Boolean Logic Class 11: Logic Gates, Truth Tables, De Morgan's Laws and Circuits

What Boolean Logic is about

Weightage: This chapter is part of Unit I, Computer Systems and Organisation, which carries 10 of the 70 theory marks in the school annual exam together with Computer System and Encoding Schemes and Number System.Source: CBSE Computer Science (083) syllabus 2026-27, Class XI distribution of marks, Unit I Computer Systems and Organisation

Key concepts

Boolean values and the NOT operation

NCERT section CBSE syllabus: Boolean logic (not in the NCERT book)

A Boolean variable can take only two values: 1 (true, on, high) or 0 (false, off, low). Digital circuits use these two states because a switch or transistor is either on or off. A logic gate is an electronic circuit that takes one or more Boolean inputs and gives one Boolean output.

NOT (inversion, complement) has one input and gives the opposite value: NOT 0 = 1, NOT 1 = 0. It is written A' (or with a bar over A). Its gate is called an inverter: a triangle with a small circle (the bubble) at its output. Applying NOT twice gives the original value: (A')' = A.

A	A'
0	1
1	0

Worked example

A night lamp switches on when it is dark. A light sensor gives L = 1 in daylight and L = 0 in the dark. Write the lamp's output in terms of L and give its truth table.

Try it first, then show the solutionSolution

1. The lamp must be on (1) exactly when L = 0, which is the opposite of L.
2. So Lamp = L'.
3. Truth table: L = 0 gives 1; L = 1 gives 0.

Answer: Lamp = L' (NOT L): 0 → 1, 1 → 0.

AND and OR

NCERT section CBSE syllabus: Boolean logic

AND (written A·B or AB) gives 1 only when all inputs are 1. It behaves like switches in series: current flows only if every switch is closed. OR (written A + B) gives 1 when at least one input is 1; it is 0 only when all inputs are 0. It behaves like switches in parallel. Note that in Boolean algebra 1 + 1 = 1, not 2.

A	B	A·B	A + B
0	0	0	0
0	1	0	1
1	0	0	1
1	1	1	1

Gate shapes: AND has a flat input side and a round (D-shaped) output side; OR has a curved input side and a pointed output side. Both can have more than two inputs: a 3-input AND is 1 only for 111, a 3-input OR is 0 only for 000.

Worked example

A lift moves only if its door is closed (D = 1) and a floor button is pressed (F = 1). An emergency bell rings if button 1 (P) or button 2 (Q) is pressed. Write both expressions and the output for D = 1, F = 0, P = 0, Q = 1.

Try it first, then show the solutionSolution

1. Both conditions needed: Move = D·F.
2. Either button is enough: Bell = P + Q.
3. Move = 1·0 = 0; Bell = 0 + 1 = 1.

Answer: Move = D·F = 0 (lift stays); Bell = P + Q = 1 (bell rings).

NAND and NOR: the universal gates

NCERT section CBSE syllabus: Boolean logic

NAND is NOT-AND: (A·B)'. It gives 0 only when all inputs are 1, and 1 otherwise. NOR is NOT-OR: (A + B)'. It gives 1 only when all inputs are 0. Their gate symbols are the AND and OR shapes with a bubble at the output.

A	B	NAND (A·B)'	NOR (A + B)'
0	0	1	1
0	1	1	0
1	0	1	0
1	1	0	0

NAND and NOR are called universal gates because NOT, AND and OR (and so any circuit) can be built from NAND gates alone, or from NOR gates alone:

• NOT A = A NAND A = A NOR A (join both inputs)
• A AND B = (A NAND B) NAND (A NAND B); A OR B = (A NAND A) NAND (B NAND B)
• A OR B = (A NOR B) NOR (A NOR B); A AND B = (A NOR A) NOR (B NOR B)

Worked example

Show by truth table that joining both inputs of a NAND gate makes it a NOT gate.

Try it first, then show the solutionSolution

1. With both inputs equal to A, the output is (A·A)'.
2. A = 0: (0·0)' = 0' = 1.
3. A = 1: (1·1)' = 1' = 0.
4. The output is always the opposite of A.

Answer: A NAND A = A', so a NAND with joined inputs is an inverter.

XOR (exclusive OR)

NCERT section CBSE syllabus: Boolean logic

XOR (A ⊕ B) gives 1 when the inputs are different and 0 when they are the same. It differs from OR only in the last row: 1 ⊕ 1 = 0. In terms of the basic operations, A ⊕ B = A'B + AB', which reads 'A is 0 and B is 1, or A is 1 and B is 0'. Its gate is the OR shape with an extra curved line on the input side.

