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
+33614 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}
+33614 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}
Thank you Alan, these are the results I came up with as well and they match. Thanks for making it clear.
+128406 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 !!
