Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A=(B\cap C) \cup (D\cap E)$ be a given set. I am looking for a different way to write this. I guess it is somehow possible to "reorder" this by using De Morgan's relations. Unfortunately, I am not successful. Does somebody see how one can write this differently?

share|improve this question
This is definitely not a duplicate of #402054 as it pertains to four different sets. –  Lord_Farin Jun 29 '13 at 11:05
I don't see a way to simplify, I think it's basically the simplest way to write down this set. –  Henno Brandsma Jun 29 '13 at 11:24
@Henno: I don't see the word "simplify" in this page prior to your comment. –  Asaf Karagila Jun 29 '13 at 11:26
This is a good question and does not deserve to be closed. –  goblin Jun 29 '13 at 11:53

2 Answers 2

up vote 3 down vote accepted

As pointed out by Asaf Karagila, there is no option to rewrite your expression in a simpler or more elegant way.

This can be demonstrated graphically:

The following is a Venn diagram for your case:

enter image description here

And here is the Karnaugh-Veitch map:

enter image description here

The two intersecting minterm blocks cannot be replaced by fewer or simpler minterm blocks.

share|improve this answer

You can't use DeMorgan laws here, but you can use the distributivity of $\cap$ over $\cup$ and vice versa. We have:

$$(B\cap C)\cup(D\cap E)=((B\cap C)\cap D)\cup((B\cap C)\cap E)$$

Using associativity of $\cap$ we get:

$$(B\cap C\cap D)\cup(B\cap C\cap E)$$

It's not simpler than the original form, though. Note that you can do the same thing by distributing $D\cap E$ over the other pair, so we get the following: $$(B\cap C)\cup(D\cap E)=(B\cap C\cap D)\cup(B\cap C\cap E)=(B\cap D\cap E)\cup(C\cap D\cap E)$$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.