Discussion
A circuit implements the logic expression \(F = (A \oplus B)' \cdot (A + C')\). The number of distinct logic levels (gate delays) in the minimal implementation is
Answer: D
The keyed answer is 3.
Simplify with the identity \((A \oplus B)' = AB + A'B'\):
\[ F = (AB + A'B')(A + C') \]\[ = AB + ABC' + A'B'A + A'B'C' \]\[ = AB + A'B'C' \]The minimal form is a two-term AND-OR circuit. Built from basic gates, the complemented inputs A', B' and C' need NOT gates first, so the signal passes through a NOT gate, an AND gate and an OR gate: three levels of gate delay. Option 4 over-counts and option 1 is impossible for a function of three variables that is not a single gate.
If complemented inputs were assumed to be available for free, the same circuit would have two levels, so the question depends on counting the input inverters; the key counts them.
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