πŸ“˜ ISC Board Β· Topic 1

Boolean Algebra
Complete Theory

Everything you need for a perfect 100/100. Propositional logic β†’ Boolean algebra β†’ K-Maps β†’ Adder circuits.

Propositional Logic
Truth Tables
Equivalence Laws
De Morgan's Theorem
SOP & POS
Karnaugh Maps
Half & Full Adder
01
Propositional Logic
WFFs, connectives, truth values, interpretation

Propositional logic deals with propositions β€” statements that are either TRUE or FALSE. We combine them using logical connectives to form compound statements called Well-Formed Formulae (WFF).

What is a Proposition?

βœ… Valid Propositions

  • "The sky is blue" β€” TRUE
  • "2 + 2 = 5" β€” FALSE
  • "It is raining" β€” TRUE/FALSE
  • "Delhi is the capital of India" β€” TRUE

❌ Not Propositions

  • "Close the door!" (command)
  • "Are you ok?" (question)
  • "x + 2 = 5" (open sentence, value of x unknown)
  • "This statement is false" (paradox)
Logical Connectives (Operators)
SymbolNameRead AsExampleDescription
~ or Β¬NegationNOT p~pReverses truth value of p
∧Conjunctionp AND qp ∧ qTrue only when BOTH are true
∨Disjunctionp OR qp ∨ qTrue when AT LEAST ONE is true
β‡’Implicationp implies qp β‡’ qFalse ONLY when p=T, q=F
⇔Biconditionalp if and only if qp ⇔ qTrue when BOTH have same value
Well-Formed Formulae (WFF)
πŸ“ Definition

A WFF is a syntactically correct expression formed using propositional variables and logical connectives following precise grammar rules.

Formation rules:

1
Atomic WFF
Any propositional variable (p, q, r, T, F) is a WFF by itself.
2
Negation Rule
If P is a WFF, then ~P is also a WFF.
3
Binary Connective Rule
If P and Q are WFFs, then (P ∧ Q), (P ∨ Q), (P β‡’ Q), (P ⇔ Q) are also WFFs.
4
Nothing Else
Only expressions formed using rules 1–3 are WFFs.
⭐ Exam Tip β€” WFF Examples

Valid: (p ∧ q) β‡’ r, ~(p ∨ ~q), ((p β‡’ q) ∧ (q β‡’ r))
Invalid: p ∧ ∨ q, β‡’ p q, p q ∧

Truth Tables for Basic Connectives
pq~pp ∧ qp ∨ q
TTFTT
TFFFT
FTTFT
FFTFF
pqp β‡’ q~p ∨ q
TTTT
TFFF
FTTT
FFTT
⚑ Key Insight

Implication (p β‡’ q) is FALSE ONLY when the premise p is TRUE and conclusion q is FALSE. All other cases are TRUE. Note: p β‡’ q ≑ ~p ∨ q (very important for simplification!)

pqp ⇔ q(pβ‡’q)∧(qβ‡’p)
TTTT
TFFF
FTFF
FFTT
⚑ Key Insight

Biconditional is TRUE when both p and q have the same truth value. It is equivalent to (pβ‡’q) ∧ (qβ‡’p).

Types of Formulae

Tautology (Valid)

A WFF that is TRUE for all possible interpretations (all rows in truth table = T).

p ∨ ~p β€” always TRUE (Law of Excluded Middle)
p β‡’ p β€” always TRUE

Contradiction (Unsatisfiable)

A WFF that is FALSE for all possible interpretations (all rows = F).

p ∧ ~p β€” always FALSE (Law of Contradiction)

Contingency (Satisfiable)

A WFF that is TRUE for at least one interpretation, but not all. Neither tautology nor contradiction.

p ∧ q β€” TRUE only when both p,q = T
Converse, Inverse, and Contrapositive

Given implication: p β‡’ q

Original
p β‡’ q
"If p, then q"
Converse
q β‡’ p
"If q, then p" β€” swap p and q
Inverse
~p β‡’ ~q
Negate both β€” equivalent to converse
Contrapositive
~q β‡’ ~p
Swap AND negate β€” EQUIVALENT to original!
⭐ Critical: p β‡’ q ≑ ~q β‡’ ~p (Contrapositive is logically equivalent!)

