Таблица истинности для функции A∨(B∨C)∧E∨D:


Промежуточные таблицы истинности:
B∨C:
BCB∨C
000
011
101
111

(B∨C)∧E:
BCEB∨C(B∨C)∧E
00000
00100
01010
01111
10010
10111
11010
11111

A∨((B∨C)∧E):
ABCEB∨C(B∨C)∧EA∨((B∨C)∧E)
0000000
0001000
0010100
0011111
0100100
0101111
0110100
0111111
1000001
1001001
1010101
1011111
1100101
1101111
1110101
1111111

(A∨((B∨C)∧E))∨D:
ABCEDB∨C(B∨C)∧EA∨((B∨C)∧E)(A∨((B∨C)∧E))∨D
000000000
000010001
000100000
000110001
001001000
001011001
001101111
001111111
010001000
010011001
010101111
010111111
011001000
011011001
011101111
011111111
100000011
100010011
100100011
100110011
101001011
101011011
101101111
101111111
110001011
110011011
110101111
110111111
111001011
111011011
111101111
111111111

Общая таблица истинности:

ABCEDB∨C(B∨C)∧EA∨((B∨C)∧E)A∨(B∨C)∧E∨D
000000000
000010001
000100000
000110001
001001000
001011001
001101111
001111111
010001000
010011001
010101111
010111111
011001000
011011001
011101111
011111111
100000011
100010011
100100011
100110011
101001011
101011011
101101111
101111111
110001011
110011011
110101111
110111111
111001011
111011011
111101111
111111111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ABCEDF
000000
000011
000100
000111
001000
001011
001101
001111
010000
010011
010101
010111
011000
011011
011101
011111
100001
100011
100101
100111
101001
101011
101101
101111
110001
110011
110101
110111
111001
111011
111101
111111
Fсднф = ¬A∧¬B∧¬C∧¬E∧D ∨ ¬A∧¬B∧¬C∧E∧D ∨ ¬A∧¬B∧C∧¬E∧D ∨ ¬A∧¬B∧C∧E∧¬D ∨ ¬A∧¬B∧C∧E∧D ∨ ¬A∧B∧¬C∧¬E∧D ∨ ¬A∧B∧¬C∧E∧¬D ∨ ¬A∧B∧¬C∧E∧D ∨ ¬A∧B∧C∧¬E∧D ∨ ¬A∧B∧C∧E∧¬D ∨ ¬A∧B∧C∧E∧D ∨ A∧¬B∧¬C∧¬E∧¬D ∨ A∧¬B∧¬C∧¬E∧D ∨ A∧¬B∧¬C∧E∧¬D ∨ A∧¬B∧¬C∧E∧D ∨ A∧¬B∧C∧¬E∧¬D ∨ A∧¬B∧C∧¬E∧D ∨ A∧¬B∧C∧E∧¬D ∨ A∧¬B∧C∧E∧D ∨ A∧B∧¬C∧¬E∧¬D ∨ A∧B∧¬C∧¬E∧D ∨ A∧B∧¬C∧E∧¬D ∨ A∧B∧¬C∧E∧D ∨ A∧B∧C∧¬E∧¬D ∨ A∧B∧C∧¬E∧D ∨ A∧B∧C∧E∧¬D ∨ A∧B∧C∧E∧D
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ABCEDF
000000
000011
000100
000111
001000
001011
001101
001111
010000
010011
010101
010111
011000
011011
011101
011111
100001
100011
100101
100111
101001
101011
101101
101111
110001
110011
110101
110111
111001
111011
111101
111111
Fскнф = (A∨B∨C∨E∨D) ∧ (A∨B∨C∨¬E∨D) ∧ (A∨B∨¬C∨E∨D) ∧ (A∨¬B∨C∨E∨D) ∧ (A∨¬B∨¬C∨E∨D)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ABCEDFж
000000
000011
000100
000111
001000
001011
001101
001111
010000
010011
010101
010111
011000
011011
011101
011111
100001
100011
100101
100111
101001
101011
101101
101111
110001
110011
110101
110111
111001
111011
111101
111111

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

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 = 1 => С10000 = 0 ⊕ 1 = 1
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 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 1 => С10100 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 1 => С10010 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 1 => С10001 = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 0 => С01100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 1 => С01010 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 1 => С11010 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 1
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 1 => С11001 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 1 => С10110 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 1
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 1 => С10101 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 1 => С10011 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 1 => С01110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 1
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 = 1 => С11110 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 = 1
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 1 => С11101 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 1 => С11011 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 1
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 1 => С10111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 1
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 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
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 = 1 => С11111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 = 1

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

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