03 Chapter 3 - Boolean Algebra & Logic Simplification (Merged)

Chapter 3: Boolean Algebra & Logic Simplification - Complete Study Notes

NotebookLM Ingestion Compilation

This merged document contains all 6 study notes for Chapter 3: Boolean Algebra & Logic Simplification from ECE 2103 Digital Electronics (Sharif Sir).

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Related Concepts: 02 Boolean Algebra Significance & Circuit Minimization | 03 SOP, POS, Canonical & Standard Forms | 06 Boolean Algebra Puzzles & Exam Proofs

3.01 Boolean Algebra Foundations & Duality Principle

Statement of the Duality Principle

The Duality Principle states that every valid Boolean algebraic relation remains mathematically valid if all OR ($+$) and AND ($\cdot$) operators are interchanged, and all identity elements ($0$ and $1$) are interchanged, provided the variables and their complements are left unchanged.

$f(x_1,\dots ,x_n,+,\cdot ,0,1)=g(x_1,\dots ,x_n,+,\cdot ,0,1)⟹f(x_1,\dots ,x_n,\cdot ,+,1,0)=g(x_1,\dots ,x_n,\cdot ,+,1,0)$

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1. Duality Transformation Rules

graph TD
    subgraph Duality Transformation Rules
        DualIn[Original Expression F] --> Rule1[Swap + and •]
        Rule1 --> Rule2[Swap 0 and 1]
        Rule2 --> Rule3[Keep Literals x and x' UNCHANGED]
        Rule3 --> DualOut[Dual Expression Dual-F]
    end
    
    subgraph Complement Transformation Rules
        CompIn[Original Expression F] --> CRule1[Swap + and •]
        CRule1 --> CRule2[Swap 0 and 1]
        CRule2 --> CRule3[INVERT Literals: x to x', x' to x]
        CRule3 --> CompOut[Complement Expression F']
    end

Dual vs. Complement

• Dual ($Dual(F)$): Swaps operators ($+\leftrightarrow \cdot$) and constants ($0\leftrightarrow 1$) ONLY. Variable complements are untouched ($x\to x,x^{\prime }\to x^{\prime }$).
• Complement ($F^{\prime }$ via De Morgan’s): Swaps operators ($+\leftrightarrow \cdot$), constants ($0\leftrightarrow 1$), AND complements every literal ($x\to x^{\prime },x^{\prime }\to x$).

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2. Key Mathematical Proofs

Proof 1: Proving Absorption Law & Its Dual

Theorem: Prove $A+AB=A$ and its dual $A\cdot (A+B)=A$.

A. Original Identity ($A+AB=A$):

$\text{L.H.S}=A+AB=A(1+B)=A\cdot 1=A=\text{R.H.S}■$

B. Dual Identity ($A\cdot (A+B)=A$):

$\text{L.H.S}=A\cdot (A+B)=A\cdot A+A\cdot B=A+AB=A(1+B)=A\cdot 1=A=\text{R.H.S}■$

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Proof 2: Major Exam Theorem — Dual of XOR Equals Its Complement

Major Term PYQ Theorem (2015 - 12 Marks, 2018 - 10 Marks)

Question: Show mathematically that the Dual of the Exclusive-OR (XOR) function is equal to its Complement (XNOR).

Let $F=A\oplus B=A^{\prime }B+A{B}^{\prime }$.

Step 1: Compute Dual of $F$ ($Dual(F)$)

Apply duality rules to $F=(A^{\prime }\cdot B)+(A\cdot B^{\prime })$: $Dual(F)=(A^{\prime }+B)\cdot (A+B^{\prime })$ Expand by distribution: $Dual(F)=A^{\prime }A+A^{\prime }{B}^{\prime }+BA+B{B}^{\prime }$ Since $A^{\prime }A=0$ and $B{B}^{\prime }=0$: $\mathbf{Dual}(\mathbf{F})=\mathbf{AB}+{\mathbf{A}}^{\prime }{\mathbf{B}}^{\prime }\text{— (Equation 1)}$

Step 2: Compute Complement of $F$ ($F^{\prime }$)