But p β‡’ q is NOT equivalent to q β‡’ p (converse) or ~p β‡’ ~q (inverse).

Chain Rule & Modus Ponens

Modus Ponens (Law of Detachment)

Premise 1: p β‡’ q
Premise 2: p     (p is TRUE)
──────────
Conclusion: q

If "if p then q" is true, and p is true, we can conclude q is true.

Chain Rule (Hypothetical Syllogism)

Premise 1: p β‡’ q
Premise 2: q β‡’ r
──────────
Conclusion: p β‡’ r

If p implies q, and q implies r, then p implies r (transitivity of implication).

02
Equivalence Laws
All laws for simplifying WFFs β€” must memorise every one!

Two WFFs P and Q are logically equivalent (P ≑ Q) if they have identical truth values for all interpretations. These laws let us simplify and transform expressions.

Complete List of Equivalence Laws

1. Identity Laws

p ∧ T ≑ p     p ∨ F ≑ p

AND with TRUE and OR with FALSE leave the variable unchanged.

2. Domination Laws (Null / Annihilation)

p ∨ T ≑ T     p ∧ F ≑ F

OR with TRUE always gives TRUE; AND with FALSE always gives FALSE.

3. Idempotent Laws

p ∧ p ≑ p     p ∨ p ≑ p

4. Commutative Laws

p ∧ q ≑ q ∧ p
p ∨ q ≑ q ∨ p

Order of operands doesn't matter for AND and OR.

5. Associative Laws

(p ∧ q) ∧ r ≑ p ∧ (q ∧ r)
(p ∨ q) ∨ r ≑ p ∨ (q ∨ r)

6. Distributive Laws

p ∧ (q ∨ r) ≑ (p ∧ q) ∨ (p ∧ r) β€” AND distributes over OR
p ∨ (q ∧ r) ≑ (p ∨ q) ∧ (p ∨ r) β€” OR distributes over AND

7. De Morgan's Laws ⭐ (Most Important!)

~(p ∧ q) ≑ ~p ∨ ~q β€” NOT of AND = OR of NOTs
~(p ∨ q) ≑ ~p ∧ ~q β€” NOT of OR = AND of NOTs

Memory trick: "Break the bar, change the operator" β€” negate each variable and flip AND↔OR.

8. Law of Implication

p β‡’ q ≑ ~p ∨ q

Used to convert implications to OR form for simplification.

9. Law of Biconditional

p ⇔ q ≑ (p β‡’ q) ∧ (q β‡’ p)
p ⇔ q ≑ (p ∧ q) ∨ (~p ∧ ~q)

10. Law of Negation (Double Negation / Involution)

~(~p) ≑ p

11. Law of Excluded Middle (Tautology)

p ∨ ~p ≑ T β€” Always TRUE

12. Law of Contradiction

p ∧ ~p ≑ F β€” Always FALSE

13. Absorption Laws

p ∨ (p ∧ q) ≑ p
p ∧ (p ∨ q) ≑ p

14. Simplification Rules (for AND)

p ∧ T ≑ p    p ∧ F ≑ F    p ∧ p ≑ p    p ∧ ~p ≑ F

15. Simplification Rules (for OR)

p ∨ T ≑ T    p ∨ F ≑ p    p ∨ p ≑ p    p ∨ ~p ≑ T
Worked Example: Simplify a WFF

Simplify: ~(p ∨ ~q) ∨ (p ∧ q)

Step 1: Apply De Morgan's to ~(p ∨ ~q)
= (~p ∧ ~~q) ∨ (p ∧ q)
Step 2: Apply Double Negation (~~q = q)
= (~p ∧ q) ∨ (p ∧ q)
Step 3: Factor out q (Distributive Law)
= q ∧ (~p ∨ p)
Step 4: Apply Law of Excluded Middle (~p ∨ p = T)
= q ∧ T
Step 5: Apply Identity Law
= q
03
Boolean Algebra
Binary values, basic postulates, AND / OR / NOT

Boolean algebra is a mathematical system dealing with binary values β€” 0 (FALSE) and 1 (TRUE) β€” and the operations performed on them. It is the foundation of digital circuits and computing.

