Computer Science
Prove the following Demorgan's laws using laws of boolean algebra:
(a) (A + B)' = A'.B'
(b) (A.B)' = A' + B'
Boolean Algebra
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Answer
The sum of a variable and its complement is 1 and their product is 0 — A+A'=1 and A.A'=0. The theorem states that A'.B' is the complement of A+B. To prove, it must satisfy:
- (A + B)(A' . B') = 0 and
- (A + B) + (A' . B') = 1
Consider the first case:
(A + B)(A' . B') = 0
L.H.S = (A + B)(A' . B')
= A.A'.B' + B.A'.B' [Distributive Law]
= 0.B' + 0.A' [∵ A.A'=0 and B.B'=0]
= 0 + 0
= 0
L.H.S = R.H.S
Hence proved.
Second Case:
(A + B) + (A' . B') = 1
L.H.S = (A + B) + (A' . B')
= (A + B + A')(A + B + B') [Distributive Law]
= (1 + B)(1 + A) [A+A'=1 and B+B'=1]
= 1.1
= 1
L.H.S = R.H.S
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