Apply De Morgan’s Law to $F=A^{\prime }B+A{B}^{\prime }$: $F^{\prime }=(A^{\prime }B+A{B}^{\prime }{)}^{\prime }=(A^{\prime }B{)}^{\prime }\cdot (A{B}^{\prime }{)}^{\prime }=(A+B^{\prime })\cdot (A^{\prime }+B)$ Expand by distribution: $F^{\prime }=A{A}^{\prime }+AB+B^{\prime }{A}^{\prime }+B^{\prime }B=0+AB+A^{\prime }{B}^{\prime }+0$ ${\mathbf{F}}^{\prime }=\mathbf{AB}+{\mathbf{A}}^{\prime }{\mathbf{B}}^{\prime }\text{— (Equation 2)}$

Conclusion:

$\mathbf{Dual}(\text{XOR})=(\text{XOR}{)}^{\prime }=\text{XNOR}■$

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Past Year Questions (PYQs)

• [PYQ 2015, 2016, 2018, 2019, 2022]: Define Duality principle.
• [PYQ 2015, 2018]: Show that the Dual of exclusive-OR is equal to its complement.

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Related Concepts: 01 Boolean Algebra Foundations & Duality Principle | 03 SOP, POS, Canonical & Standard Forms | 05 Tabular Method (Quine-McCluskey) & Prime Implicants

3.02 Boolean Algebra Significance & Circuit Minimization

What is Boolean Algebra?

Boolean Algebra {Switching Algebra} is the mathematical system used to analyze, model, and simplify binary digital circuits. Variables are strictly restricted to two discrete binary states: 0 (LOW/OFF) and 1 (HIGH/ON).

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1. Why Do We Simplify Boolean Expressions?

graph TD
    Min[Boolean Expression Minimization] --> C1[1. Reduces Hardware Cost]
    Min --> C2[2. Decreases Gate Fan-in]
    Min --> C3[3. Lowers Power Consumption]
    Min --> C4[4. Reduces Propagation Delay]
    Min --> C5[5. Saves Silicon / PCB Space]
    Min --> C6[6. Improves System Reliability]
    
    C1 --> G1[Fewer Logic Gates]
    C2 --> G2[Fewer Inputs per Gate]
    C3 --> G3[Less Current & Heat]
    C4 --> G4[Fewer Gate Levels = Faster]
    C5 --> G5[Smaller IC Package]
    C6 --> G6[Fewer Wiring Failure Points]

Six Key Engineering Reasons for Simplification:

1. Reduces Hardware Cost: Direct reduction in total logic gate count on ICs.
2. Decreases Gate Fan-in: Reduces input pin requirements per logic gate.
3. Lowers Power Consumption: Reduces dynamic switching current and heat dissipation.
4. Increases Speed (Reduces Propagation Delay): Decreases gate levels {cascading depth}, enabling higher clock frequencies.
5. Saves Silicon Area: Reduces physical PCB footprint and silicon die size.
6. Improves System Reliability: Fewer interconnects directly lower statistical failure rates.

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2. Summary of Circuit Cost Metrics

Circuit Cost Metric	Definition & Engineering Impact
Gate Count	Total number of physical logic gates required.
Literal Count (Fan-in)	Total number of variable appearances in the expression.
Number of Gate Levels	Depth of the longest path from input to output {propagation delay}.
Interconnection Complexity	Total number of wiring connections between logic gates.

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Past Year Questions (PYQs)

• [PYQ 2020, 2024, 2025]: Significance of studying digital electronics and Boolean algebra.
• [PYQ 2016, 2018, 2022, 2023]: Why do we simplify Boolean expressions? (06 to 10 Marks)

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Related Concepts: 04 Mathematical Conversions & Expansion of SOP and POS | 01 Boolean Algebra Foundations & Duality Principle | 02 Boolean Algebra Significance & Circuit Minimization

3.03 SOP, POS, Canonical & Standard Forms

Introduction to Boolean Logic Forms

Any Boolean function can be expressed in two standard algebraic structures: Sum of Products (SOP) or Product of Sums (POS).

graph TD
    BF[Boolean Expressions] --> SOP[Sum of Products SOP]
    BF --> POS[Product of Sums POS]
    
    SOP -->|OR-ing AND terms| CSOP[Canonical SOP / Sum of Minterms Σm]
    SOP -->|Simplified| SSOP[Standard SOP e.g. AB + BC]
    
    POS -->|AND-ing OR terms| CPOS[Canonical POS / Product of Maxterms ΠM]
    POS -->|Simplified| SPOS[Standard POS e.g. A+B B+C]

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1. Canonical Forms vs. Standard Forms

A. The Canonical Form (Strictly Unique)

Definition: Canonical Form

A Boolean expression is in Canonical Form if every term contains all domain variables of the function exactly once, in either complemented or uncomplemented form. Canonical forms are strictly unique.

