Computer ScienceISC Class 12 · Computer Hardware

Logic Gates and NAND, NOR as Universal Gates (ISC Class 12 Computer Science)

Updated 17 October 2026 · Draft, not yet fact-checked

The seven elementary gates are NOT, AND, OR, NAND, NOR, XOR and XNOR. NAND and NOR are universal: NOT, AND and OR can each be built from NAND gates alone (or NOR gates alone), so any Boolean circuit can be redrawn using only one of them. Apply De Morgan's theorem to convert a circuit.

Understand Logic Gates and NAND, NOR as Universal Gates

The seven gates. For inputs A and B:

A	B	AND	OR	NAND	NOR	XOR	XNOR
0	0	0	0	1	1	0	1
0	1	0	1	1	0	1	0
1	0	0	1	1	0	1	0
1	1	1	1	0	0	0	1

NOT gives A'. AND gives A·B and OR gives A + B. NAND is (A·B)' and NOR is (A + B)'. XOR is A'B + AB' (1 when the inputs differ). XNOR is AB + A'B' (1 when the inputs agree), which is (A ⊕ B)'.

Circuit from an expression. Each product becomes an AND gate, each sum an OR gate, each complement a NOT gate. Assume complements are available as inputs when the question says so.

Universal gates. A gate set is universal if every Boolean function can be built from it. With NAND only:

NOT: A NAND A = (A·A)' = A'.
AND: (A NAND B) fed to a NAND used as NOT gives AB. Two gates.
OR: A' NAND B' = (A'·B')' = A + B (De Morgan). Three gates.

With NOR only:

NOT: A NOR A = A'.
OR: (A NOR B) fed to a NOR used as NOT gives A + B. Two gates.
AND: A' NOR B' = (A' + B')' = AB. Three gates.

Sum of products in NAND-NAND form. To realise F = AB + CD, double-complement and use De Morgan: F = ((AB + CD)')' = ((AB)'·(CD)')'. That is two NAND gates for the products feeding a third NAND. In general an SOP circuit of AND-OR gates becomes a NAND-NAND circuit with the same structure. Similarly a POS circuit becomes NOR-NOR.

Half and full adders, encoders, decoders and multiplexers are the next topic.

Logic Gates and NAND, NOR as Universal Gates: key facts

NAND and NOR: NAND: (A·B)'
NOR: (A + B)'
NAND is 0 only when all inputs are 1
NOR is 1 only when all inputs are 0; Both are universal gates.
XOR and XNOR: XOR: A⊕B = A'B + AB'
XNOR: (A⊕B)' = AB + A'B'
XOR is 1 when inputs differ
XNOR is 1 when inputs agree; XOR of three inputs is odd parity.
NOT, AND, OR from NAND: NOT A NAND A
AND NAND followed by NAND as NOT (2 gates)
OR NOT A and NOT B into a NAND (3 gates); Each uses De Morgan's theorem.
NOT, AND, OR from NOR: NOT A NOR A
OR NOR followed by NOR as NOT (2 gates)
AND NOT A and NOT B into a NOR (3 gates); The dual of the NAND constructions.

NOT	A NAND A
AND	NAND followed by NAND as NOT (2 gates)
OR	NOT A and NOT B into a NAND (3 gates)

NOT	A NOR A
OR	NOR followed by NOR as NOT (2 gates)
AND	NOT A and NOT B into a NOR (3 gates)

How to answer Logic Gates and NAND, NOR as Universal Gates questions

Use this routine to convert a circuit or expression to NAND-only (or NOR-only).

1Write the expression in SOP form for NAND (POS form for NOR).
2Put a double bar over the whole expression.
3Apply De Morgan once to the inner bar: the sum of products becomes a product of NANDed terms.
4Each term (AB)' is one NAND gate; the outer bar is one more NAND gate.
5Replace each lone complement by a NAND with both inputs tied together.
6Count the gates and verify two or three input rows.

Logic Gates and NAND, NOR as Universal Gates in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Two-switch staircase light

Suppose a staircase light has one switch at the top (A) and one at the bottom (B), and flipping either switch changes the state of the light.

Which gate describes the light, and what is its output when both switches are up (A = B = 1)?

Try it first, then show the solutionSolution

The light changes when exactly one switch differs from the other, so it is XOR.
For A = 1 and B = 1 the inputs agree, so XOR = 0.

Answer: XOR gate; the output is 0 (light off).

Memory hook: XOR means "different".

