Let A, B, C, and D be sets. Prove or disprove the following: $(A ∩ B) ∪ (C ∩ D)= (A ∩ D) ∪ (C ∩ B)$

Question

Let A, B, C, and D be sets. Prove or disprove the following:

    (A ∩ B) ∪ (C ∩ D)= (A ∩ D) ∪ (C ∩ B)

I am just wondering can i simply prove it using a membership table ( seems to easy ) or do i have to use setbuilder notation?

Thank you!

Answer 1

Let $$A=B=\{1,2,3,4,5\}$$ and $$C=D=\{6,7,8,9,10\}$$ We have $$(A ∩ B) ∪ (C ∩ D)= \{1,2,3,4,5,6,7,8,9,10\}$$

while $$ (A ∩ D) ∪ (C ∩ B) =\emptyset $$

Answer 2

This is false. Take A = {1,2,3} B = {3,4,5} C = {7,8,9} D = {9,10,11} LHS = {3,9} RHS = phi Hence disproved. Hope it helps:)

Answer 3

Well it's sometimes true. Example, let all of the sets be $\{1\}$.

To show it doesn't always hold we could consider $A,B=\{1\}$ and $C,D=\{0\}$. Then $A\cap B=\{1\}$ and $C\cap D=\{0\}$, so $(A\cap B)\cup(C\cap D)=\{0,1\}$. Notice however that $A\cap D=C\cap B=\emptyset$, therefore $(A\cap D)\cup(C\cap B)=\emptyset$. Since $\emptyset\not=\{0,1\}$ we have a counterexample.