# Proving that $B \cap ((A \cup B) \cap (B' \cap A')') = B$ using set algebra

Problem: Use set operation laws to prove the following set equality, and clearly indicate which law(s) you use in each step: $$B ∩ ((A ∪ B) ∩ (B' ∩ A')') = B.$$

Answer: \begin{align} B ∩ ((A ∪ B) ∩ (B' ∩ A')') &= B\\ &= B ∩ ((A ∪ B) ∩ B'' ∩ A'') &&\text{DeMorgan}\\ &= B ∩ (A ∪ B) ∩ B ∪ A &&\text{Double complement}\\ &= B ∩ (B ∪ A) ∩ B ∪ A &&\text{Commutativity}\\ &= B ∩ (B ∪ A) &&\text{Absorption}\\ &= B &&\text{Absorption}\\ &= B &&\text{Identity}\\ \end{align}

Is this correct?

• Your use of De Morgan's law on step 1 was incorrect but somehow corrected on the next line. (B' ∩ A')' = B'' ∩ A'' makes no sense – Dleep Aug 9 '15 at 23:13
• In the line with De Morgen's Law, change $B'' \cap A''$ to $B'' \cup A''$. I suspect a typo since you were correct beyond this point. – user134593 Aug 9 '15 at 23:13
• Can you add parentheses for the $B \cup A$ terms? They should be calculated first before the outer $\cap$ terms. – peterwhy Aug 9 '15 at 23:15

## 2 Answers

The way you have written it up is very odd. You start with what you are trying to prove and expand from there. Here is how I would write up your proof (feel free to comment if a step does not make sense): \begin{align} B\cap[(A\cup B)\cap(B^C\cap A^C)^C]&= B\cap[(A\cup B)\cap(B\cup A)]\\[0.5em] &= B\cap(A\cup B)\\[0.5em] &= (B\cap A)\cup(B\cap B)\\[0.5em] &=(B\cap A)\cup B\\[0.5em] &= B. \end{align} Can you see where DeMorgan's law was used? How about distributivity and the like?

By De Morgan's law: $A' \cup B' = (A \cap B)'$ So: $(A' \cap B')' = ((A \cup B)')' = A \cup B$. Thus: $((A \cup B) \cap (A' \cap B')') = ((A \cup B) \cap (A \cup B)) = A \cup B$. But $B \subset A \cup B \implies B \cap (A \cup B) = B$.

I might have misplaced some $\cap$ or $\cup$ because I got dizzy writing "cap" and "cup".