# Min and Max Terms Notation in Boolean Algebra

In this tutorial, we are going to learn about the **Min and Max Terms in Boolean Algebra in Digital Electronics**.

Submitted by Saurabh Gupta, on November 28, 2019

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 **m _{0}**,

**m**,

_{1}**m**, etc. We generally use the

_{2}**∑ (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 **M _{0}**,

**M**,

_{2}**M**, etc. We generally use

_{2}**∏ (pi) notation to represent the max terms**.

**Note:** For **n-variable** logic function there are **2 ^{n} minterms and 2^{n} 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 | m_{0} | A+B | M_{0} |

0 | 1 | A.B | m1 | A+B | M_{1} |

1 | 0 | A.B | m_{2} | A+B | M_{2} |

1 | 1 | A.B | m_{3} | A+B | M_{3} |

### 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 | m_{0} | A+B+C | M_{0} |

0 | 0 | 1 | A.B.C | m_{1} | A+B+C | M_{1} |

0 | 1 | 0 | A.B.C | m_{2} | A+B+C | M_{2} |

0 | 1 | 1 | A.B.C | m_{3} | A+B+C | M_{3} |

1 | 0 | 0 | A.B.C | m_{4} | A+B+C | M_{4} |

1 | 0 | 1 | A.B.C | m_{5} | A+B+C | M_{5} |

1 | 1 | 0 | A.B C | m_{6} | A+B+C | M_{6} |

1 | 1 | 1 | A.B.C | m_{7} | A+B+C | M_{7} |

**Example: Express the following in corresponding minterm and maxterm expression**

- Y = ABC + A.B.C + A.B. C + A. B. C
- 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 = m_{7}+ m_{3}+ m_{5}+ m_{4}= ∑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) = M_{0}+ M_{5}+ M_{7}= ∏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 m_{2}, m_{5}, m_{6} and m_{7} i.e.,

F = ∑m (2, 5,6, 7)

and maxterms will be M_{0}, M_{1}, M_{3} and M_{4} 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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