Alarm that needs both sensors off

Suppose a store lamp turns on only when both the motion sensor (A) and the door sensor (B) are quiet (0).

Name the gate and give the output for A = 0, B = 0.

Try it first, then show the solutionSolution

Output is 1 only when all inputs are 0, which is NOR.
For 0, 0: (0 + 0)' = 1.

Answer: NOR gate; the lamp is on (1).

Memory hook: NOR: nothing on, so output on.

Tricks to remember Logic Gates and NAND, NOR as Universal Gates

Short memory aids. Say each one aloud once, then use it on a question the same day.

N means flip
NAND and NOR are AND and OR with the output flipped.
Use it when: Filling the NAND and NOR columns.
XOR = different
XOR is 1 when the inputs are different; XNOR is 1 when they are the same.
Use it when: Quick truth tables.
Tie the inputs
Joining both inputs of a NAND or NOR gives a NOT gate.
Use it when: Building NOT from universal gates.
Double bar, break once
Put two bars on the function and break the lower one with De Morgan.
Use it when: NAND-NAND conversion.

Logic Gates and NAND, NOR as Universal Gates: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: XNOR output · 1 mark · MCQBoard level

What is the output of an XNOR gate when A = 1 and B = 0?
(a) 0 (b) 1 (c) undefined (d) A

Try it first, then show the solutionSolution

XNOR is 1 when the inputs agree.
The inputs 1 and 0 differ, so the output is 0.

Answer: (a) 0

Example 2: AND, OR and NOT from NAND · 3 marks · Short answerHarder board level

Show how NOT, AND and OR gates are built using NAND gates only. State the number of NAND gates for each.

Try it first, then show the solutionSolution

NOT: tie both inputs, (A·A)' = A'. One gate. (1 mark)
AND: X = (A·B)', then X NAND X = (X)' = AB. Two gates. (1 mark)
OR: A NAND A = A', B NAND B = B', then (A'·B')' = A + B by De Morgan. Three gates. (1 mark)

Answer: NOT: 1 gate, AND: 2 gates, OR: 3 gates.

Example 3: SOP to NAND-NAND · 3 marks · Short answerHarder board level

Realise F = AB + CD using NAND gates only. Show the algebra and count the gates.

Try it first, then show the solutionSolution

Double complement: F = ((AB + CD)')'. (1 mark)
De Morgan on the inner bar: F = ((AB)'·(CD)')'. (1 mark)
Two NAND gates give (AB)' and (CD)'; a third NAND combines them. Total 3 NAND gates. (1 mark)

Answer: F = ((AB)'·(CD)')'; three 2-input NAND gates.

Example 4: XOR using four NAND gates · 4 marks · Short answerHarder board level

N1 = (A·B)', N2 = (A·N1)', N3 = (B·N1)' and F = (N2·N3)'. Find F for (A, B) = (0, 1) and (1, 1) and say which gate F realises.

Try it first, then show the solutionSolution

(0, 1): N1 = (0·1)' = 1; N2 = (0·1)' = 1; N3 = (1·1)' = 0; F = (1·0)' = 1. (1.5 marks)
(1, 1): N1 = (1·1)' = 0; N2 = (1·0)' = 1; N3 = (1·0)' = 1; F = (1·1)' = 0. (1.5 marks)
Checking (0, 0) gives 0 and (1, 0) gives 1, so F = 1 exactly when A and B differ. (1 mark)

Answer: F = 1 for (0, 1) and 0 for (1, 1); the circuit is an XOR gate.

Example 5: Case: staircase light · 4 marks · Case-basedHOTS / competency

Suppose a staircase light is controlled by two switches A and B and the light is ON when the two switch positions are different.
(i) Name the gate and write F in SOP form. (1 mark)
(ii) Give the output for the inputs (1, 0) and (1, 1). (1 mark)
(iii) Which single gate gives the opposite output for all inputs? (1 mark)
(iv) How many NAND gates are needed for F in the best known form? (1 mark)

Try it first, then show the solutionSolution

(i) XOR, F = A'B + AB'.
(ii) (1, 0) gives 1; (1, 1) gives 0.
(iii) XNOR, which is the complement of XOR.
(iv) Four NAND gates, as in the previous example.

Answer: (i) XOR, A'B + AB' (ii) 1 and 0 (iii) XNOR (iv) 4 NAND gates

Where students lose marks in Logic Gates and NAND, NOR as Universal Gates

The slips examiners see most often, and what to write instead.

