Homework 4 Solutions

Homework 4 Solutions - EECS 203 HOMEWORK 4 SOLUTIONS 1. (M)...

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EECS 203 HOMEWORK 4 SOLUTIONS 1. (M) Prove that the union of all elements of a power set of a set is the set itself. Let S be the original set and P(S) be the power set and UPS be the union of all elements of a power set. To Prove: S = UPS We prove the above statement by showing each side is a subset of the other side. If an element X exists in S, then P(S) will contain the set containing X, namely, {X} as an element. (This is by the definition of a power set .) Then, by definition of union, the union of all the elements of P(S) will contain all the elements in S. Thus, S is a subset of the union of all elements of P(S). Let x UPS, we prove x S by contradiction. Assume x S, then there is no element (or set) which contains x in P(S), hence x UPS which is a contradiction. Hence, UPS is a subset of S. Therefore, S = UPS. 2. (M) 18 c) (A B) C A C A B C A B (A B) C A C 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 1 1 1 1 0 1 1 0 0 1 1 0 0 0 1 1 1 1 0 0 0 d) (A C) (C B) = A B C A C C B (A C) (C B) 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 0 1 1 1 0 0 0
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e) (B A) (C A) = (B C) A To prove equality, we are going to show inclusion in both directions. To prove that (B A) ∪( C-A) (B C) A , suppose that x (B-A) ∪( C-A). Then either x (B-A) or x (C-A). Without loss of generality, assume the former (the proof in the latter case is exactly parallel) Then x B and x A. From the first of these assertions, it follows that x B C and hence x (B C)- A. For the converse, suppose that x
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This note was uploaded on 02/12/2012 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.

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Homework 4 Solutions - EECS 203 HOMEWORK 4 SOLUTIONS 1. (M)...

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