# 8 - C implies x ∈ A ∪ B ∩ A ∪ C To prove part 2 we...

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arbitrary sets. The proof is given by: A ( B C ) = { x : x A x ( B C ) } = { x : x A ( x B x C ) } = { x : ( x A x B ) ( x A x C ) } = { x : ( x A B ) ( x A C ) } = { x : x ( A B ) ( A C ) } Notice that this proof depends on the distributivity of the logical connective over introduced in the logic course. This style of proof can be confusing to those who are unsure of the logical notation. We can also give another style of proof, which is more wordy but basically reasons in the same way. Let A , B and C be arbitrary sets. We will prove the following two results: 1. A ( B C ) ( A B ) ( A C ) 2. ( A B ) ( A C ) A ( B C ) To prove part 1, assume x A ( B C ) for an arbitrary element x . By the deFnition of set union, this means that x A or x B C . By the deFnition of set intersection, this means that either x A , or x is in both B and C . By the distributivity of ‘or’ over ‘and’, it follows that x A or x B , and also that x A or x C . This means that x A B and x A C , and hence x ( A B ) ( A C ) . We have shown that x A ( B
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Unformatted text preview: C ) implies x ∈ ( A ∪ B ) ∩ ( A ∪ C ) . To prove part 2, we must prove that x ∈ ( A ∪ B ) ∩ ( A ∪ C ) implies x ∈ A ∪ ( B ∩ C ) . In this case the proof is simple, since it just follows the above proof in reverse. The details are left as an exercise. ± The above proof is an example of the generality often required to prove a property about sets: it uses arbitrary sets and arbitrary elements of such a set. In contrast, to show that a property is false, it is enough to Fnd one counter-example. Such counter-examples should be as simple as possible, to illustrate that a statement is not true with minimum effort to the reader. P ROPOSITION 2.7 The following statements are not true: 1. A ∪ ( B ∩ C ) = ( A ∩ B ) ∪ C ; 2. A ∪ ( B ∩ C ) = ( A ∩ B ) ∪ ( A ∩ C ) . 9...
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