Test: Boolean Logic Operations - Electrical Engineering (EE) MCQ

The expression for Associative property is given by A + (B + C) = (A + B) + C & A*(B*C) = (A*B)*C.
The expression for Commutative property is given by A + B = B + A & A*B = B*A.
The expression for Distributive property is given by A + BC = (A + B)(A + C) & A(B + C) = AB + AC.

A + 1 = 1, as per 1’s Property.

A(A + B) = AA + AB (By Distributive Property) = A + AB (A.A = A By Commutative Property) = A(1 + B) = A*1 (1 + B = 1 by 1’s Property) = A.

The DeMorgan’s law states that (AB)’ = A’ + B’ & (A + B)’ = A’ * B’, as per the Dual Property.

Y = AB’ + (A’ + B)C = AB’ + (AB’)’C = (AB’ + C)( AB’ + AB’) = (AB’ + C).1 = (AB’ + C).

The expression for Absorption Law is given by: A+AB = A.
Proof: A + AB = A(1 + B) = A (Since 1 + B = 1 as per 1’s Property).

The involution of A means double inversion of A (i.e. A”) and is equal to A.
Proof: ((A)’)’ = A

The DeMorgan’s law states that (AB)’ = A’ + B’ & (A + B)’ = A’ * B’, as per the Dual Property.

(A’B + CD’)’ = (A’B)'(CD’)’ (By DeMorgan’s Theorem) = (A” + B’)(C’ + D”) (By DeMorgan’s Theorem) = (A + B’)(C’ + D).

(A + B)(A + C) = AA + AC + AB + BC = A + AC + AB + BC (By Commutative Property) = A(1 + C + B) + BC = A + BC (1 + B + C =1 By 1’s Property).