Binary-Valued Quantities

Value 0 (FALSE)

  • Switch is OFF
  • Voltage is LOW
  • Condition is False
  • Represented as 0

Value 1 (TRUE)

  • Switch is ON
  • Voltage is HIGH
  • Condition is True
  • Represented as 1
Basic Postulates of Boolean Algebra
AND with 0
0 Β· 0 = 0   1 Β· 0 = 0
AND with 1
0 Β· 1 = 0   1 Β· 1 = 1
OR with 0
0 + 0 = 0   1 + 0 = 1
OR with 1
0 + 1 = 1   1 + 1 = 1
NOT (Complement)
0Μ„ = 1     1Μ„ = 0
Note
Β· = AND, + = OR, overbar = NOT
Truth Tables for Basic Boolean Operations

AND Gate (Β·)

ABA Β· B
000
010
100
111

OR Gate (+)

ABA + B
000
011
101
111

NOT Gate (overbar)

AΔ€ (NOT A)
01
10
04
Theorems of Boolean Algebra
All theorems with proofs β€” duality, idempotence, De Morgan's, absorption...
Principle of Duality
πŸ“ Duality Principle

Every algebraic expression deducible from the postulates of Boolean algebra remains valid if the operators AND and OR are interchanged and the identity elements 0 and 1 are interchanged.

Original
A + 0 = A
Dual
A Β· 1 = A
Original
A + A = A
Dual
A Β· A = A
Complete Theorem Reference

1. Identity Theorems

A + 0 = A     A Β· 1 = A
A + 1 = 1     A Β· 0 = 0

Proof (A + 0 = A): When A=0: 0+0=0 βœ“; When A=1: 1+0=1 βœ“

2. Idempotent Laws

A + A = A     A Β· A = A

Proof: OR table: 0+0=0=0, 1+1=1=1 βœ“ | AND table: 0Β·0=0, 1Β·1=1 βœ“

3. Commutative Laws

A + B = B + A
A Β· B = B Β· A

4. Associative Laws

A + (B + C) = (A + B) + C
A Β· (B Β· C) = (A Β· B) Β· C

5. Distributive Laws

A Β· (B + C) = AΒ·B + AΒ·C β€” AND distributes over OR
A + (B Β· C) = (A + B) Β· (A + C) β€” OR distributes over AND
⭐ Note: Second law has NO analogy in ordinary algebra!

In normal algebra, a + (bΒ·c) β‰  (a+b)Β·(a+c), but in Boolean algebra it is valid.

6. Complement Laws (Operations with 0 and 1)

A + Δ€ = 1 β€” Complement OR = 1 (Tautology)
A Β· Δ€ = 0 β€” Complement AND = 0 (Contradiction)
Δ€Μ„ = A β€” Double complement = original (Involution)

7. Absorption Laws

A + AΒ·B = A
A Β· (A + B) = A

Proof of A + AΒ·B = A:
= AΒ·1 + AΒ·B        (Identity: A = AΒ·1)
= AΒ·(1 + B)        (Distributive)
= AΒ·1               (1 + B = 1)
= A                 (Identity) βœ“

8. De Morgan's Theorems ⭐⭐⭐

Μ„(AΒ·B) = Δ€ + BΜ„ β€” NOT(A AND B) = NOTA OR NOTB
Μ„(A+B) = Δ€ Β· BΜ„ β€” NOT(A OR B) = NOTA AND NOTB

Proof by Truth Table: Μ„(AΒ·B) = Δ€ + BΜ„

ABAΒ·BΜ„(AΒ·B)Δ€BΜ„Δ€ + BΜ„Same?
0001111βœ“
0101101βœ“
1001011βœ“
1110000βœ“
05
SOP, POS, Minterms & Maxterms
Canonical & cardinal forms, sum of products, product of sums
Minterms and Maxterms

Minterm (mβ‚™)

A product (AND) term containing all n variables in true or complemented form. Evaluates to 1 for exactly one input combination.