• Minterm ($m_j$): An AND product term containing all $n$ domain variables. Evaluates to 1 for exactly one input state. Used in Canonical SOP ($\Sigma m$).
• Maxterm ($M_j$): An OR sum term containing all $n$ domain variables. Evaluates to 0 for exactly one input state. Used in Canonical POS ($\Pi M$).

B. The Standard Form (Simplified / Non-Unique)

Definition: Standard Form

A Boolean expression is in Standard Form if it follows SOP or POS structure, but terms are simplified and do not need to contain all domain variables. Standard forms are non-unique.

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2. Comparison Table: SOP vs. POS

Parameter	Sum of Products (SOP)	Product of Sums (POS)
Basic Structure	Formed by OR-ing multiple AND product terms.	Formed by AND-ing multiple OR sum terms.
Truth Table Source	Derived from rows where output $F=1$.	Derived from rows where output $F=0$.
Canonical Component	Sum of Minterms ($\Sigma m$).	Product of Maxterms ($\Pi M$).
Native Gate Realization	2-level AND-OR or NAND-NAND logic.	2-level OR-AND or NOR-NOR logic.

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Past Year Questions (PYQs)

• [PYQ 2015, 2016, 2019, 2021, 2022, 2024]: Define Canonical form vs. Standard form.
• [PYQ 2021, 2022]: Distinguish between canonical form and standard form of a Boolean function. (08 to 10 Marks)

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Related Concepts: 03 SOP, POS, Canonical & Standard Forms | 01 Boolean Algebra Foundations & Duality Principle | 05 Tabular Method (Quine-McCluskey) & Prime Implicants

3.04 Mathematical Conversions & Expansion of SOP and POS

Concept Overview

Converting standard Boolean expressions into canonical Sum of Products (SOP) or Product of Sums (POS) form is accomplished algebraically via Missing-Variable Expansion.

graph TD
    Expression[Non-Canonical Expression] --> Type{Expansion Target}
    
    Type -->|Expand to SOP| SOPRule[Multiply term by x + x' = 1]
    SOPRule --> ExpandSOP[Apply Distributive Law: AB x+x' = ABx + ABx']
    ExpandSOP --> ListMinterms[List Sum of Minterms Σm]
    
    Type -->|Expand to POS| POSRule[Add xx' = 0 to sum term]
    POSRule --> ExpandPOS[Apply Distributive Law: X + YZ = X+Y X+Z]
    ExpandPOS --> ListMaxterms[List Product of Maxterms ΠM]

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1. Algebraic Expansion Rules

• SOP Expansion Rule: Multiply each non-canonical AND product term by $(x+x^{\prime })=1$ for every missing variable $x$.
• POS Expansion Rule: Add $(x\cdot x^{\prime })=0$ to each non-canonical OR sum term for every missing variable $x$, then expand via distribution: $A+BC=(A+B)(A+C)$.

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2. 4-Variable Canonical Expansion (Major PYQ)

Major PYQ Problem (2015, 2017, 2021 - 12 Marks)

Question: Express $F(A,B,C,D)=(A+B^{\prime }+C)(B+D^{\prime })$ in Product of Maxterms ($\Pi M$) and Sum of Minterms ($\Sigma m$) notation.

Solution:

Part 1: Product of Maxterms ($\Pi M$)

• Term 1 $(A+B^{\prime }+C)$ is missing $D$: $(A+B^{\prime }+C+D{D}^{\prime })=(A+B^{\prime }+C+D)(A+B^{\prime }+C+D^{\prime })$ $\text{Binary: }0100_2=M_4,0101_2=M_5$

• Term 2 $(B+D^{\prime })$ is missing $A$ and $C$: $B+D^{\prime }+A{A}^{\prime }+C{C}^{\prime }$ Expanding via distribution yields 4 maxterms:

  • $(A+B+C+D^{\prime })\to 0001_2=M_1$
  • $(A+B+C^{\prime }+D^{\prime })\to 0011_2=M_3$
  • $(A^{\prime }+B+C+D^{\prime })\to 1001_2=M_9$
  • $(A^{\prime }+B+C^{\prime }+D^{\prime })\to 1011_2=M_{11}$

$\text{Product of Maxterms: }\mathbf{F}(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})=\prod \mathbf{M}(1,3,4,5,9,11)$

Part 2: Sum of Minterms ($\Sigma m$)

