Таблица истинности для функции A→¬B≡C∧D∨E:


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

C∧D:
CDC∧D
000
010
100
111

(C∧D)∨E:
CDEC∧D(C∧D)∨E
00000
00101
01000
01101
10000
10101
11011
11111

A→(¬B):
AB¬BA→(¬B)
0011
0101
1011
1100

(A→(¬B))≡((C∧D)∨E):
ABCDE¬BA→(¬B)C∧D(C∧D)∨E(A→(¬B))≡((C∧D)∨E)
0000011000
0000111011
0001011000
0001111011
0010011000
0010111011
0011011111
0011111111
0100001000
0100101011
0101001000
0101101011
0110001000
0110101011
0111001111
0111101111
1000011000
1000111011
1001011000
1001111011
1010011000
1010111011
1011011111
1011111111
1100000001
1100100010
1101000001
1101100010
1110000001
1110100010
1111000110
1111100110

Общая таблица истинности:

ABCDE¬BC∧D(C∧D)∨EA→(¬B)A→¬B≡C∧D∨E
0000010010
0000110111
0001010010
0001110111
0010010010
0010110111
0011011111
0011111111
0100000010
0100100111
0101000010
0101100111
0110000010
0110100111
0111001111
0111101111
1000010010
1000110111
1001010010
1001110111
1010010010
1010110111
1011011111
1011111111
1100000001
1100100100
1101000001
1101100100
1110000001
1110100100
1111001100
1111101100

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ABCDEF
000000
000011
000100
000111
001000
001011
001101
001111
010000
010011
010100
010111
011000
011011
011101
011111
100000
100011
100100
100111
101000
101011
101101
101111
110001
110010
110101
110110
111001
111010
111100
111110
Fсднф = ¬A∧¬B∧¬C∧¬D∧E ∨ ¬A∧¬B∧¬C∧D∧E ∨ ¬A∧¬B∧C∧¬D∧E ∨ ¬A∧¬B∧C∧D∧¬E ∨ ¬A∧¬B∧C∧D∧E ∨ ¬A∧B∧¬C∧¬D∧E ∨ ¬A∧B∧¬C∧D∧E ∨ ¬A∧B∧C∧¬D∧E ∨ ¬A∧B∧C∧D∧¬E ∨ ¬A∧B∧C∧D∧E ∨ A∧¬B∧¬C∧¬D∧E ∨ A∧¬B∧¬C∧D∧E ∨ A∧¬B∧C∧¬D∧E ∨ A∧¬B∧C∧D∧¬E ∨ A∧¬B∧C∧D∧E ∨ A∧B∧¬C∧¬D∧¬E ∨ A∧B∧¬C∧D∧¬E ∨ A∧B∧C∧¬D∧¬E
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ABCDEF
000000
000011
000100
000111
001000
001011
001101
001111
010000
010011
010100
010111
011000
011011
011101
011111
100000
100011
100100
100111
101000
101011
101101
101111
110001
110010
110101
110110
111001
111010
111100
111110
Fскнф = (A∨B∨C∨D∨E) ∧ (A∨B∨C∨¬D∨E) ∧ (A∨B∨¬C∨D∨E) ∧ (A∨¬B∨C∨D∨E) ∧ (A∨¬B∨C∨¬D∨E) ∧ (A∨¬B∨¬C∨D∨E) ∧ (¬A∨B∨C∨D∨E) ∧ (¬A∨B∨C∨¬D∨E) ∧ (¬A∨B∨¬C∨D∨E) ∧ (¬A∨¬B∨C∨D∨¬E) ∧ (¬A∨¬B∨C∨¬D∨¬E) ∧ (¬A∨¬B∨¬C∨D∨¬E) ∧ (¬A∨¬B∨¬C∨¬D∨E) ∧ (¬A∨¬B∨¬C∨¬D∨¬E)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ABCDEFж
000000
000011
000100
000111
001000
001011
001101
001111
010000
010011
010100
010111
011000
011011
011101
011111
100000
100011
100100
100111
101000
101011
101101
101111
110001
110010
110101
110110
111001
111010
111100
111110

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

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 = 1 => С00001 = 0 ⊕ 1 = 1
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 1 => С11000 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
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 ⊕ 1 ⊕ 1 = 0
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 = 1 => С01001 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 1 => С00110 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 1 => С00101 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 1 => С00011 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 1 => С11100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 1 => С11010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 1 => С10110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 1 => С10101 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 1 => С10011 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 1 => С01110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 1 => С01101 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 1 => С01011 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 1 => С00111 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 1
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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 0 => С11011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 1 => С10111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 1 => С01111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 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 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0

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

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