Min Term and Max Term | AL ICT Sri Lanka | Unit 4 | Boolean Logic and Digital Circuit | in English and Tamil | தமிழில்

[Boolean Logic] Min Terms & Max Terms Explained

GCE A/L ICT – Sri Lanka | Beginner Friendly Guide with Examples & Exercises

💡 Teacher's Tip: Min Terms and Max Terms are two sides of the same coin. Master them once — and K-Maps, SOP/POS forms will become easy!

1. What are Min Terms and Max Terms?

Min Term (Product Term)

• A Min Term is a Boolean expression where ALL variables appear exactly once, connected by AND (·) operators.
• Also called a Product Term because AND is like multiplication (e.g., X·Y = XY).
• Each Min Term produces output 1 (HIGH) for exactly ONE combination of inputs.
• Represented by lowercase m with subscript (e.g., m₀, m₁).

Example (2 variables X, Y):
Possible Min Terms: XY, X'Y, XY', X'Y'
→ Total = 2² = 4 Min Terms (m₀ to m₃)

Max Term (Sum Term)

• A Max Term is a Boolean expression where ALL variables appear exactly once, connected by OR (+) operators.
• Also called a Sum Term because OR is like addition.
• Each Max Term produces output 0 (LOW) for exactly ONE combination of inputs.
• Represented by uppercase M with subscript (e.g., M₀, M₁).

Example (2 variables X, Y):
Possible Max Terms: X+Y, X'+Y, X+Y', X'+Y'
→ Total = 2² = 4 Max Terms (M₀ to M₃)

2. Why "Min" Term and "Max" Term?

🧠 Memory Trick:
• Min Term = Minimal condition to make output = 1
• Max Term = Maximal condition to make output = 0

• Min Term: It's the "smallest" (minimal) combination of inputs that forces F=1. Any other input change makes it 0.
• Max Term: It's the "largest" (maximal) combination of inputs that forces F=0. Any other input change makes it 1.

3. Truth Tables for 2 Variables (X, Y)

Min Terms Table

X	Y	Min Term	Name	Output = 1 when?
0	0	X'Y'	m₀	Only when X=0, Y=0
0	1	X'Y	m₁	Only when X=0, Y=1
1	0	XY'	m₂	Only when X=1, Y=0
1	1	XY	m₃	Only when X=1, Y=1

Max Terms Table

X	Y	Max Term	Name	Output = 0 when?
0	0	X+Y	M₀	Only when X=0, Y=0
0	1	X+Y'	M₁	Only when X=0, Y=1
1	0	X'+Y	M₂	Only when X=1, Y=0
1	1	X'+Y'	M₃	Only when X=1, Y=1

🔑 Key Insight: For the SAME input row:
• Min Term = 1
• Max Term = 0
→ They are complements: mᵢ = (Mᵢ)'

4. SOP (Sum of Products) vs POS (Product of Sums)

Feature	SOP (Min Term Form)	POS (Max Term Form)
Basis	Uses Min Terms	Uses Max Terms
Structure	OR of AND terms (e.g., XY + X'Y)	AND of OR terms (e.g., (X+Y)(X'+Y))
Output focus	Describes when F = 1	Describes when F = 0
Canonical form	F = Σm(1,3)	F = ΠM(0,2)

5. Step-by-Step Examples

Example 1: F = XY + XZ (SOP Form)

Variables: X, Y, Z (3 variables → 2³ = 8 possible Min Terms)

Step 1: Expand each term to include ALL variables
• XY = XY(Z + Z') = XYZ + XYZ'
• XZ = XZ(Y + Y') = XYZ + XY'Z

Step 2: Combine and remove duplicates
F = XYZ + XYZ' + XYZ + XY'Z = XYZ + XYZ' + XY'Z

Step 3: Map to Min Term numbers (X=MSB, Z=LSB)
• XYZ → 111 → m₇
• XYZ' → 110 → m₆
• XY'Z → 101 → m₅

Final Answer: F = Σm(5,6,7) = m₅ + m₆ + m₇

Example 2: F = (X+Y)(X+Z) (POS Form)

Note: Your query said "(X+Y)+(X+Z)" — but POS uses AND (·) between OR terms, not OR (+).

