Question

Reduce the following expression using boolean algebra: abc'+bc'd'+a'bd+c'd into the form b+c'd

Answer 1

Given
abc'+bc'd'+a'bd+c'd
abc'(1)+bc'd'(1)+a'bd(1)+c'd { We know that P(1) = P }
abc'(d+d')+bc'd'(a+a')+a'bd(c+c')+c'd { We know that P+P'=1 }
(abc'd+abc'd')+(abc'd'+a'bc'd')+(a'bcd+a'bc'd)+c'd  { By Distributive law P(Q+R) = PQ+PR }
abc'd+abc'd'+abc'd'+a'bc'd'+a'bcd+a'bc'd+c'd { By commutative law PR+PQ = PQ+PR }
(abc'd+abc'd')+bc'd'(a+a')+a'bcd+a'bc'd+c'd  { By Distributive law PQ+PR= P(Q+R) }
(abc'd)+bc'd'(1)+a'bcd+a'bc'd+c'd  { We know that P+P'=1 }
abc'd+bc'd'+a'bcd+a'bc'd+c'd  
abc'd+a'bc'd+bc'd'+a'bcd+c'd  { By commutative law PR+PQ = PQ+PR }
bc'd(a+a')+bc'd'+a'bcd+c'd  { By Distributive law PQ+PR= P(Q+R) }
bc'd(1)+bc'd'+a'bcd+c'd  { We know that P+P'=1 }
bc'(d+d')+a'bcd+c'd  { By Distributive law PQ+PR= P(Q+R) }
bc'(1)+a'bcd+c'd  { We know that P+P'=1 }
b(c'+a'cd)+c'd
b[(c'+a')(c'+c)(c'+d)]+c'd   { By Distributive law P+QR = (P+Q)(P+R) }
b[(c'+a')(1)(c'+d)]+c'd  { We know that P+P'=1 }
b[(c'+a'd)]+c'd
bc'+a'bd+c'd   { By Distributive law P(Q+R) = PQ+PR }

b[(c'+a'd)]+c'd

Which is Required Simplified Expression

Above Simplified Expression in K-map as follows

bc'+a'bd+c'd