Discussion
The output of the following logic circuit is :
In the circuit, inputs A and B feed an OR gate; the OR output and input C feed a NAND gate; inputs D and E feed an AND gate whose output passes through a NOT gate; the outputs of the first NAND gate and the NOT gate feed a final NAND gate, whose output is the output of the circuit.
Answer: B
The output is (A + B)C + DE.
Trace the gates. The OR gate gives A + B. The first NAND gate gives X = NOT((A + B)C). The AND gate and inverter give Y = NOT(DE). The final NAND gate gives:
\[ \begin{aligned} \overline{X \cdot Y} &= \overline{X} + \overline{Y} \\ &= (A + B)C + DE \end{aligned} \]
by De Morgan's theorem, since each inverted term is inverted back. The option (A + B)C + D + E forgets that D and E pass through an AND gate, and the other options mix up the OR and AND stages.
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