For 2 variables A, B:
mβ‚€ = Δ€Β·BΜ„ (when A=0, B=0)
m₁ = Δ€Β·B (when A=0, B=1)
mβ‚‚ = AΒ·BΜ„ (when A=1, B=0)
m₃ = AΒ·B (when A=1, B=1)

Maxterm (Mβ‚™)

A sum (OR) term containing all n variables. Evaluates to 0 for exactly one input combination.

For 2 variables A, B:
Mβ‚€ = A+B (0 when A=0, B=0)
M₁ = A+BΜ„ (0 when A=0, B=1)
Mβ‚‚ = Δ€+B (0 when A=1, B=0)
M₃ = Δ€+BΜ„ (0 when A=1, B=1)
⚑ Key Relationship: Mβ‚™ = complement of mβ‚™

For the same index n: Mβ‚™ = mΜ„β‚™. Example: mβ‚€ = Δ€Β·BΜ„, so Mβ‚€ = A+B βœ“

SOP β€” Sum of Products

A Boolean expression where product (AND) terms are summed (OR). The Canonical SOP uses only minterms.

Example: f(A,B,C) = βˆ‘m(1,4,7) = m₁ + mβ‚„ + m₇
= Δ€Β·BΜ„Β·C + AΒ·BΜ„Β·CΜ„ + AΒ·BΒ·C
βœ… How to find SOP from a truth table

1. Find all rows where output = 1
2. Write the minterm for each (0β†’complement, 1β†’true variable)
3. OR all the minterms together

POS β€” Product of Sums

A Boolean expression where sum (OR) terms are multiplied (AND). The Canonical POS uses only maxterms.

Example: f(A,B,C) = Ξ M(0,2,3,5,6) = Mβ‚€Β·Mβ‚‚Β·M₃·Mβ‚…Β·M₆
= (A+B+C)Β·(A+BΜ„+C)Β·(A+BΜ„+CΜ„)Β·(Δ€+B+CΜ„)Β·(Δ€+BΜ„+C)
βœ… How to find POS from a truth table

1. Find all rows where output = 0
2. Write the maxterm for each (0β†’true variable, 1β†’complement)
3. AND all the maxterms together

Complete Example: 3-variable Function

Truth table for f(A,B,C)

RowABCfMintermMaxterm
00000mβ‚€=Δ€Β·BΜ„Β·CΜ„Mβ‚€=A+B+C ←POS
10011m₁=Δ€Β·BΜ„Β·C ←SOPM₁=A+B+CΜ„
20100mβ‚‚=Δ€Β·BΒ·CΜ„Mβ‚‚=A+BΜ„+C ←POS
30111m₃=Δ€Β·BΒ·C ←SOPM₃=A+BΜ„+CΜ„
41001mβ‚„=AΒ·BΜ„Β·CΜ„ ←SOPMβ‚„=Δ€+B+C
51010mβ‚…=AΒ·BΜ„Β·CMβ‚…=Δ€+B+CΜ„ ←POS
61100m₆=AΒ·BΒ·CΜ„M₆=Δ€+BΜ„+C ←POS
71111m₇=AΒ·BΒ·C ←SOPM₇=Δ€+BΜ„+CΜ„
Canonical SOP (output=1 rows: 1,3,4,7):
f = βˆ‘m(1,3,4,7) = Δ€Β·BΜ„Β·C + Δ€Β·BΒ·C + AΒ·BΜ„Β·CΜ„ + AΒ·BΒ·C
Canonical POS (output=0 rows: 0,2,5,6):
f = Ξ M(0,2,5,6) = (A+B+C)Β·(A+BΜ„+C)Β·(Δ€+B+CΜ„)Β·(Δ€+BΜ„+C)
06
Karnaugh Maps (K-Maps)
Systematic minimisation β€” 2, 3, and 4 variable maps

A Karnaugh Map is a graphical method for simplifying Boolean expressions. Adjacent cells differ by exactly one variable (Gray code ordering), allowing visual identification of groupings.

