lect6 - Prove or disprove the following proposition For any...

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1 For any three sets A, B and C ( A B ) C = A ( B C ) Prove or disprove the following proposition To disprove that the proposition is true for any sets its sufficient to find a single example of three sets, such that the proposition is false. First you need to decide what to do…If the proposition looks like to be true, prove it for arbitrary sets. But suppose you suspect that it not always true… It is sufficient to find a single counterexample to disprove it!
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2 A B C A B ( A B ) C = A ( B C ) ? B C A B C ( A B ) C A B C A ( B C ) A B C
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3 We are going to disprove by counterexample . A = {1, 2, 3, 4} B = {1, 3, 5, 6} C = {1, 2, 5, 7} A B C 1 3 4 2 5 7 6 ( A B ) C = A ( B C ) A B = {1, 3}, ( A B ) C = {1, 2, 3, 5, 7} B C = {1, 2, 3, 5, 6, 7}, A ( B C ) = {1, 2, 3}
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4 You can use a membership method here. A B C A B ( A B ) C B C A ( B C ) 1 1 1 1 1 1 1 1 2 1 1 0 1 1 1 1 3 1 0 1 0 1 1 1 4 1 0 0 0 0 0 0 5 0 1 1 0 1 1 0 6 0 1 0 0 0 1 0 7 0 0 1 0 1 1 0 8 0 0 0 0 0 0 0
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5 Theorem 5. Let A , B and C be arbitrary sets. Prove that if A B = C B and A B = C B, then A = C . What this theorem says? A B = C B A = C / / A B = C B A = C But A B = C B and A B = C B A = C Counterexample: A = {1}, C = {1, 2}, B ={1, 2} Counterexample: A = {1}, C = {1, 2}, B ={1}
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6 A B = C B A C A C
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7 A B = C B A C A C
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8 Direct proof of p q Assume p : A B = C B and A B = C B to prove q: A = C The equality of two sets can be proved as two subset relations (‘double inclusion’ ): A C C A A = C So, there should be two parts of the proof. 1). A B = C B and A B = C B A C 2). A B = C B and A B = C B C A x A x A B x C B x C x B x A B x C B x C
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9 Part 1: proof of A C Assume A B = C B (1) and A B = C B (2) .
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lect6 - Prove or disprove the following proposition For any...

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