Consider the universal set

U = {a, b, c, d, e, f}

and the subsets

A = {a, b, d}, B = {b, d, e, f}, C = {c, f}

Write down each of the following:

A ∩ (B U C)

(B’ ∩ C)’ - A

Guest May 13, 2014

#1**+8 **

B∪C is the set of all the items in B and C put together. {b, c, d, e, f}

A∩(B∪C) is the set of items that are common to A and to B∪C. In this case that is {b, d}

B` is everything in the universal set that is not in B, namely {a, c}

B`∩C is the set of items common to B` and C, namely {c}

(B`∩C)` is everything in the universal set not in B`∩C. So {a, b, d, e, f}

(B`∩C)`-A is the set of items left when you take those in A away from (B`∩C)`, so you are left with {e, f}

Alan
May 13, 2014

#1**+8 **

Best Answer

B∪C is the set of all the items in B and C put together. {b, c, d, e, f}

A∩(B∪C) is the set of items that are common to A and to B∪C. In this case that is {b, d}

B` is everything in the universal set that is not in B, namely {a, c}

B`∩C is the set of items common to B` and C, namely {c}

(B`∩C)` is everything in the universal set not in B`∩C. So {a, b, d, e, f}

(B`∩C)`-A is the set of items left when you take those in A away from (B`∩C)`, so you are left with {e, f}

Alan
May 13, 2014

#2**+5 **

Thank you Alan, these are the results I came up with as well and they match. Thanks for making it clear.

Guest May 13, 2014

#3**+5 **

Let me see how much of this I remember.......!!

The first one is A ∩ (B U C), which, by Distributive Laws can be writtten as:

(A ∩ B) U (A ∩ C) =

({a, b, d} ∩ {b, d, e, f }) U { } =

{b, d } U { } =

{b, d} {By, defnition, the empty set is a subset of every set}

The second one is (B’ ∩ C)’ - A ... which, by using De Morgan's Laws can be written as

(B U C') - A =

({b, d, e, f} U {a, b, d, e }) - {a, b, d } =

({a, b, d, e, f} - {a, b, d}) = {e, f }

I believe that's it........but if someone else can check my answer , it would be great !!

CPhill
May 13, 2014