Question

given y = A'B+ A'BC' + A'BCD + A'BC'D'E

- create a logic schematic

- simplify the boolean expression

- create a new logic schematic of the simplified version

Answer 1

Given y = A'B+ A'BC' + A'BCD + A'BC'D'E

logic schematic:

Explanation:
Negation(¬) of any logical Identity P is Nothing but Opposite Truth Value for P. i.e If P is T then (¬P) is F
And gate is used to find the Product of Two literals P AND Gate Q Output is PQ
OR gate is used to find the Sum of Two literals P OR Gate Q Output is P+Q

simplify the boolean expression:

Given y = A'B+ A'BC' + A'BCD + A'BC'D'E

y = A'B+ A'B(C' + CD) + A'BC'D'E { By Distributive law PQ+PR = P(Q+R) }

y = A'B+ A'B(C' + C)(C' + D) + A'BC'D'E   { By Distributive law P+QR = (P+Q)(P+R) }

y = A'B+ A'B(1)(C' + D) + A'BC'D'E  { We know that P+P'= 1 }

y = A'B+ (A'BC' + A'BD) + A'BC'D'E  { By Distributive law P(Q+R) = PQ+PR }

y = A'B(1+ C') + A'B(D +C'D'E) { By Distributive law PQ+PR = P(Q+R) }

y = A'B(1) + A'B(D +C')(D +D')(D +E)  { By Distributive law P+QR = (P+Q)(P+R) }

y = A'B(1) + A'B(D +C')(1)(D +E) { We know that P+P'= 1 }

y = A'B(1) + A'B(D +C'E)  { By Distributive law (P+Q)(P+R) = P+QR }

y = A'B+ (A'BD +A'BC'E) { By Distributive law P(Q+R) = PQ+PR }

y = A'B(1+D) +A'BC'E { By Distributive law PQ+PR = P(Q+R) }

y = A'B(1) +A'BC'E  { We know that P+1= 1 }

y = A'B(1 + C'E)   { By Distributive law PQ+PR = P(Q+R) }

y = A'B(1 + C')(1 + E) { By Distributive law P+QR = (P+Q)(P+R) }

y = A'B(1)(1) { We know that P+1= 1 }

y = A'B

Which is Required Simplified Expression

New logic schematic of the simplified version

Explanation:
Negation(¬) of any logical Identity P is Nothing but Opposite Truth Value for P. i.e If P is T then (¬P) is F
And gate is used to find the Product of Two literals P AND Gate Q Output is PQ