(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove the following identities involving complementation of sets.

1. A ∩ [(A ∩ B)^{C}] = A - B

2. A ∩ [(A ∩ B^{C})^{C}] = A ∩ B

3. (A ∪ B) ∩ A^{C}= B - A

2. Relevant equations

A^{C}= A complement. If A is contained in some understood universal set U, then the complement of A is the set A^{C}= {x ∈ U: x ∉ A}.

A - B = { x ∈ A and x ∉ B}

B - A = { x∈ B and x ∉ A}

A ∪ B = {x ∈ A or x ∈ B}

A ∩ B = { x ∈ A and x ∈ B}

3. The attempt at a solution

Since I am dealing with set equality and trying to show when one set equals another, I have been trying to show that x ∈ the first set if and only if x ∈ the second set in each of these cases.

I am not actually going to put this in legit proof form here but I will just summarize my steps for you and know that I plan on doing it better later once I figure it out entirely.

For part 1, I started off by choosing an arbitrary x in A ∩ [(A ∩ B)^{C}]. Then, if x is in this set, x ∈ A and x ∉ A ∩ B. This means that x ∈ A and x ∉ A or x ∉ B. Well I am trying to show that this is equal to A - B, where x ∈ A and x ∉ B so I have the parts I need, however, I don't know what to do with the part where it says x ∉ A.

For part 2, I started off by choosing an arbitrary x in A ∩ [(A ∩ B^{C/SUP])C/SUP]]. Then, if x is in this set, x∈ A and x ∉ A ∩ BC/SUP]. This means that x ∈ A and x ∉ A or x ∉ B^C. Once again, this means that x∈A and x∉A or x∈ B. Well I am trying to show that this set is equal to A ∩ B, where x ∈ A and x ∈ B. But, once again, I have that extra part in the middle where x ∉ A that I don't know what to do about. For part 3, I chose an arbitrary x in (A ∪ B) ∩ AC. Then, if x is in this set, x ∈ A or x ∈ B and x ∈ AC. Well, this means x ∈ A or x ∈ B and x ∉ A. This time, I have that other term in the front saying x ∈ A when I really need to get to B - A, where x∈ B and x ∉ A. Am I completely missing something here? Thanks.}

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# Homework Help: Proving set equality

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