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Min and Max Terms Notation in Boolean Algebra

In this tutorial, we will learn about the min and max terms in Boolean algebra in digital electronics. By Saurabh Gupta Last updated : May 10, 2023

In short or for convenience purposes, we represent canonical SOP/POS form in min/max terms.

Minterm

Each of the product terms in the canonical SOP form is called a minterm. Minterm are represented as binary numbers in terms of 0s and 1s. The binary words are formed by representing each non-complemented variable by 1 and each complemented variable by 0, and the decimal equivalent of this binary word is represented as a subscript of m as m₀, m₁, m₂, etc. We generally use the ∑ (sigma) notation to represent minterms.

Maxterm

Each of the sum terms in the canonical POS form is called a maxterm. Maxterm can also be represented using binary numbers where each non-complemented variable is represented using 0 and complemented variable using 1, and the decimal equivalent of this binary word is represented as a subscript of M as M₀, M₂, M₂, etc. We generally use ∏ (pi) notation to represent the max terms.

Note: For n-variable logic function there are 2ⁿ minterms and 2ⁿ maxterms.

Min and Max terms for two literal binary expressions

Input Variable (A)	Input Variable (B)	Minterm	Minterm notation	Maxterm	Maxterm notation
0	0	A.B	m0	A+B	M0
0	1	A.B	m1	A+B	M1
1	0	A.B	m2	A+B	M2
1	1	A.B	m3	A+B	M3

Min and Max terms for three literal binary expressions

Input Variable (A)	Input Variable (B)	Input Variable (C)	Minterm	Minterm notation	Maxterm	Maxterm notation
0	0	0	A.B.C	m0	A+B+C	M0
0	0	1	A.B.C	m1	A+B+C	M1
0	1	0	A.B.C	m2	A+B+C	M2
0	1	1	A.B.C	m3	A+B+C	M3
1	0	0	A.B.C	m4	A+B+C	M4
1	0	1	A.B.C	m5	A+B+C	M5
1	1	0	A.B C	m6	A+B+C	M6
1	1	1	A.B.C	m7	A+B+C	M7

Example

Express the following in corresponding minterm and maxterm expression/

1. Y = ABC + A.B.C + A.B. C + A. B. C
2. Y= (A+B+C) (A+ B+C) (A+ B+ C)

Solution (a)

Y = ABC + A. B.C + A.B. C + A. B. C , is an example of canonical SOP expression, so its each term can be represented in minterm notation. Therefore,

Y = ABC + A. B.C + A.B. C + A. B. C
  = m7 + m3 + m5 + m4
  = ∑m (3, 4, 5, 7)		[ ∑ is used to denote CSOP]

Solution (b)

Y= (A+B+C) (A+ B+ C) (A+ B+ C), is an example of canonical POS expression, so its each term can be represented in maxterm notation.

Y = (A+B+C) (A+ B+ C) (A+ B+ C)
  = M0 + M5 + M7
  = ∏M (0, 5, 7)		[ ∏ is used to denote CPOS]

Note: If a truth table is given, and if the output is 1 then it corresponds to minterm and in case the output is 0 then it corresponds to 0.

Input Variable (A)	Input Variable (B)	Input Variable (C)	Output (Y)
0	0	0	0
0	0	1	0
0	1	0	1
0	1	1	0
1	0	0	0
1	0	1	1
1	1	0	1
1	1	1	1

In the above truth table, the minterms will be m₂, m₅, m₆ and m₇ i.e.,

F = ∑m (2, 5,6, 7)

and maxterms will be M₀, M₁, M₃ and M₄ i.e.,

F = ∏M (0, 1, 3, 4)

Hence, from this we can conclude there is a complementary relationship between CSOP and CPOS.

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