$Remark:$ the complement of m0 is M0.

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+22 votes

+6 votes

(P+Q'+R').(P+Q'+R).(P+Q+R')

=(P+R`)(Q+Q`)(P+Q`+R)

=P+PR+PQ`+R`P+R`Q`

=P(1+Q`)+R`P+R`Q`

=P(1+R`)+R`Q`

=P+R`Q`

=(P+R`)(Q+Q`)(P+Q`+R)

=P+PR+PQ`+R`P+R`Q`

=P(1+Q`)+R`P+R`Q`

=P(1+R`)+R`Q`

=P+R`Q`

+1 vote

Given Boolean Expression:

(P+Q'+R').(P+Q'+R).(P+Q+R')

we know that dual of a boolean expression is equivalent. Take dual of it:

PQ'R'+PQ'R+PQR'

Now simplify it.(I am taking dual because solving the given expression will be difficult hence, take dual and solve it)

=PQ'(R'+R)+PQR'

=PQ'+PQR'

=P(Q'+QR')

=P(Q'+R').

Now if we see option b) and take dual of it then we will be getting the above simplified expression.

(P+Q'+R').(P+Q'+R).(P+Q+R')

we know that dual of a boolean expression is equivalent. Take dual of it:

PQ'R'+PQ'R+PQR'

Now simplify it.(I am taking dual because solving the given expression will be difficult hence, take dual and solve it)

=PQ'(R'+R)+PQR'

=PQ'+PQR'

=P(Q'+QR')

=P(Q'+R').

Now if we see option b) and take dual of it then we will be getting the above simplified expression.

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