Consider the Boolean function: f(A,B,C,D)=(A′+D′)(B+C) If the function is…

Answer: 4

Reasoning (concise):

• Recognize that A' + D' = NAND(A, D). This uses one 2-input NAND gate to produce the term (A' + D').

• Recognize that B + C = NAND(B', C'). If the complements of B and C are available as inputs, one 2-input NAND gate produces (B + C).

• Use a 2-input NAND to combine the two intermediate results: NAND(NAND(A,D), NAND(B',C')) produces the complement of the function, i.e. f'.

• Invert that result with a 2-input NAND used as an inverter (tie its inputs): NAND(previous, previous) gives f. This is the fourth NAND.

Gate count summary (when complemented inputs B' and C' are available):

• 1 gate: NAND(A, D) → produces (A' + D')

• 1 gate: NAND(B', C') → produces (B + C)

• 1 gate: NAND(the two results) → produces f'

• 1 gate: NAND(f', f') → produces f

Note: If complemented inputs B' and C' are NOT provided, you must generate them with two additional 2-input NAND gates (NAND(B,B) and NAND(C,C)), bringing the total to 6 NAND gates. The minimal count is 4 under the common assumption that complemented inputs are available; otherwise the minimal practical implementation requires 6 two-input NAND gates.