Rules for Grouping (Implicants)
  • Group only 1s (for SOP minimisation) or only 0s (for POS minimisation)
  • Group sizes must be powers of 2: 1, 2, 4, 8, 16...
  • Groups must be rectangular (including wrapping around edges)
  • Make groups as large as possible (larger groups = fewer literals)
  • Each group eliminates one variable per doubling in size
  • Every 1 must be covered by at least one group
  • Overlapping groups are allowed (same 1 can be in multiple groups)
  • Corners wrap around (top-bottom and left-right)
2-Variable K-Map
f(A,B) β€” Layout
B=0
B=1
A=0
1
0
A=1
1
0
Group of 2 β†’ f = BΜ„
3-Variable K-Map
⚑ Important: Column order in 3/4-variable K-Maps

Columns/rows must follow Gray code order: 00, 01, 11, 10 (NOT 00, 01, 10, 11). This ensures adjacent cells differ by only one bit.

3-variable K-Map (A, BC)
A\BC
00
01
11
10
0
0
1
1
0
1
1
0
1
1
Green group (4 cells) β†’ Δ€Β·C ... wait: Δ€Β·(B... )
Amber group: AΒ·BΜ„ (wraps left-right corners)
4-Variable K-Map β€” Detailed Example

f(A,B,C,D) = βˆ‘m(0,1,4,5,6,7,8,9,14,15)

4-variable K-Map (AB rows, CD columns)
AB\CD
00
01
11
10
00
1
1
0
0
01
1
1
1
1
11
0
0
1
0
10
1
1
0
0
Group 1 (4 cells rows 00,01 cols 00,01): BΜ„Β·DΜ„
Group 2 (4 cells rows 01,11 cols 11): BΒ·C
Group 3 (row 10 cols 00,01): AΒ·BΜ„Β·DΜ„ ... wrap check
⭐ Step-by-step K-Map solving process

1. Plot all 1s in the K-Map from the minterm list or truth table
2. Identify all essential prime implicants (1s covered by only one group)
3. Form the largest possible groups (prefer 8 > 4 > 2 > 1)
4. Cover all 1s using minimum number of groups
5. Write one term per group (variables that don't change within a group are eliminated)
6. OR all the terms together (SOP form)

How to Read a Group (Extract the Term)

For each group of 1s, look at which variables stay constant across all cells in the group:

  • If variable stays at 0 throughout the group β†’ include it complemented (e.g., Δ€)
  • If variable stays at 1 throughout the group β†’ include it true (e.g., A)
  • If variable changes (0 and 1 both appear) β†’ eliminate that variable
Group of 4 spanning all values of D, but A=0, B=0, C=1:
Result = Δ€ Β· BΜ„ Β· C (D is eliminated)
07
Digital Circuits
Half adder, full adder, majority circuit β€” SOP realization
Half Adder
πŸ“ Definition

A Half Adder adds two 1-bit binary numbers A and B, producing a Sum (S) and a Carry (C). It cannot accept a carry-in from a previous stage.

Half Adder Truth Table

ABSum (S)Carry (C)
0000
0110
1010
1101

Boolean Expressions

Sum (S) β€” XOR
S = A βŠ• B = Δ€Β·B + AΒ·BΜ„
Carry (C) β€” AND
C = A Β· B
⭐ Logic gates needed

Sum: 1 XOR gate (or 2 AND + 1 OR + 2 NOT)
Carry: 1 AND gate

Full Adder
πŸ“ Definition

A Full Adder adds three 1-bit inputs: A, B, and Carry-in (Cα΅’β‚™), producing Sum (S) and Carry-out (Cβ‚’α΅€β‚œ). It can be cascaded to add multi-bit numbers.

Full Adder Truth Table

ABCα΅’β‚™Sum (S)Carry-out (Cβ‚’α΅€β‚œ)
00000
00110
01010
01101
10010
10101
11001
11111

SOP Expressions

Sum (S) = βˆ‘m(1,2,4,7)
S = Δ€Β·BΜ„Β·Cα΅’β‚™ + Δ€Β·BΒ·CΜ„α΅’β‚™ + AΒ·BΜ„Β·CΜ„α΅’β‚™ + AΒ·BΒ·Cα΅’β‚™
S = A βŠ• B βŠ• Cα΅’β‚™ (simplified)
Carry-out = βˆ‘m(3,5,6,7)
Cβ‚’α΅€β‚œ = AΒ·B + BΒ·Cα΅’β‚™ + AΒ·Cα΅’β‚™
= AΒ·B + Cα΅’β‚™Β·(A βŠ• B)