A	B	A ⊕ B
0	0	0
0	1	1
1	0	1
1	1	0

Useful facts: A ⊕ 0 = A, A ⊕ 1 = A', A ⊕ A = 0. For more inputs, XOR gives 1 when the number of 1s is odd, which is why it is used to make parity bits for error checking. In Python, a ^ b on 0 and 1 gives XOR.

Worked example

A staircase light has one switch at the bottom (A) and one at the top (B). Flipping either switch changes the light. With both switches down (0, 0) the light is off. Which gate gives the light's state?

Try it first, then show the solutionSolution

1. (0,0): off. Flip one switch: (0,1) or (1,0): on.
2. Flip the other as well: (1,1): off again.
3. The light is on exactly when the switches differ: XOR.

Answer: Light = A ⊕ B (XOR).

Truth tables for expressions

NCERT section CBSE syllabus: truth tables

A truth table lists the output for every combination of inputs. With n variables there are 2^n rows: 4 rows for two variables, 8 for three. Write the input rows in binary counting order (000, 001, 010, ..., 111) so none is missed.

Method for an expression:
1. Make one column per input variable.
2. Add a column for each sub-expression in the order it is evaluated: brackets first, then NOT, then AND, then OR.
3. Fill the final column from the sub-expression columns.

Two expressions are equivalent if their final columns match in every row. This is how laws are proved by truth table.

Worked example

Draw the truth table of Y = (A + B)·C'.

Try it first, then show the solutionSolution

1. Three variables, so 8 rows. Columns: A, B, C, A + B, C', Y.

2. A	B	C	A + B	C'	Y
   0	0	0	0	1	0
   0	0	1	0	0	0
   0	1	0	1	1	1
   0	1	1	1	0	0
   1	0	0	1	1	1
   1	0	1	1	0	0
   1	1	0	1	1	1
   1	1	1	1	0	0

3. Y is 1 only where A + B = 1 and C = 0.

Answer: Y = 1 for ABC = 010, 100, 110; Y = 0 for the other five rows.

De Morgan's laws

NCERT section CBSE syllabus: De Morgan's laws

De Morgan's two laws say how NOT acts on a whole AND or OR:

• First law: (A·B)' = A' + B'. The complement of a product is the sum of the complements.
• Second law: (A + B)' = A'·B'. The complement of a sum is the product of the complements.

Rule to remember: break the bar, change the sign (AND becomes OR, OR becomes AND, each variable gets its own NOT). The laws extend to more variables: (A·B·C)' = A' + B' + C' and (A + B + C)' = A'·B'·C'.

In gate terms, the first law says a NAND gate is the same as an OR gate with bubbles on its inputs; the second says a NOR gate is the same as an AND gate with inverted inputs.

Proof of the first law by truth table:

A	B	A·B	(A·B)'	A'	B'	A' + B'
0	0	0	1	1	1	1
0	1	0	1	1	0	1
1	0	0	1	0	1	1
1	1	1	0	0	0	0

The columns (A·B)' and A' + B' are the same in all four rows, so the law holds.

Worked example

Use De Morgan's laws to simplify Y = (A' + B')'.

Try it first, then show the solutionSolution

1. Break the outer bar and change + to ·: Y = (A')'·(B')'.
2. A double NOT cancels: (A')' = A, (B')' = B.
3. So Y = A·B.

Answer: Y = A·B (an AND gate).

Logic circuits: from circuit to expression and back

NCERT section CBSE syllabus: logic circuits

A logic circuit joins gates so that the output of one is the input of another. To write the expression of a circuit, label the output of every gate from the inputs towards the output, then combine. For example, if A and B go into an OR gate, and its output and C go into an AND gate, the output is Y = (A + B)·C.

To draw the circuit for an expression, work inside-out: draw a NOT gate for each complemented variable, then the AND gates for the products, then the OR gate that joins them. For Y = A'B + AB': two NOT gates make A' and B'; one AND gate takes A' and B, another takes A and B'; an OR gate joins the two AND outputs. (This circuit is an XOR built from basic gates.)

To find the output for given inputs, put the values into each gate in turn, or write the expression and evaluate it. A full truth table of the circuit is the safest way to see what it does.

Worked example

Inputs A and B go into a NOR gate. Its output and input C go into an AND gate, whose output goes through a NOT gate to give Y. Write Y and simplify it with De Morgan's laws.

Try it first, then show the solutionSolution

1. NOR output: (A + B)'.
2. AND output: (A + B)'·C.
3. NOT output: Y = ((A + B)'·C)'.
4. First law: Y = ((A + B)')' + C' = A + B + C'.

Answer: Y = ((A + B)'·C)' = A + B + C'. Y is 0 only when A = 0, B = 0 and C = 1.

