CODEWITHSUNDEEP
boolean-expressionsimplification
Sesese
Sesese

Reputation: 33

Boolean Expression with Laws

Hello can somebody help me? How this boolean expression simplified?

abcd + d

it simplified as:

d

im trying to use the Laws I don't understand at all

Here's the Laws

Basic Boolean Laws
Idempotent Law
    A * A = A
    A + A = A 
Associative Law
    (A * B) * C = A * (B * C)
    (A + B) + C = A + (B + C) 
Commutative Law
    A * B = B * A
    A + B = B + A 
Distributive Law
    A * (B + C) = A * B + A * C
    A + (B * C) = (A + B) * (A + C) 
Identity Law
    A * 0 = 0     A * 1 = A
    A + 1 = 1     A + 0 = A 
Complement Law
    A * ~A = 0
    A + ~A = 1 
Involution Law
    ~(~A) = A 
DeMorgan's Law
    ~(A * B) = ~A + ~B
    ~(A + B) = ~A * ~B 
Redundancy Laws
Absorption
    A + (A * B) = A
    A * (A + B) = A 

    (A * B) + (A * ~B) = A
    (A + B) * (A + ~B) = A 

    A + (~A * B) = A + B
    A * (~A + B) = A * B

Thanks in advance!

Upvotes: 0

Views: 46

Answers (2)

Krishna Vyas
Krishna Vyas

Reputation: 1028

Boolean Expression: abcd + d can be simplified as - 

LHS = abcd + d    [Assume: abcd + d*1  as A * 1 = A ]
    = d(abc + 1)  [Distributive Law]
    = d(1 + abc)
    = d(1) [Identity Law (A + 1 = 1)]
    = d    [Identity Law (A * 1 = A)]
    = RHS

Upvotes: 0

Nir Levy
Nir Levy

Reputation: 12953

It's indeed D, by the following:

abcd+d -> (a+d)*(b+d)*(c+d)*(d+d) // Distributive Law  
(a+d)*(b+d)*(c+d)*(d+d) -> (a+d)*(b+d)*(c+d)*d // Idempotent Law - d+d=d  
(a+d)*(b+d)*(c+d)*d -> (a+d)*(b+d)*d  // Redundancy Laws - (c+d)*d = d  
(a+d)*(b+d)*d -> (a+d)*d  // Redundancy Laws - (b+d)*d = d  
(a+d)*d -> d  // Redundancy Laws - (a+d)*d = d

Upvotes: 1

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