What is the simplified expression for the Boolean function F(A, B, C, D) =…

Concept: On a four-variable K-map, a group of 2k adjacent 1-cells removes k variables from the product term that covers them. When the cells where a function equals 0 themselves form one such group, minimising the 0-set is far cheaper: build the simplest expression for F′ and then complement it with DeMorgan's law, (X·Y)′ = X′ + Y′.

Application: Four variables give 16 map cells. This function names 12 of them as 1, so only four cells hold 0 — minterms 3, 7, 11 and 15. Writing those four in binary shows what they have in common:

1. Across these four zero-cells A and B take all four combinations 00, 01, 10 and 11, so both A and B drop out of the term.

2. C = 1 and D = 1 in every one of the four zero-cells, so only those two literals survive: the quad of zeros is the single product term C·D, that is F′ = CD.

3. Complement both sides and apply DeMorgan's law: F = (F′)′ = (CD)′ = C′ + D′.

Cross-check:

• Counting: C′ + D′ is 0 only where C = 1 and D = 1, which is 4 of the 16 cells, so it is 1 on the remaining 12 — exactly as many 1-cells as the question lists.

• Minterm 10 = 1010 has C = 1 but D = 0, so D′ = 1 and C′ + D′ evaluates to 1, matching its presence in the list.

• Minterm 7 = 0111 has C = 1 and D = 1, so both complements are 0 and C′ + D′ evaluates to 0, matching its absence from the list.

Result: A sum of full four-variable product terms covers exactly one map cell per term, so four or five such terms can cover only four or five of the sixteen cells, and no grouping has been carried out at all. The K-map simplification of this function is F = C′ + D′.