Mistake: Confusing NOR with XNOR
NOR is 1 only for 00; XNOR is 1 for 00 and 11.
Wrong working
NOR(1, 1) = 1
Right working
NOR(1, 1) = (1 + 1)' = 0
Do this: Add the inputs first, then flip.
Usually costs: 1 mark
Mistake: Skipping the double bar
Without it the De Morgan step has nothing to break and the NAND form is not justified.
Wrong working
AB + CD = (AB)' + (CD)'
Right working
AB + CD = ((AB)'·(CD)')'
Do this: Write two bars, break the lower one.
Usually costs: 1 mark
Mistake: Using a single NAND for OR
OR from NAND needs two NOT-type gates and one NAND, not a direct single gate.
Wrong working
A NAND B = A + B
Right working
A' NAND B' = A + B (3 gates)
Do this: Complement the inputs first.
Usually costs: 1 mark

Exam tips

Draw the standard symbols; a small circle on the output means inversion.
Label every internal wire (N1, N2) so the examiner can follow the trace.
State the number of gates in each NAND or NOR construction.
Verify the final circuit for at least two input rows.
For XOR and XNOR give both the table and the expression.

Practice questions on Logic Gates and NAND, NOR as Universal Gates

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A gate outputs 0 for inputs (0, 0) and (1, 1) but 1 for (0, 1) and (1, 0). Which gate is it?
(a) XNOR (b) XOR (c) NAND (d) OR
Hint
Different inputs give 1.
Answer
(b) XOR
Q2 · 2 marks · Short answerBoard level
Write the truth table of a 2-input NAND gate and say why it is called a universal gate.
Hint
It can build the three basic gates.
Answer
Outputs for (A, B) = 00, 01, 10, 11 are 1, 1, 1, 0. It is universal because NOT, AND and OR, and hence every circuit, can be made from NAND gates alone.
Q3 · 3 marks · Short answerHarder board level
Realise Y = A + B' using NOR gates only. Show the gates, count them and verify for A = 0, B = 1.
Hint
NOR with tied inputs is NOT; a NOR followed by a NOT is OR.
Answer
N1 = B NOR B = B'. N2 = A NOR N1 = (A + B')'. Y = N2 NOR N2 = A + B'. Three NOR gates. For A = 0, B = 1: N1 = 0; N2 = (0 + 0)' = 1; Y = (1 + 1)' = 0, and A + B' = 0 + 0 = 0.
Q4 · 2 marks · Short answerHarder board level
A circuit has X = A NAND B and F = X NAND X. Write F in terms of A and B, and find F for A = 1, B = 1.
Hint
Tied NAND is NOT.
Answer
X = (AB)'; F = (X·X)' = X' = AB, an AND gate. For A = 1, B = 1: X = 0 and F = (0·0)' = 1.
Q5 · 3 marks · Case-basedHOTS / competency
F = (A ⊕ B) ⊕ C, where ⊕ is XOR. Find F for (A, B, C) = (1, 1, 0), (1, 1, 1), (0, 0, 0) and (1, 0, 0). What does F detect?
Hint
Compute the inner XOR first.
Answer
(1, 1, 0): 1⊕1 = 0, then 0⊕0 = 0. (1, 1, 1): 0⊕1 = 1. (0, 0, 0): 0. (1, 0, 0): 1⊕0 = 1, then 1⊕0 = 1. F is 1 when an odd number of inputs are 1, so it detects odd parity.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.When is a NAND gate output 0?Show answer
Only when all inputs are 1.
2.When is a XNOR gate output 1?Show answer
When the inputs are equal.
3.How do you make a NOT gate from a NAND gate?Show answer
Join both inputs together.
4.Which two gates are universal?Show answer
NAND and NOR.
5.What is the Boolean expression of XOR?Show answer
A'B + AB'.
6.How many NAND gates make an OR gate?Show answer
Three.

Logic Gates and NAND, NOR as Universal Gates: frequently asked questions

Why are NAND and NOR called universal gates?

Because the NOT, AND and OR operations, and so every Boolean function, can be built using only NAND gates or only NOR gates.

What is the difference between XOR and OR?

OR is 1 when at least one input is 1; XOR is 1 only when exactly one input is 1, so for 1 and 1 OR gives 1 but XOR gives 0.

How is a circuit converted to NAND gates only?

Write the function as SOP, double-complement it, apply De Morgan once, and replace every AND-OR structure by NAND-NAND with NOT as a tied NAND.

Do the logic symbols matter in the ISC theory paper?

Questions often ask for a logic gate diagram of the reduced expression, so the standard symbols and correct connections matter.