The minterm indices are the remaining numbers out of $0\dots 15$: $\text{Sum of Minterms: }\mathbf{F}(\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D})=\sum \mathbf{m}(0,2,6,7,8,10,12,13,14,15)■$

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Past Year Questions (PYQs)

• [PYQ 2015, 2017, 2021, 2024]: Express equations into full SOP (Sum of Minterms) and POS (Product of Maxterms) forms. (10 to 12 Marks)

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Related Concepts: 02 Boolean Algebra Significance & Circuit Minimization | 04 Mathematical Conversions & Expansion of SOP and POS | 06 Boolean Algebra Puzzles & Exam Proofs

3.05 Tabular Method (Quine-McCluskey) & Prime Implicants

Concept Overview: Tabular Method

While Karnaugh maps are ideal for up to 4 variables, they become visually impractical for 6 or more variables. The Tabular Method (Quine-McCluskey) is an algorithmic, tabular minimization method that guarantees the exact minimal standard-form expression for any number of variables.

graph TD
    Minterms[Input Minterms Σm] --> Phase1[Phase 1: Determine Prime Implicants]
    Phase1 --> Group1s[Group Minterms by Count of 1s]
    Group1s --> MatchPairs[Match Adjacent Groups Differing by Power of 2]
    MatchPairs --> MarkUnticked[Unticked Terms = Prime Implicants]
    
    MarkUnticked --> Phase2[Phase 2: Selection Table]
    Phase2 --> FindSingleX[Identify Columns with Single X]
    FindSingleX --> SelectEPI[Select Essential Prime Implicants]
    SelectEPI --> CoverRest[Select Minimum Remaining PIs to Cover All Minterms]
    CoverRest --> MinExpr[Minimal Boolean Function]

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1. Core Definitions

• Prime Implicant (PI): A product term obtained by combining the maximum possible number of adjacent minterm squares. If a product term cannot be combined with any other term to eliminate a variable, it is a Prime Implicant.
• Essential Prime Implicant (EPI): A prime implicant that covers at least one minterm that is not covered by any other prime implicant. All EPIs must be included in the final minimal expression.

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2. Step-by-Step Methodology

Phase 1: Determination of Prime Implicants

1. Group Minterms: Group decimal minterms based on the number of 1s in their binary equivalent.
2. Compare Adjacent Groups: Compare numbers in Group $k$ with Group $k+1$.
3. Power of 2 Rule: Two numbers combine if their difference is a power of $2$ ($1,2,4,8,16,32$) and the smaller number belongs to the upper group.
4. Record & Check: Write the combined pair with their difference in parentheses {e.g., 0,8 (8)} and tick ($✓$) original terms.
5. Repeat: Continue cascading to quads and octets until no further terms combine. Un-ticked terms are Prime Implicants.

Phase 2: Selection Table

1. Construct a grid with columns as original minterms and rows as Prime Implicants.
2. Place an X in row $P{I}_i$ under every minterm column it covers.
3. Identify columns containing a single X. Circle that X—its row is an Essential Prime Implicant.
4. Include all EPIs, then pick the minimal set of remaining PIs to cover all unchecked columns.

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3. PYQ Solution: 4-Variable Tabular Minimization

Major PYQ Problem (2023, 2025 - 10 Marks)

Question: Simplify $F(w,x,y,z)=\sum m(0,1,2,8,10,11,14,15)$ using the Tabular Method.

Solution:

Phase 1: Prime Implicant Determination Table

Group	Column 1 (Minterms)	Column 2 (Pairs)	Column 3 (Quads)
G0	$0✓$	$0,1(1)$	$0,2,8,10(2,8)[x^{\prime }{z}^{\prime }]$
G1	$1✓$	$0,2(2)✓$	$10,11,14,15(1,4)[wy]$
	$2✓$	$0,8(8)✓$
	$8✓$	$2,10(8)✓$
G2	$10✓$	$8,10(2)✓$
G3	$11✓$	$10,11(1)✓$
	$14✓$	$10,14(4)✓$
G4	$15✓$	$11,15(4)✓$
		$14,15(1)✓$

• Un-ticked Prime Implicants:
  1. 0,1 (1) $⟹w^{\prime }{x}^{\prime }{y}^{\prime }$
  2. 0,2,8,10 (2,8) $⟹x^{\prime }{z}^{\prime }$
  3. 10,11,14,15 (1,4) $⟹wy$

