SOLUTION: Simplify following functions using Boolean Algebra a. F(X, Y, Z) =(XY) + ( X + Y +Z)’X + YZ b. F(X, Y, Z) = (XY)’ + (X+Y +Z)’ c. F(X,Y,Z) = YZ + (X+Y)’ + (XYZ)’ d

Question 1191034: Simplify following functions using Boolean Algebra
a. F(X, Y, Z) =(XY) + ( X + Y +Z)’X + YZ
b. F(X, Y, Z) = (XY)’ + (X+Y +Z)’
c. F(X,Y,Z) = YZ + (X+Y)’ + (XYZ)’
d. F(X,Y,Z) = (X+Y +Z)’ ( X+Y)

Could you please explain me this homework question step-by-step? Thank you!

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's the simplification of the given Boolean functions using Boolean algebra:
**a. F(X, Y, Z) = (XY) + (X + Y + Z)'X + YZ**
1. Distribute the X:
F = XY + X(X + Y + Z)' + YZ
2. Note that X(X + Y + Z)' = X(X'Y'Z') = XX'Y'Z' = 0 (because XX' = 0)
F = XY + 0 + YZ
F = XY + YZ
**b. F(X, Y, Z) = (XY)' + (X + Y + Z)'**
1. Apply De Morgan's Law:
F = X' + Y' + X'Y'Z'
2. Notice that X'Y'Z' is redundant as it's contained within X' + Y'
F = X' + Y'
**c. F(X, Y, Z) = YZ + (X + Y)' + (XYZ)'**
1. Apply De Morgan's Law:
F = YZ + X'Y' + X' + Y' + Z'
2. Notice that X'Y' is redundant as it's contained in X' and Y'.
F = YZ + X' + Y' + Z'
**d. F(X, Y, Z) = (X + Y + Z)'(X + Y)**
1. Apply De Morgan's Law:
F = (X'Y'Z')(X + Y)
2. Distribute X'Y'Z':
F = X'Y'Z'X + X'Y'Z'Y
3. Note that X'Y'Z'X = XX'Y'Z' = 0 and X'Y'Z'Y = X'YY'Z' = 0 (since XX' = 0 and YY' = 0)
F = 0 + 0
F = 0

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