Question

Realize the basic gates (AND, OR, NOT) using only NAND or NOR universal gates.

Hint:

Remember: NAND → AND with double inversion, NOR → OR with double inversion. Both use De Morgan’s laws.

Answer 1

Concept: NAND and NOR are called universal gates because:

• Any Boolean function can be implemented using only NAND gates.
• Any Boolean function can also be implemented using only NOR gates.

Part A: Using NAND Gates Only
Step 1: NOT gate using NAND.
Tie both inputs together: \[ Y = (A \cdot A)' = A' \] Thus, a NAND gate acts as a NOT gate.
Step 2: AND gate using NAND.
First NAND gives: \[ (A \cdot B)' \] Then pass through another NAND (as NOT): \[ Y = ((A \cdot B)')' = A \cdot B \]
Step 3: OR gate using NAND.
Using De Morgan’s law: \[ A + B = (A' \cdot B')' \] Implementation:

• Use two NAND gates as NOT gates to get \(A'\) and \(B'\).
• Feed them into a NAND gate to get OR output.

Part B: Using NOR Gates Only
Step 4: NOT gate using NOR.
Tie inputs together: \[ Y = (A + A)' = A' \]
Step 5: OR gate using NOR.
First NOR gives: \[ (A + B)' \] Apply NOR again as inverter: \[ Y = ((A + B)')' = A + B \]
Step 6: AND gate using NOR.
Using De Morgan’s law: \[ A \cdot B = (A' + B')' \] Implementation:

• Use two NOR gates as NOT gates to get \(A'\) and \(B'\).
• Feed into another NOR gate to obtain AND output.

Conclusion:
Both NAND and NOR gates can independently realize NOT, AND, and OR gates, proving they are universal gates.