hw2_solutions_2011

hw2_solutions_2011 - Homework 2 Solutions ECE 15A Winter...

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Homework 2 Solutions ECE 15A Winter 2011 1. Prove that (a+b)’ = a’b’ using any of P1-P4 and Theorem 2-8. Referring to slide 35 of lecture #3: (ab)(a’+b’) = aba’+abb’ P3 = 0b+0a = 0 P1, P2, P4, Theorem 3 (ab)+(a’+b’) = a’+b’+ab P1 = (a’+b’+a)(a’+b’+b) P3 = (1+b’)(1+a’) P4, P1 = 1 Theorem 3, P2 We can see that from P4 and Theorem 6, these conditions imply that (ab)’ = a’+b’, and so it follows that (a+b)’ = a’b’ also holds. 2. Prove that in every Boolean algebra a(a+b)=a for every pair of elements a and b, using any of P1-P4, Theorems 2-3. =a(a+b) given =aa+ab P3 (distributive law) =a+ab Theorem 2, XX=X =a(1+b) P3 (distributive law), P2 X*1=X =a(1) Theorem 3, X+1=1 =a P2, X*1=X 3. Prove that if a + x = b + x and a + x’ = b + x’ , then a = b . a = a + ax + ax’ Thm 4 b = b + bx + bx’ Thm 4 a = a + ax + ax’ Given = aa + ax + ax’ + xx’ P2, P4 = a ( a + x) + x’( a + x) P3 = ( a + x ) ( a + x’) P3 = ( b + x ) ( b + x’) Given = bb + bx’ + xb + xx’
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This note was uploaded on 02/26/2011 for the course ECE 15A taught by Professor M during the Winter '08 term at UCSB.

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hw2_solutions_2011 - Homework 2 Solutions ECE 15A Winter...

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