Let f(w, x, y, z) = ∑(0, 4, 5, 7, 8, 9, 13, 15). Which of the following expressions are NOT equivalent to f?
(A) x\’y\’z\’ + w\’xy\’ + wy\’z + xz
(B) w\’y\’z\’ + wx\’y\’ + xz
(C) w\’y\’z\’ + wx\’y\’ + xyz + xy\’z
(D) x\’y\’z\’ + wx\’y\’ + w\’y


Answer: (D)

Explanation:

Solving this k-map we get x\’y\’z\’ + w\’xy\’ + wy\’z + xz which is (A) part

Solving this k-map we get w\’y\’z\’ + wx\’y\’ + xz which is (B) part.

Solving this k-map we get w\’y\’z\’ + wx\’y\’ + xyz + xy\’z which is ( C) part .

But we can’t get (D) part from any combination so Ans is (D) part.

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