Phase 2: Selection Table

Prime Implicant	0	1	2	8	10	11	14	15	Status
$w^{\prime }{x}^{\prime }{y}^{\prime }(0,1)$	X	(X)							EPI (Covers 1)
$x^{\prime }{z}^{\prime }(0,2,8,10)$	X		(X)	(X)	X				EPI (Covers 2, 8)
$wy(10,11,14,15)$					X	(X)	(X)	(X)	EPI (Covers 11, 14, 15)

$\text{Minimal Boolean Function: }\mathbf{F}={\mathbf{w}}^{\prime }{\mathbf{x}}^{\prime }{\mathbf{y}}^{\prime }+{\mathbf{x}}^{\prime }{\mathbf{z}}^{\prime }+\mathbf{wy}■$

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4. 6-Variable Tabular Setup (Exam Monster)

Major PYQ Monster Problem (2024 - 14 Marks)

Question: Determine prime implicants using Tabular Method for $F(A,B,C,D,E,F)=\sum m(6,9,13,18,19,25,27,29,41,45,57,61)$.

Setup Analysis:

• Variables: $6⟹$ Possible pair differences are powers of 2: $1,2,4,8,16,32$.
• Group 2 (Two 1s): $6(000110_2),9(001001_2),18(010010_2)$
• Group 3 (Three 1s): $13(001101_2),19(010011_2),25(011001_2),41(101001_2)$
• Group 4 (Four 1s): $27(011011_2),29(011101_2),45(101101_2),57(111001_2)$
• Group 5 (Five 1s): $61(111101_2)$

(Note: Minterm 6 cannot combine with any Group 3 term $⟹6$ is an immediate Prime Implicant!)

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Past Year Questions (PYQs)

• [PYQ 2018]: Define Prime Implicants with example. (03 Marks)
• [PYQ 2023, 2025]: 4-variable Tabular simplification. (10 Marks)
• [PYQ 2024]: 6-variable 14M Tabular prime implicant determination. (14 Marks)

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Related Concepts: 01 Boolean Algebra Foundations & Duality Principle | 04 Mathematical Conversions & Expansion of SOP and POS | 05 Tabular Method (Quine-McCluskey) & Prime Implicants

3.06 Boolean Algebra Puzzles & Exam Proofs

Exam Edge Cases & Mathematical Puzzles

High-yield exam questions often include 9 to 11-mark mathematical “puzzles” testing reverse-engineering, factoring constraints, and multi-variable gate expansions.

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1. The 5-Variable “Reverse Don’t Care” Puzzle

Major PYQ Problem (2017, 2020 - 11 Marks)

Question: The expression $BE+B^{\prime }D{E}^{\prime }$ is a simplified version of $A^{\prime }BE+BCDE+B{C}^{\prime }{D}^{\prime }E+A^{\prime }{B}^{\prime }D{E}^{\prime }+B^{\prime }{C}^{\prime }D{E}^{\prime }$. Are there any don’t care conditions? If so, what are they?

Solution:

Step 1: Minterms of Unsimplified Expression ($F_{\text{un}}$)

Expand each 5-variable term ($ABCDE$):

• $A^{\prime }BE\to 01XX1⟹m_9,m_{11},m_{13},m_{15}$
• $BCDE\to X1111⟹m_{15},m_{31}$
• $B{C}^{\prime }{D}^{\prime }E\to X1001⟹m_9,m_{25}$
• $A^{\prime }{B}^{\prime }D{E}^{\prime }\to 00X10⟹m_2,m_6$
• $B^{\prime }{C}^{\prime }D{E}^{\prime }\to X0010⟹m_2,m_{18}$ $\text{Unsimplified Minterms: }\{2,6,9,11,13,15,18,25,31\}$

Step 2: Minterms of Simplified Expression ($F_{\text{simp}}$)

• $BE\to X1XX1⟹m_9,m_{11},m_{13},m_{15},m_{25},m_{27},m_{29},m_{31}$
• $B^{\prime }D{E}^{\prime }\to X0X10⟹m_2,m_6,m_{18},m_{22}$ $\text{Simplified Minterms: }\{2,6,9,11,13,15,18,22,25,27,29,31\}$

Step 3: Isolate Don’t Care Minterms

$\text{Don’t Cares }d=F_{\text{simp}}\setminus F_{\text{un}}=\{22,27,29\}$

• $m_{22}(10110_2)=\mathbf{A}{\mathbf{B}}^{\prime }\mathbf{CD}{\mathbf{E}}^{\prime }$
• $m_{27}(11011_2)=\mathbf{AB}{\mathbf{C}}^{\prime }\mathbf{DE}$
• $m_{29}(11101_2)=\mathbf{ABC}{\mathbf{D}}^{\prime }\mathbf{E}■$

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2. The 8-Literal Limit Factoring Puzzle

Major PYQ Problem (2022 - 9 Marks)

Question: Express $F=B^{\prime }{C}^{\prime }{D}^{\prime }+A{B}^{\prime }C{D}^{\prime }+B{C}^{\prime }D+A^{\prime }BCD$ in three different ways using 8 or fewer literals.