Implementation

A Full Adder can be built using:

  • 2 Half Adders + 1 OR gate (most common)
  • Direct gate implementation using SOP
Using two half adders:
HA1: S₁ = A βŠ• B, C₁ = AΒ·B
HA2: S = S₁ βŠ• Cα΅’β‚™
Cβ‚’α΅€β‚œ = C₁ + (S₁ Β· Cα΅’β‚™)
Majority Circuit
πŸ“ Definition

A Majority circuit outputs 1 when the majority of inputs are 1. For 3 inputs (A, B, C), output is 1 when 2 or more inputs are 1.

Majority Circuit (3-input)

ABCOutput (M)
0000
0010
0100
0111
1000
1011
1101
1111

SOP Expression & Simplification

Canonical SOP (minterms 3,5,6,7):
M = Δ€Β·BΒ·C + AΒ·BΜ„Β·C + AΒ·BΒ·CΜ„ + AΒ·BΒ·C
After K-Map simplification:
M = AΒ·B + BΒ·C + AΒ·C

The simplified form needs only 3 AND gates and 1 three-input OR gate.

08
Quick Quiz
Test yourself β€” ISC exam-level questions
Score 0 / 0
Q1. Which of the following is logically equivalent to p β‡’ q?
Law of Implication: p β‡’ q ≑ ~p ∨ q. This is a fundamental equivalence used in simplification.
Q2. De Morgan's theorem states: ~(A + B) = ?
~(A + B) = Δ€ Β· BΜ„. De Morgan's: NOT of OR = AND of NOTs. "Break the bar, change the operator."
Q3. Which type of formula is TRUE for ALL possible interpretations?
A Tautology is a WFF that is TRUE under all interpretations. Example: p ∨ ~p. A Contradiction is always FALSE. A Contingency can be either.
Q4. The canonical SOP of f(A,B) with minterms {1,3} is:
m₁ = Δ€Β·B (A=0,B=1), m₃ = AΒ·B (A=1,B=1). f = Δ€Β·B + AΒ·B = BΒ·(Δ€ + A) = BΒ·1 = B.
Q5. In a Half Adder, what are the Boolean expressions for Sum (S) and Carry (C)?
Sum = A XOR B = AβŠ•B = Δ€Β·B + AΒ·BΜ„ (output 1 when inputs differ). Carry = A AND B (output 1 only when both inputs are 1).
Q6. Which law states: A + AΒ·B = A?
A + AΒ·B = A is the Absorption Law. Proof: AΒ·1 + AΒ·B = AΒ·(1+B) = AΒ·1 = A. The "absorbed" term AΒ·B disappears because A alone already captures it.
⚑
Master Cheat Sheet
All key formulas at a glance β€” print and revise!

Boolean Algebra β€” All Laws in One Place

Identity
A+0=A   AΒ·1=A
Null/Domination
A+1=1   AΒ·0=0
Idempotent
A+A=A   AΒ·A=A
Complement
A+Δ€=1   AΒ·Δ€=0
Involution
Δ€Μ„ = A
Commutative
A+B=B+A   AΒ·B=BΒ·A
Associative
(A+B)+C=A+(B+C)
Distributive
AΒ·(B+C)=AΒ·B+AΒ·C
De Morgan 1
Μ„(AΒ·B) = Δ€+BΜ„
De Morgan 2
Μ„(A+B) = Δ€Β·BΜ„
Absorption 1
A+AΒ·B = A
Absorption 2
AΒ·(A+B) = A
Implication
pβ‡’q ≑ ~p∨q
Contrapositive
pβ‡’q ≑ ~qβ‡’~p
Biconditional
p⇔q ≑ (pβ‡’q)∧(qβ‡’p)
Chain Rule
pβ‡’q, qβ‡’r ⊒ pβ‡’r
🎯 Top 5 Things That Appear Most in ISC Exams

1. De Morgan's theorem β€” simplification problems and truth table proofs
2. K-Map simplification β€” always include grouping visualization
3. SOP from truth table β€” construct canonical form then simplify
4. Full adder truth table + expressions β€” circuit design questions
5. WFF simplification using equivalence laws step-by-step