Boolean function (X + Y + Z).(X + Y + Z').(X + Z') can be minimized to

2013

Boolean function (X + Y + Z).(X + Y + Z').(X + Z') can be minimized to

  1. A.

    X + Y

  2. B.

    X + (Y.Z')

  3. C.

    X + (Y' + Z)

  4. D.

    X' + (Y' + Z)

Attempted by 11 students.

Show answer & explanation

Correct answer: B

Concept

Two product-of-sums factors that agree in every literal except one variable that appears complemented in one and uncomplemented in the other combine by the rule (A + B)(A + B') = A. Distribution then gives (X + A)(X + B) = X + A.B. The rule A + A.B = A (absorption) only removes a product term that already contains A as a factor.

Application

  1. Group the first two factors, which share (X + Y) and differ only in Z vs Z': (X + Y + Z)(X + Y + Z') = (X + Y), by (A + B)(A + B') = A with A = X + Y, B = Z.

  2. The expression becomes (X + Y)(X + Z').

  3. Apply (X + A)(X + B) = X + A.B with A = Y, B = Z': (X + Y)(X + Z') = X + Y.Z'.

  4. Y.Z' is NOT absorbed into Y, because absorption needs the bigger term to contain the smaller as a factor; here X + Y.Z' is already minimal.

So the minimized sum-of-products form is X + Y.Z'.

Cross-check

Truth-table check: setting X = 0 collapses the original product to (Y + Z)(Y + Z')(Z') = Y.Z' (the same (A + B)(A + B') = A rule, with A = Y, B = Z), so it is 1 only when Y = 1 and Z = 0, and 0 at the other three X = 0 rows. At Y = 1, Z = 1 this gives Y.Z' = 1.0 = 0, and X + Y.Z' reproduces that same 0, while X + Y would give 0 + 1 = 1 -- so X + Y fails to match the function there, whereas X + Y.Z' matches it at all eight input combinations.

Explore the full course: Up Police Computer Operator

Loading lesson…