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18 September, 13:51

Given the Boolean function: F (x, y, z) = x' y + xyz', derive an algebraic expression for the complement of F. Express in sum-of-products form.

a. (x'y + xyz') ' = xy' + xz + x'y' + y' + y'z'

b. (x'y + xyz') ' = x'y' + xz + x'y' + y' + y'z

c. (x'y + xyz') ' = xy' + xz + x'y' + yy' + y'z

d. (x'y + xyz') ' = xy' + xz + x'y' + y' + y'z

+1
Answers (1)
  1. 18 September, 17:50
    0
    d. (x'y + xyz') ' = xy' + xz + x'y' + y' + y'z

    Explanation:

    F = x'y + xyz'

    F' = (x'y + xyz') ', DeMorgan's

    = (x'y) ' (xyz') '

    = (x+y') (x'+y'+z), Distributive Property

    = xx' + xy' + xz + y'x' + y'y' + y'z, Redundancy Law: AA' = 0

    = 0 + xy' + xz + y'x' + y' + y'z, Redundancy Law: AA = A

    = xy' + xz + y'x' + y' + y'z
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Home » Engineering » Given the Boolean function: F (x, y, z) = x' y + xyz', derive an algebraic expression for the complement of F. Express in sum-of-products form. a. (x'y + xyz') ' = xy' + xz + x'y' + y' + y'z' b. (x'y + xyz') ' = x'y' + xz + x'y' + y' + y'z c.
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