Boolean Logic in real life

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

Topics in this chapter

• De Morgan's Laws, Universal Gates and Logic Circuits
• Truth Tables for Boolean Expressions with Two and Three Variables

Boolean Logic: syntax and key facts

Basic operations

NOT: A'; AND: A·B (1 only if both 1); OR: A + B (0 only if both 0)

1 + 1 = 1 in Boolean algebra.

Derived operations

NAND = (A·B)'; NOR = (A + B)'; XOR: A ⊕ B = A'B + AB'

XOR is 1 when the inputs differ; XNOR (A ⊕ B)' = AB + A'B' is 1 when they match.

De Morgan's laws

(A·B)' = A' + B'; (A + B)' = A'·B'

Break the bar, change the sign; works for any number of variables.

Rows in a truth table

n inputs → 2^n rows

2 inputs: 4 rows; 3 inputs: 8 rows.

Universal gates

A' = A NAND A = A NOR A; A·B = (A NAND B)'; A + B = (A NOR B)'

The complement of NAND is AND; of NOR is OR.

Useful identities

(A')' = A; A + A' = 1; A·A' = 0; A ⊕ A = 0; A ⊕ 1 = A'

Check any of them with a two-row or four-row truth table.

Boolean Logic: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Boolean Logic

The question patterns to expect, and the method that works for each.

Tricks to remember Boolean Logic

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

• AND = all, OR = any

  AND needs ALL inputs 1; OR needs ANY one input 1.

  Use it when: Filling truth tables quickly.

• Bubble means NOT

  A small circle at a gate's output turns AND into NAND and OR into NOR.

  Use it when: Reading gate symbols.

• Break the bar, change the sign

  When a long bar over a product or sum is split, · becomes + and + becomes ·.

  Use it when: Applying De Morgan's laws.

• XOR = odd one out

  XOR is 1 when the number of 1s is odd; for two inputs, when they are different.

  Use it when: XOR tables and parity.

• NAND and NOR are universal

  'N' gates can do aNything: join the inputs for NOT, add another for AND or OR.

  Use it when: Universal gate questions.

Boolean Logic: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Boolean Logic

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

• ✗Mistake: Writing 1 + 1 = 2 or 1 + 1 = 0 in an OR column

  In Boolean algebra OR of 1 and 1 is 1; 0 is the XOR result.

  Wrong working

  A = 1, B = 1: A + B = 0

  Right working

  A = 1, B = 1: A + B = 1 (and A ⊕ B = 0)

  ✓Do this: OR is 0 only for all-zero inputs.

  Usually costs: 1 mark per table

• ✗Mistake: Applying De Morgan without changing the operator

  Only the operator change makes the law true; (A·B)' is not A'·B'.

  Wrong working

  (A·B)' = A'·B'

  Right working

  (A·B)' = A' + B'

  ✓Do this: Break the bar AND change the sign; check with the row A = 1, B = 0.

  Usually costs: 1-2 marks

• ✗Mistake: Missing rows in a three-variable truth table

  Three variables need 8 rows; writing 4 or 6 leaves cases out.

  Wrong working

  Rows 000, 011, 101, 111 only

  Right working

  All 8 rows from 000 to 111 in counting order

  ✓Do this: Count in binary from 0 to 2^n - 1.

  Usually costs: 1 mark

• ✗Mistake: Losing brackets when writing a circuit's expression

  An OR output feeding an AND gate must stay in brackets; without them the precedence changes the meaning.

  Wrong working

  OR(A, B) then AND with C: Y = A + B·C

  Right working

  Y = (A + B)·C

  ✓Do this: Put each gate's output in brackets when it feeds another gate.

  Usually costs: 1 mark

• ✗Mistake: Confusing NOR with XOR, or NAND with NOT AND of each input

  NOR is the complement of OR; (A·B)' is different from A'·B'.

  Wrong working

  NOR table: 00→0, 01→1, 10→1, 11→0

  Right working

  NOR table: 00→1, 01→0, 10→0, 11→0

  ✓Do this: Write the OR or AND column first, then complement it.

  Usually costs: 1 mark

NCERT exercise guide: Boolean Logic

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Boolean Logic

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

1. Q1 · 1 mark · MCQExam level

   Which gate gives output 1 only when all its inputs are 0? (a) AND (b) NAND (c) NOR (d) XOR

   Hint

   Complement of OR.

   Answer

   (c) NOR

2. Q2 · 2 marks · Short answerExam level

   Write the truth table of a 3-input AND gate.

   Hint

   8 rows.

   Answer

   Rows 000 to 110 give 0; only 111 gives 1.

