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Information about Homework 2 2-24

This is a detailed solution to problem 2.2.24 from Discrete Mathematics With Graph Theory, 3rd Ed, by Goodaire and Parmenter.

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So x ∈ A and x ∈ B c ∩ C . So x ∈ A and x ∈ B c and x ∈ C. So x ∈ A and x ∈ B. / So x ∈ A ∩ B c . Also, x ∈ A and x ∈ C. So x ∈ A ∩ C. So x ∈ (A ∩ B c )orx ∈ (A ∩ C) is true. (note that we know it is in both. But we want to show that it is in the union, which is an ”or” statement. This is one of those creative steps in realizing what you want versus what you have.) So x ∈ (A ∩ B c ) ∪ (A ∩ C), which is equivalent to (AB) ∪ (AC c ). ————————————————————————————————————This is the ﬁrst direction. It only took a little bit of out of the box thinking to get where we needed to go. Now let’s prove the other direction and see what happens. ————————————————————————————————————– Proof. (⇐) Let x ∈ (AB) ∪ (AC c ). c This means that x ∈ (A ∩ B c ) ∪ (A ∩ C c ). Simplifying, this means that x ∈ (A ∩ B c ) ∪ (A ∩ C). So x ∈ (A ∩ B c ) or x ∈ (A ∩ C). Now we have two cases. Case 1: Assume x ∈ A ∩ B c . So x ∈ A and x ∈ B c . x ∈ A ∩ Bc. x ∈ A and x ∈ B c . x ∈ A and x ∈ B. x ∈ A and x ∈ B ∩ C c . (Think about why this is true and why we can just add the C c here.) x ∈ A and x ∈ BC. x ∈ A and x ∈ (BC)c . x ∈ A ∩ (BC)c . x ∈ A(BC). Case 2: Assume x ∈ A ∩ C. So, x ∈ A and x ∈ C. So, x ∈ A and x ∈ C c . / So, x ∈ A and x ∈ (C c ∩ B). / So, x ∈ A and x ∈ (B ∩ C c ). / So, x ∈ A and x ∈ (BC). / 2

So, x ∈ A and x ∈ (BC)c . So, x ∈ A ∩ (BC)c . So, x ∈ A(BC). Some things to notice: 1. How to structure a proof. 2. When you get stuck, think about what you want to get to and ask yourself how you can bring things into your proof that will help you get there. 3. Pay attention to all situations. Most proofs will have more than one case to deal with. 4. All proofs where you show that one thing is equal to another has two directions. 3

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