Anention Hene se nsed frsom he stun s t e helshe uses to solve the problem by wr
ID: 3750651 • Letter: A
Question
Anention Hene se nsed frsom he stun s t e helshe uses to solve the problem by wrting the simplification law or its name. Agreat weight will be given to the analysis but a sole answer without any proof for the solution worth nothing, Question.1 Reduce the following to minimum number of Boolean variables A. ABC+A C"AC B. (A' C(A+C(A +B+CD) to 3 literals to 4 literals Question.2 Simplify to minimum number of Boolean variables. B. (CD+AB)(AD BC) Question.3 Get the compleent lowingoolean expressions B. E +E(CD +AB)Explanation / Answer
Answer 1 . a. A’C’ + AC’ + ABC
-> C’(A’+ A) + ABC (taking common)
-> C’1 + ABC (we know that this property's value is 1)
-> C’ + ABC
-> (C’+ AB)(C’+C) (here applying distributive law)
-> AB + C’ (final answer)
Answer 1. B . (A’+C)(A’+C’)(A+B+C’D)
-> (A’ + CC’)(A + B + C’D) (normal opening of bracket)
-> A’(A + B + C’D) (by property CC’= 0)
-> A’A + A’B + A’C’D (normal opening of bracket)
-> A’B + A’C’D (by property AA’= 0)
-> A’(B + C’D) (final answer)
Answer 2 . A . XYZ + XZ
-> Z(XY+X) (taking common)
-> Z(X+X)(X+Y) (applying property)
-> Z(X+Y) (final answer)
Answer 2 . B .
(CD+AB)(AD+BC) (opening bracket)
-> ABAD + CDBC + ABBC + CDAD (here due to property AA'=0 everything would be zero)
-> 0
Answer 3 .A .
[XY+XY]
-> (X+Y).(X+Y) (by boolean property (AB)'= (A'+B') )
-> XX + YY +XY + X’Y’ (by property XX'=0)
-> XY + X’Y’
Answer 3.B
[E + E'(A'B+CD)]'
-> E' * [E'(A'B+CD)]' (by boolean property (AB)'= (A'+B') )
-> E' * [E+(A'B+CD)'] (boolean property (AB)'= (A'+B') )
-> E' * [E+ (A+B')*(C'+D')]