3. Q3 · 2 marks · Short answerExam level

   State both De Morgan's laws in words.

   Hint

   Complement of a product; complement of a sum.

   Answer

   The complement of a product equals the sum of the complements: (A·B)' = A' + B'. The complement of a sum equals the product of the complements: (A + B)' = A'·B'.

4. Q4 · 1 mark · Very short answerHarder exam level

   Simplify (A' · B')' using De Morgan's law and name the single gate that gives it.

   Hint

   Break the bar.

   Answer

   (A')' + (B')' = A + B: an OR gate.

5. Q5 · 3 marks · Short answerHarder exam level

   Draw the truth table of Y = A'B + BC and say in words when Y is 1.

   Hint

   Both terms need B = 1.

   Answer

   Y = 1 for ABC = 010, 011, 111; 0 otherwise. Y is 1 when B = 1 and either A = 0 or C = 1.

6. Q6 · 2 marks · Short answerHarder exam level

   Predict the output:

   for a in (0, 1):
       for b in (0, 1):
           print(a ^ b, int(not (a and b)), end=' | ')

   Hint

   XOR and NAND.

   Answer

   0 1 | 1 1 | 1 1 | 0 0 |

7. Q7 · 1 mark · MCQHarder exam level

   Assertion (A): A NAND gate can be used to build any logic circuit. Reason (R): NOT, AND and OR can all be made using only NAND gates. Choose (a) both true, R explains A (b) both true, R does not explain A (c) A true, R false (d) A false, R true.

   Hint

   Why is NAND universal?

   Answer

   (a)

8. Q8 · 2 marks · Short answerHarder exam level

   Inputs A and B go to an XOR gate; its output and C go to an AND gate giving Y. Write Y and find all inputs for which Y = 1.

   Hint

   C must be 1 and A, B must differ.

   Answer

   Y = (A ⊕ B)·C; Y = 1 for ABC = 011 and 101.

9. Q9 · 2 marks · Short answerHOTS / competency

   A car's warning beep W sounds if the engine is on (E) and either the seat belt is not fastened (B = 0) or a door is open (D = 1). Write W and find it for E = 1, B = 1, D = 0.

   Hint

   Not fastened means B'.

   Answer

   W = E·(B' + D); for 1, 1, 0: W = 1·(0 + 0) = 0, no beep.

10. Q10 · 2 marks · Short answerHarder exam level

    Prove by truth table that A + A·B = A.

    Hint

    Two variables, four rows.

    Answer

    A·B column 0, 0, 0, 1; A + A·B column 0, 0, 1, 1, which equals the A column (0, 0, 1, 1). Proved.

11. Q11 · 2 marks · Short answerHOTS / competency

    Is (A + B)' the same as A' + B'? Justify with one row of the truth table.

    Hint

    Try A = 1, B = 0.

    Answer

    No. For A = 1, B = 0: (A + B)' = 1' = 0 but A' + B' = 0 + 1 = 1. The correct law is (A + B)' = A'·B'.

12. Q12 · 3 marks · Short answerHOTS / competency

    Draw the logic circuit for Y = (A + B)·(A·B)' and show that it behaves as an XOR gate.

    Hint

    Compare with the XOR table row by row.

    Answer

    Circuit: an OR gate on A, B; a NAND gate on A, B; an AND gate joining both outputs. Rows: 00: 0·1 = 0; 01: 1·1 = 1; 10: 1·1 = 1; 11: 1·0 = 0, which is the XOR table.

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. 1.How many rows has the truth table of a 3-input gate?Show answer

   2^3 = 8.

2. 2.For which inputs is a NAND gate's output 0?Show answer

   Only when all inputs are 1.

3. 3.For which inputs is a NOR gate's output 1?Show answer

   Only when all inputs are 0.

4. 4.State De Morgan's second law.Show answer

   (A + B)' = A'·B'.

5. 5.What is 1 ⊕ 1?Show answer

   0.

6. 6.Name the two universal gates.Show answer

   NAND and NOR.

7. 7.What is (A')' equal to?Show answer

   A.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Boolean values: 1 (true) and 0 (false).
• ☐NOT flips; AND is 1 only for all 1s; OR is 0 only for all 0s.
• ☐NAND = NOT AND (0 only for all 1s); NOR = NOT OR (1 only for all 0s).
• ☐XOR is 1 when inputs differ: A ⊕ B = A'B + AB'.
• ☐n variables: 2^n rows in the truth table.
• ☐De Morgan: (A·B)' = A' + B'; (A + B)' = A'·B'.
• ☐NAND and NOR are universal gates.
• ☐Circuit to expression: label each gate's output, keep brackets.

Boolean Logic: frequently asked questions