Step 1: Expand each Max Term to include ALL variables
• (X+Y) = (X+Y+ZZ') = (X+Y+Z)(X+Y+Z')
• (X+Z) = (X+Z+YY') = (X+Y+Z)(X+Y'+Z)

Step 2: Combine and remove duplicates
F = (X+Y+Z)(X+Y+Z')(X+Y+Z)(X+Y'+Z) = (X+Y+Z)(X+Y+Z')(X+Y'+Z)

Step 3: Map to Max Term numbers (X=MSB, Z=LSB)
• X+Y+Z → 000 → M₀
• X+Y+Z' → 001 → M₁
• X+Y'+Z → 010 → M₂

Final Answer: F = ΠM(0,1,2) = M₀ · M₁ · M₂

6. Converting Between Min Terms and Max Terms

Golden Rule: For n variables:
• Min Terms where F=1 → Max Terms where F=0 are the remaining terms
• If F = Σm(1,3,5) for 3 variables (0-7), then F = ΠM(0,2,4,6,7)

Example: F = Σm(1,2) for 2 variables (X,Y)

• Total terms = 4 (0 to 3)
• F=1 for m₁, m₂ → F=0 for m₀, m₃
• ∴ F = ΠM(0,3)

Boolean Proof: X' + Y' = (XY)' ← De Morgan's Theorem

• Left side: Max Term (OR form)
• Right side: Complement of Min Term (AND form)

7. Complete 3-Variable Example (X, Y, Z)

Row	X	Y	Z	Min Term	m#	Max Term	M#
0	0	0	0	X'Y'Z'	m₀	X+Y+Z	M₀
1	0	0	1	X'Y'Z	m₁	X+Y+Z'	M₁
2	0	1	0	X'YZ'	m₂	X+Y'+Z	M₂
3	0	1	1	X'YZ	m₃	X+Y'+Z'	M₃
4	1	0	0	XY'Z'	m₄	X'+Y+Z	M₄
5	1	0	1	XY'Z	m₅	X'+Y+Z'	M₅
6	1	1	0	XYZ'	m₆	X'+Y'+Z	M₆
7	1	1	1	XYZ	m₇	X'+Y'+Z'	M₇

8. Practice Exercises (With Answers)

Exercise 1

For function F(X,Y) = X'Y + XY', write:

(a) Min Term list (Σm)

(b) Max Term list (ΠM)

✅ Click to see answer

Solution:

(a) F = X'Y + XY' = m₁ + m₂ → Σm(1,2)

(b) Total terms = 4. F=0 for rows 0 and 3 → ΠM(0,3)

Exercise 2

Convert F = Σm(0,3,5,6) for 3 variables to POS form.

✅ Click to see answer

Solution:

Total Min Terms for 3 variables = 8 (0 to 7)

F=1 for m₀,m₃,m₅,m₆ → F=0 for m₁,m₂,m₄,m₇

∴ F = ΠM(1,2,4,7)

Exercise 3

Prove using truth table: X' + Y' = (XY)'

✅ Click to see answer

X	Y	XY	(XY)'	X'	Y'	X'+Y'
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

Columns (XY)' and X'+Y' are identical → Proved! (This is De Morgan's Theorem)

9. Quick Summary Cheat Sheet

• ✓ Min Term = AND of all variables → Output = 1 for ONE row → Notation: m₀, m₁...
• ✓ Max Term = OR of all variables → Output = 0 for ONE row → Notation: M₀, M₁...
• ✓ SOP = Sum (OR) of Min Terms → Focus on F=1 rows
• ✓ POS = Product (AND) of Max Terms → Focus on F=0 rows
• ✓ For n variables: Total terms = 2ⁿ
• ✓ Min/Max conversion: Missing terms swap between Σm and ΠM
• ✓ mᵢ = (Mᵢ)' → They are complements!

📚 A/L Exam Tip: When asked to "express in canonical form":

• If given SOP → Expand to Min Terms → Write Σm(...)
• If given POS → Expand to Max Terms → Write ΠM(...)

Prepared with care for Sri Lankan A/L ICT Students | Practice Truth Tables Daily!