Таблица истинности для функции A∧¬B∧¬C∧¬D∧F:


Промежуточные таблицы истинности:
¬B:
B¬B
01
10

¬C:
C¬C
01
10

¬D:
D¬D
01
10

A∧(¬B):
AB¬BA∧(¬B)
0010
0100
1011
1100

(A∧(¬B))∧(¬C):
ABC¬BA∧(¬B)¬C(A∧(¬B))∧(¬C)
0001010
0011000
0100010
0110000
1001111
1011100
1100010
1110000

((A∧(¬B))∧(¬C))∧(¬D):
ABCD¬BA∧(¬B)¬C(A∧(¬B))∧(¬C)¬D((A∧(¬B))∧(¬C))∧(¬D)
0000101010
0001101000
0010100010
0011100000
0100001010
0101001000
0110000010
0111000000
1000111111
1001111100
1010110010
1011110000
1100001010
1101001000
1110000010
1111000000

(((A∧(¬B))∧(¬C))∧(¬D))∧F:
ABCDF¬BA∧(¬B)¬C(A∧(¬B))∧(¬C)¬D((A∧(¬B))∧(¬C))∧(¬D)(((A∧(¬B))∧(¬C))∧(¬D))∧F
000001010100
000011010100
000101010000
000111010000
001001000100
001011000100
001101000000
001111000000
010000010100
010010010100
010100010000
010110010000
011000000100
011010000100
011100000000
011110000000
100001111110
100011111111
100101111000
100111111000
101001100100
101011100100
101101100000
101111100000
110000010100
110010010100
110100010000
110110010000
111000000100
111010000100
111100000000
111110000000

Общая таблица истинности:

ABCDF¬B¬C¬DA∧(¬B)(A∧(¬B))∧(¬C)((A∧(¬B))∧(¬C))∧(¬D)A∧¬B∧¬C∧¬D∧F
000001110000
000011110000
000101100000
000111100000
001001010000
001011010000
001101000000
001111000000
010000110000
010010110000
010100100000
010110100000
011000010000
011010010000
011100000000
011110000000
100001111110
100011111111
100101101100
100111101100
101001011000
101011011000
101101001000
101111001000
110000110000
110010110000
110100100000
110110100000
111000010000
111010010000
111100000000
111110000000

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ABCDFF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100011
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110
Fсднф = A∧¬B∧¬C∧¬D∧F
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ABCDFF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100011
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110
Fскнф = (A∨B∨C∨D∨F) ∧ (A∨B∨C∨D∨¬F) ∧ (A∨B∨C∨¬D∨F) ∧ (A∨B∨C∨¬D∨¬F) ∧ (A∨B∨¬C∨D∨F) ∧ (A∨B∨¬C∨D∨¬F) ∧ (A∨B∨¬C∨¬D∨F) ∧ (A∨B∨¬C∨¬D∨¬F) ∧ (A∨¬B∨C∨D∨F) ∧ (A∨¬B∨C∨D∨¬F) ∧ (A∨¬B∨C∨¬D∨F) ∧ (A∨¬B∨C∨¬D∨¬F) ∧ (A∨¬B∨¬C∨D∨F) ∧ (A∨¬B∨¬C∨D∨¬F) ∧ (A∨¬B∨¬C∨¬D∨F) ∧ (A∨¬B∨¬C∨¬D∨¬F) ∧ (¬A∨B∨C∨D∨F) ∧ (¬A∨B∨C∨¬D∨F) ∧ (¬A∨B∨C∨¬D∨¬F) ∧ (¬A∨B∨¬C∨D∨F) ∧ (¬A∨B∨¬C∨D∨¬F) ∧ (¬A∨B∨¬C∨¬D∨F) ∧ (¬A∨B∨¬C∨¬D∨¬F) ∧ (¬A∨¬B∨C∨D∨F) ∧ (¬A∨¬B∨C∨D∨¬F) ∧ (¬A∨¬B∨C∨¬D∨F) ∧ (¬A∨¬B∨C∨¬D∨¬F) ∧ (¬A∨¬B∨¬C∨D∨F) ∧ (¬A∨¬B∨¬C∨D∨¬F) ∧ (¬A∨¬B∨¬C∨¬D∨F) ∧ (¬A∨¬B∨¬C∨¬D∨¬F)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ABCDFFж
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100011
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111110

Построим полином Жегалкина:
Fж = C00000 ⊕ C10000∧A ⊕ C01000∧B ⊕ C00100∧C ⊕ C00010∧D ⊕ C00001∧F ⊕ C11000∧A∧B ⊕ C10100∧A∧C ⊕ C10010∧A∧D ⊕ C10001∧A∧F ⊕ C01100∧B∧C ⊕ C01010∧B∧D ⊕ C01001∧B∧F ⊕ C00110∧C∧D ⊕ C00101∧C∧F ⊕ C00011∧D∧F ⊕ C11100∧A∧B∧C ⊕ C11010∧A∧B∧D ⊕ C11001∧A∧B∧F ⊕ C10110∧A∧C∧D ⊕ C10101∧A∧C∧F ⊕ C10011∧A∧D∧F ⊕ C01110∧B∧C∧D ⊕ C01101∧B∧C∧F ⊕ C01011∧B∧D∧F ⊕ C00111∧C∧D∧F ⊕ C11110∧A∧B∧C∧D ⊕ C11101∧A∧B∧C∧F ⊕ C11011∧A∧B∧D∧F ⊕ C10111∧A∧C∧D∧F ⊕ C01111∧B∧C∧D∧F ⊕ C11111∧A∧B∧C∧D∧F

Так как Fж(00000) = 0, то С00000 = 0.

Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
Fж(10000) = С00000 ⊕ С10000 = 0 => С10000 = 0 ⊕ 0 = 0
Fж(01000) = С00000 ⊕ С01000 = 0 => С01000 = 0 ⊕ 0 = 0
Fж(00100) = С00000 ⊕ С00100 = 0 => С00100 = 0 ⊕ 0 = 0
Fж(00010) = С00000 ⊕ С00010 = 0 => С00010 = 0 ⊕ 0 = 0
Fж(00001) = С00000 ⊕ С00001 = 0 => С00001 = 0 ⊕ 0 = 0
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 0 => С11000 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 0 => С10100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 0 => С10010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 1 => С10001 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 0 => С01100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 0 => С01010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 0 => С01001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 0 => С00110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 0 => С00101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 0 => С00011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 0 => С11100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 0 => С10110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 0 => С10101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 0 => С01110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 0 => С01101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 0 => С11110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 0 => С11101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 0 => С11011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 0 => С10111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 0 => С01111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11111) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С11100 ⊕ С11010 ⊕ С11001 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С11110 ⊕ С11101 ⊕ С11011 ⊕ С10111 ⊕ С01111 ⊕ С11111 = 0 => С11111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 = 1

Таким образом, полином Жегалкина будет равен:
Fж = A∧F ⊕ A∧B∧F ⊕ A∧C∧F ⊕ A∧D∧F ⊕ A∧B∧C∧F ⊕ A∧B∧D∧F ⊕ A∧C∧D∧F ⊕ A∧B∧C∧D∧F
Логическая схема, соответствующая полиному Жегалкина:

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