A'(A+B)
= A'A + A'B
= A'B as A'A = 0 for any A
(B+AA) (A+B')
= ( B + A) (A + B') as AA = A
= ( B A + AA + BB' + AB'
= BA + A + AB' as AA = A and BB' = 0
= A + BA as A + AB' = A
so A'(A+B) + (B+AA) (A+B')
= A'B + A + BA
= A + B(A'+A)
= A + B as A' + A = 1
2 comments:
How is A+AB'=A
A+AB'=A(1+B') = A * 1 = A (this is a standard result)
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