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f= A(C+D'B)+A' Simplify the boolean expression. Thanks
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F=A(C+D’B) + A’ =A(C(D+D’)+D’B(C+C’)) + A’ [ Since, A+A’=1] =A(CD+CD’+BCD’+ BC’D’) + A’ =A(CD+CD’(1+B)+BC’D’) + A’ =A(CD+CD’+BC’D’) + A’ [ Since, A+1=1] =A((1-C’)(1-D’)+CD’+BC’D’) + A’ [ Since, A+A’=1] =A(1-C’-D’+C’D’ +CD’+BC’D’) + A’ =A(1-C’-D’+C’D’(1+B)+CD’) + A’ [ Since, A+1=1] =A(1-C’-D’+C’D’+CD’) + A’ =A(1-C’-D’+D’(C+C’)) + A’ [ Since, A+A’=1] =A(1-C’-D’+D’) + A’ =A(1-C’) + A’ =AC +A’ [ Since, A+A’=1] =AC+1-A [ Since, A+A’=1] =1+A(C-1) =1-AC’ [ Since, A+A’=1] therefore, Ans: 1-AC'
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