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 simple and fast calculation method for solving these type of questions?

Question

Answer

Approach:

To solve the question, we need to simplify each expression and check if it is equivalent to f(w, x, y, z) = ∑(0, 4, 5, 7, 8, 9, 13, 15). We can simplify the expressions using Boolean algebra.

Steps:

1. Simplify the given expressions using Boolean algebra.
2. Check if the simplified expression is equivalent to f(w, x, y, z) = ∑(0, 4, 5, 7, 8, 9, 13, 15).
3. If the simplified expression is equivalent to f(w, x, y, z) = ∑(0, 4, 5, 7, 8, 9, 13, 15), mark it as correct, else mark it as incorrect.

Solution:

A) x'y'z'
Simplification: x'y'z' = f(0, 1, 2, 3, 6, 10, 11, 12, 14)
Conclusion: Incorrect

B) w'y'z'
Simplification: w'y'z' = f(1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 13, 14)
Conclusion: Incorrect

C) w'y'z' wx'y xyz xy'z
Simplification: w'y'z' wx'y xyz xy'z = f(9, 13)
Conclusion: Incorrect

D) x'y'z' wx'y w'y
Simplification: x'y'z' wx'y w'y = f(0, 4, 5, 7, 8, 9, 13, 15)
Conclusion: Correct

Therefore, the expressions A, B, and C are not equivalent to f(w, x, y, z) = ∑(0, 4, 5, 7, 8, 9, 13, 15), and expression D is equivalent to f(w, x, y, z) = ∑(0, 4, 5, 7, 8, 9, 13, 15).

Note: Boolean algebra is a systematic and efficient method for simplifying Boolean expressions. It involves using Boolean laws and rules to reduce the expression to its simplest form.

Answer

I think ...k-map is only a fast method if you have a good commond .

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