Implement the following with NAND and NOR logic \(F=(A+\overline{B})(\overline{…

A NAND gate and a NOR gate are each functionally complete (“universal”) gates: tying both inputs of a NAND gate together gives NOT, and by De Morgan’s theorem a Sum-of-Products (AND-OR) expression realizes directly as a two-level NAND–NAND network, while a Product-of-Sums (OR-AND) expression realizes directly as a two-level NOR–NOR network. Converting the given expression into the canonical form that matches the target gate (SOP for NAND, POS for NOR) is the key step before the circuit can be drawn.

The given function \(F=(A+\overline{B})(\overline{C}+D)\) is already written as a product of two sum terms — it is already in POS form — so the NOR realization is immediate. For the NAND realization the same function must first be expanded into SOP form.

NOR-gate realization (direct, since F is already in POS form):

1. Generate the complements the expression needs using NOR gates with both inputs tied together: \(\overline{B}=NOR(B,B)\) and \(\overline{C}=NOR(C,C)\).

2. Form the complement of each bracket at the first level: \(G_1=NOR(A,\overline{B})=\overline{(A+\overline{B})}\) and \(G_2=NOR(\overline{C},D)=\overline{(\overline{C}+D)}\).

3. Combine the two first-level outputs at the second level: \(F=NOR(G_1,G_2)=\overline{(G_1+G_2)}=(A+\overline{B})(\overline{C}+D)\), which is F itself by De Morgan’s theorem.

This uses 5 NOR gates in total (2 inverters + 2 first-level gates + 1 output gate).

NAND-gate realization (requires the SOP form first):

Expand the product into a sum of four product terms by distributing: \(F=(A+\overline{B})(\overline{C}+D)=A\overline{C}+AD+\overline{B}\,\overline{C}+\overline{B}D\).

1. Generate the complements using NAND gates with both inputs tied together: \(\overline{B}=NAND(B,B)\) and \(\overline{C}=NAND(C,C)\).

2. Form the complement of each product term at the first level: \(P_1'=NAND(A,\overline{C})\), \(P_2'=NAND(A,D)\), \(P_3'=NAND(\overline{B},\overline{C})\), \(P_4'=NAND(\overline{B},D)\).

3. Combine all four first-level outputs at the second level: \(F=NAND(P_1',P_2',P_3',P_4')=\overline{P_1'\,P_2'\,P_3'\,P_4'}=P_1+P_2+P_3+P_4\), which is F by De Morgan’s theorem.

This uses 7 NAND gates in total (2 inverters + 4 first-level gates + 1 output gate).

Cross-check: substitute A=1, B=1, C=0, D=0 and verify all three routes independently.

1. Direct evaluation of F: \(A+\overline{B}=1+0=1\) and \(\overline{C}+D=1+0=1\), so \(F=1\cdot1=1\).

2. Trace the NOR network: \(\overline{B}=0,\overline{C}=1,G_1=NOR(1,0)=0,G_2=NOR(1,0)=0,F=NOR(0,0)=1\).

3. Trace the NAND network: \(\overline{B}=0,\overline{C}=1,P_1'=NAND(1,1)=0,P_2'=NAND(1,0)=1,P_3'=NAND(0,1)=1,P_4'=NAND(0,0)=1,F=NAND(0,1,1,1)=1\).

4. All three routes agree (\(F=1\)), confirming both gate-level implementations are correct.