F(A, B, C, D) = Σ(0, 2, 6, 11, 13, 14)

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SN | A | B | C | D | Min-terms |

0 | 0 | 0 | 0 | 0 | A’B’C’D’ |

1 | 0 | 0 | 0 | 1 | A’B’C’D |

2 | 0 | 0 | 1 | 0 | A’BCD’ |

3 | 0 | 0 | 1 | 1 | A’B’CD |

4 | 0 | 1 | 0 | 0 | A’BC’D’ |

5 | 0 | 1 | 0 | 1 | A’BC’D |

6 | 0 | 1 | 1 | 0 | A’BCD’ |

7 | 0 | 1 | 1 | 1 | A’BCD |

8 | 1 | 0 | 0 | 0 | AB’C’D’ |

9 | 1 | 0 | 0 | 1 | AB’C’D |

10 | 1 | 0 | 1 | 0 | AB’CD’ |

11 | 1 | 0 | 1 | 1 | AB’CD |

12 | 1 | 1 | 0 | 0 | ABC’D’ |

13 | 1 | 1 | 0 | 1 | ABC’D |

14 | 1 | 1 | 1 | 0 | ABCD’ |

15 | 1 | 1 | 1 | 1 | ABCD |

Now, complement of following function is

F'(A, B, C, D) = Σ(1, 3, 4, 5, 7, 8, 9, 10, 12, 15)

So, sum of min-terms is

= A’B’C’D + A’B’CD + A’BC’D’ + A’BC’D + A’BCD + AB’C’D’ + AB’C’D + AB’CD’ + AB’CD’ + ABC’D’ + ABCD

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