Discussion
What is the minimum number of 2 input NAND gates required to implement the function
F = (x'+y') (z+w)
Answer: C
Correct answer: 4
Write \( F=(x'+y')(z+w) \) and let \( A = x\,\text{NAND}\,y = x'+y' \). Then
\[ F = A(z+w) = Az + Aw = \big((Az)'\cdot(Aw)'\big)' \]
so \( (Az)' \) and \( (Aw)' \) are two more NAND gates and a fourth NAND combines them. Four gates in all: \( \text{NAND}(x,y) \), \( \text{NAND}(A,z) \), \( \text{NAND}(A,w) \) and \( \text{NAND} \) of the last two.
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