Ask Question
20 February, 04:43

The sum of all the minterms of a boolean function of n variables is equal to 1.

(a) Prove the above statement for n=3.

(b) Suggest a procedure for a general proof.,

+3
Answers (1)
  1. 20 February, 07:36
    0
    The answers are as follows:

    a) F (A, B, C) = A'B'C' + A'B'C + A'BC' + A'BC + AB'C' + AB'C + ABC' + ABC

    = A' (B'C' + B'C + BC' + BC) + A ((B'C' + B'C + BC' + BC)

    = (A' + A) (B'C' + B'C + BC' + BC) = B'C' + B'C + BC' + BC

    = B' (C' + C) + B (C' + C) = B' + B = 1

    b) F (x1, x2, x3, ..., xn) = ∑mi has 2n/2 minterms with x1 and 2n/2 minterms

    with x'1, which can be factored and removed as in (a). The remaining 2n1

    product terms will have 2n-1/2 minterms with x2 and 2n-1/2 minterms

    with x'2, which and be factored to remove x2 and x'2, continue this

    process until the last term is left and xn + x'n = 1
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “The sum of all the minterms of a boolean function of n variables is equal to 1. (a) Prove the above statement for n=3. (b) Suggest a ...” in 📙 Computers & Technology if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers