B')\$.