Question

AB+AbarCbar+ABbarC (AB+C) =1


​

Answer 1

Explanation:

How do I prove that AB+(AC)' +AB'C (AB + C) = 1?

Taking L.H.S we have,

=AB+(AC)’+AB'C(AB+C)

=AB+A’+C’+AB’CAB+AB’CC {Applying De Morgan’s law on (AC)’}

=AB+A’+C’+0+AB’C {AB’CAB=0 because B.B’=0}

=AB+A’+C’+AB’C

=A(B+B’C)+A’+C’

=A(B+B’)(B+C)+A’+C’ {B+B’=0}

=A(B+C)+A’+C’

=AB+AC+A’+C’

=(A’+AB)+(C’+AC)

=(A’+A)(A’+B)+(C’+C)(C’+A) {A+A’=1 and C+C’=1}

=A’+B+A+C’

=A’+A+B+C’

=1+B+C’ {A’+A=1}

=1 {1+B+C’=1 because any variable OR-ING with 1 gives the result 1}

=R.H.S Hence Proved .