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# am_lect_12 - B ⊆ C … BUT such an x may not exist(if A =...

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A Direct Proof with (Transitivity of ): If A B and B C then A C Proof : Let x A Since A B , then by definition y , y A y B So x B Again since B C , then by definition z , z B z C So x C Thus x , x A x C Hence by definition, A C

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Another Direct Proof with Claim : If A B and C D then A × C B × D Proof : Let ( x , y ) A × C Then by definition, x A and y C Since A B , x B and since C D , y D Hence by definition, ( x , y ) B × D Thus ( x , y ), ( x , y ) A × C ( x , y ) B × D Hence by definition, A × C B × D
A False Proof with Claim ”: If A × B A × C then B C “Proof” : Let y B Consider any x A Then by definition ( x , y ) A × B So ( x , y ) A × C (since A × B A × C ) Then by definition x A and y C Thus y , y B y C Hence by definition,

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Unformatted text preview: B ⊆ C … BUT such an x may not exist (if A = φ ) Proving two sets are equal In general to show that X = Y we need to show two things: X ⊆ Y and Y ⊆ X Occasionally, however, the proof follows directly by logical equivalence: Example (DeMorgan’s Law): A ∪ B = A ∩ B Proof : A ∪ B = { x ∈ U x ∉ A ∪ B } = { x ∈ U ¬ ( x ∈ A ∪ B ) } = { x ∈ U ¬ ( x ∈ A ∨ x ∈ B ) } = { x ∈ U ¬ ( x ∈ A ) ∧ ¬ ( x ∈ B ) } = { x ∈ U ( x ∉ A ) ∧ ( x ∉ B ) } = { x ∈ U ( x ∈ A ) ∧ ( x ∈ B ) } = A ∩ B...
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am_lect_12 - B ⊆ C … BUT such an x may not exist(if A =...

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