Solution:

Way 1: Group by $B^{\prime }{D}^{\prime }$ and $BD$

$F=B^{\prime }{D}^{\prime }(C^{\prime }+AC)+BD(C^{\prime }+A^{\prime }C)$ Using absorption identity $(X^{\prime }+XY=X^{\prime }+Y)$: $\mathbf{F}={\mathbf{B}}^{\prime }{\mathbf{D}}^{\prime }({\mathbf{C}}^{\prime }+\mathbf{A})+\mathbf{BD}({\mathbf{C}}^{\prime }+{\mathbf{A}}^{\prime })\text{(8 literals)}$

Way 2: Group by $C^{\prime }$ and $C$

Expand Way 1: $F=A{B}^{\prime }{D}^{\prime }+B^{\prime }{C}^{\prime }{D}^{\prime }+A^{\prime }BD+B{C}^{\prime }D$ $\mathbf{F}={\mathbf{C}}^{\prime }({\mathbf{B}}^{\prime }{\mathbf{D}}^{\prime }+\mathbf{BD})+\mathbf{C}(\mathbf{A}{\mathbf{B}}^{\prime }{\mathbf{D}}^{\prime }+{\mathbf{A}}^{\prime }\mathbf{BD})\text{(8 literals)}$

Way 3: Group by $D^{\prime }$ and $D$

$\mathbf{F}={\mathbf{D}}^{\prime }(\mathbf{A}{\mathbf{B}}^{\prime }+{\mathbf{B}}^{\prime }{\mathbf{C}}^{\prime })+\mathbf{D}({\mathbf{A}}^{\prime }\mathbf{B}+\mathbf{B}{\mathbf{C}}^{\prime })\text{(8 literals)}■$

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3. 4-Variable XNOR Expansion Proof

OCR Exam Scanner Typo Correction

PYQ 2017 text writes $A\oplus B\oplus C\oplus D=\Sigma m(0,3,5,6,9,10,12,15)$. This is an OCR typo. Minterm 0 ($0000$) evaluates to 0 in XOR. The true textbook question (Morris Mano 4-23) uses XNOR ($\odot$).

Major PYQ Proof (2017 - 10 Marks)

Question: Show that $A\odot B\odot C\odot D=\sum m(0,3,5,6,9,10,12,15)$.

Proof Solution:

Let $X=A\odot B=AB+A^{\prime }{B}^{\prime }$ and $Y=C\odot D=CD+C^{\prime }{D}^{\prime }$. $F=X\odot Y=XY+X^{\prime }{Y}^{\prime }$

• $X^{\prime }=(A\odot B{)}^{\prime }=A\oplus B=A^{\prime }B+A{B}^{\prime }$
• $Y^{\prime }=(C\odot D{)}^{\prime }=C\oplus D=C^{\prime }D+C{D}^{\prime }$

Substitute into $F$: $F=(AB+A^{\prime }{B}^{\prime })(CD+C^{\prime }{D}^{\prime })+(A^{\prime }B+A{B}^{\prime })(C^{\prime }D+C{D}^{\prime })$ $=ABCD+AB{C}^{\prime }{D}^{\prime }+A^{\prime }{B}^{\prime }CD+A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }+A^{\prime }B{C}^{\prime }D+A^{\prime }BC{D}^{\prime }+A{B}^{\prime }{C}^{\prime }D+A{B}^{\prime }C{D}^{\prime }$ Evaluating these 8 product terms: $\mathbf{F}=\sum \mathbf{m}(0,3,5,6,9,10,12,15)■$

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Past Year Questions (PYQs)

• [PYQ 2017, 2020]: 5-variable Reverse Don’t Care condition puzzle. (11 Marks)
• [PYQ 2022]: Express Boolean function with 8 or fewer literals. (09 Marks)
• [PYQ 2017]: Show that $A\odot B\odot C\odot D=\Sigma m(0,3,5,6,9,10,12,15)$. (10 Marks)

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