Consider the following Boolean function of four variables f(A, B, C, D) = Σ(2,…

Key idea: the minterms come in pairs that differ only in D, so f does not depend on D.

• A=0, B=0, C=1: minterms 2 and 3 are both present (f = 1 for D = 0 and D = 1).

• A=1, B=0, C=0: minterms 8 and 9 are both present (f = 1 for both D).

• A=1, B=1, C=1: minterms 14 and 15 are both absent (f = 0 for both D).

Therefore D is redundant and we can write g(A,B,C) = f(A,B,C,D). The g minterms (for ABC) are Σ(1,3,4,5,6).

1. Combine minterms 1 and 3: A' C

2. Combine minterms 4 and 5: A B'

3. Combine minterms 4 and 6: A C'

Putting these together:

g(A,B,C) = A' C + A B' + A C' = A' C + A(B' + C')

Final answer: the function is independent of one variable (D).