**Standard Forms**

**each term in the function****can have any number of literals.**

**Example, F1 = a +b’c + cde **

**There are 5 variables in F1 (a,b,c,d,e)**- Canonical form terms should be written as m0, m1… (sum of Products) or M0, M1, M2… (Product of sums).
- Each term in the canonical form should have all the literals.

**Example: F1 = ab’c’d’e’ + abcd’e’ + abcde**

__More about Canonical Forms__

**Minterms**

- It is
**Sum of Products** - Canonical form is Sum of Minterms
- three variable minterms are shown below

x | a | b | c | minterms |

0 | 0 | 0 | 0 | m0=a’.b’.c’ |

1 | 0 | 0 | 1 | m1=a’.b’.c |

2 | 0 | 1 | 0 | m2=a’.b.c’ |

3 | 0 | 1 | 1 | m3=a’.b.c |

4 | 1 | 0 | 0 | m4=a.b’.c’ |

5 | 1 | 0 | 1 | m5=a.b’.c |

6 | 1 | 1 | 0 | m6=a.b.c’ |

7 | 1 | 1 | 1 | m7=a.b.c |

**Maxterms**

- Product of Sum (PoS)
- Canonical form is Product of MaxTerms
- three variable maxterms are shown below

x | a | b | c | minterms |

0 | 0 | 0 | 0 | M0=(a+b+c) |

1 | 0 | 0 | 1 | M1=(a+b+c’) |

2 | 0 | 1 | 0 | M2=(a+b’+c) |

3 | 0 | 1 | 1 | M3=(a+b’+c’) |

4 | 1 | 0 | 0 | M4=(a’+b+c) |

5 | 1 | 0 | 1 | M5=(a’+b+c’) |

6 | 1 | 1 | 0 | M6=(a’+b’+c) |

7 | 1 | 1 | 1 | M7=(a’+b’+c’) |

usually

M_{i} = (m_{j})’ |

Express the boolean function F = A + BC in a sum of minterms.

The function has three variables,

so * F = A + BC* will be

**F = A(B + B’) + (A +A’) BC ***[since, x + x’ =1]*

**F = AB + AB’ + ABC + A’BC**

**F = AB(C+C’) + AB’(C+C’) + ABC + A’BC**

**F = ABC + ABC’ + AB’C + AB’C’ + ABC + A’BC**

**F = ABC + ABC’ + AB’C + AB’C’ + A’BC ***[since x + x = x]*

**F= m0 + m6 + m5 + m4 + m3**