Таблица истинности для функции (¬(A∧B))≡(A∨¬B)∧X∧O∧R∧A:


Промежуточные таблицы истинности:
A∧B:
ABA∧B
000
010
100
111

¬(A∧B):
ABA∧B¬(A∧B)
0001
0101
1001
1110

¬B:
B¬B
01
10

A∨(¬B):
AB¬BA∨(¬B)
0011
0100
1011
1101

(A∨(¬B))∧X:
ABX¬BA∨(¬B)(A∨(¬B))∧X
000110
001111
010000
011000
100110
101111
110010
111011

((A∨(¬B))∧X)∧O:
ABXO¬BA∨(¬B)(A∨(¬B))∧X((A∨(¬B))∧X)∧O
00001100
00011100
00101110
00111111
01000000
01010000
01100000
01110000
10001100
10011100
10101110
10111111
11000100
11010100
11100110
11110111

(((A∨(¬B))∧X)∧O)∧R:
ABXOR¬BA∨(¬B)(A∨(¬B))∧X((A∨(¬B))∧X)∧O(((A∨(¬B))∧X)∧O)∧R
0000011000
0000111000
0001011000
0001111000
0010011100
0010111100
0011011110
0011111111
0100000000
0100100000
0101000000
0101100000
0110000000
0110100000
0111000000
0111100000
1000011000
1000111000
1001011000
1001111000
1010011100
1010111100
1011011110
1011111111
1100001000
1100101000
1101001000
1101101000
1110001100
1110101100
1111001110
1111101111

((((A∨(¬B))∧X)∧O)∧R)∧A:
ABXOR¬BA∨(¬B)(A∨(¬B))∧X((A∨(¬B))∧X)∧O(((A∨(¬B))∧X)∧O)∧R((((A∨(¬B))∧X)∧O)∧R)∧A
00000110000
00001110000
00010110000
00011110000
00100111000
00101111000
00110111100
00111111110
01000000000
01001000000
01010000000
01011000000
01100000000
01101000000
01110000000
01111000000
10000110000
10001110000
10010110000
10011110000
10100111000
10101111000
10110111100
10111111111
11000010000
11001010000
11010010000
11011010000
11100011000
11101011000
11110011100
11111011111

(¬(A∧B))≡(((((A∨(¬B))∧X)∧O)∧R)∧A):
ABXORA∧B¬(A∧B)¬BA∨(¬B)(A∨(¬B))∧X((A∨(¬B))∧X)∧O(((A∨(¬B))∧X)∧O)∧R((((A∨(¬B))∧X)∧O)∧R)∧A(¬(A∧B))≡(((((A∨(¬B))∧X)∧O)∧R)∧A)
00000011100000
00001011100000
00010011100000
00011011100000
00100011110000
00101011110000
00110011111000
00111011111100
01000010000000
01001010000000
01010010000000
01011010000000
01100010000000
01101010000000
01110010000000
01111010000000
10000011100000
10001011100000
10010011100000
10011011100000
10100011110000
10101011110000
10110011111000
10111011111111
11000100100001
11001100100001
11010100100001
11011100100001
11100100110001
11101100110001
11110100111001
11111100111110

Общая таблица истинности:

ABXORA∧B¬(A∧B)¬BA∨(¬B)(A∨(¬B))∧X((A∨(¬B))∧X)∧O(((A∨(¬B))∧X)∧O)∧R((((A∨(¬B))∧X)∧O)∧R)∧A(¬(A∧B))≡(A∨¬B)∧X∧O∧R∧A
00000011100000
00001011100000
00010011100000
00011011100000
00100011110000
00101011110000
00110011111000
00111011111100
01000010000000
01001010000000
01010010000000
01011010000000
01100010000000
01101010000000
01110010000000
01111010000000
10000011100000
10001011100000
10010011100000
10011011100000
10100011110000
10101011110000
10110011111000
10111011111111
11000100100001
11001100100001
11010100100001
11011100100001
11100100110001
11101100110001
11110100111001
11111100111110

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ABXORF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101111
110001
110011
110101
110111
111001
111011
111101
111110
Fсднф = A∧¬B∧X∧O∧R ∨ A∧B∧¬X∧¬O∧¬R ∨ A∧B∧¬X∧¬O∧R ∨ A∧B∧¬X∧O∧¬R ∨ A∧B∧¬X∧O∧R ∨ A∧B∧X∧¬O∧¬R ∨ A∧B∧X∧¬O∧R ∨ A∧B∧X∧O∧¬R
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ABXORF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101111
110001
110011
110101
110111
111001
111011
111101
111110
Fскнф = (A∨B∨X∨O∨R) ∧ (A∨B∨X∨O∨¬R) ∧ (A∨B∨X∨¬O∨R) ∧ (A∨B∨X∨¬O∨¬R) ∧ (A∨B∨¬X∨O∨R) ∧ (A∨B∨¬X∨O∨¬R) ∧ (A∨B∨¬X∨¬O∨R) ∧ (A∨B∨¬X∨¬O∨¬R) ∧ (A∨¬B∨X∨O∨R) ∧ (A∨¬B∨X∨O∨¬R) ∧ (A∨¬B∨X∨¬O∨R) ∧ (A∨¬B∨X∨¬O∨¬R) ∧ (A∨¬B∨¬X∨O∨R) ∧ (A∨¬B∨¬X∨O∨¬R) ∧ (A∨¬B∨¬X∨¬O∨R) ∧ (A∨¬B∨¬X∨¬O∨¬R) ∧ (¬A∨B∨X∨O∨R) ∧ (¬A∨B∨X∨O∨¬R) ∧ (¬A∨B∨X∨¬O∨R) ∧ (¬A∨B∨X∨¬O∨¬R) ∧ (¬A∨B∨¬X∨O∨R) ∧ (¬A∨B∨¬X∨O∨¬R) ∧ (¬A∨B∨¬X∨¬O∨R) ∧ (¬A∨¬B∨¬X∨¬O∨¬R)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ABXORFж
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101111
110001
110011
110101
110111
111001
111011
111101
111110

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

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 = 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 = 0 => С10001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 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 = 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 = 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 = 1 => С11001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
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 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
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 = 1 => С11110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 1 => С11101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 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 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 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 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 0

Таким образом, полином Жегалкина будет равен:
Fж = A∧B ⊕ A∧X∧O∧R
Логическая схема, соответствующая полиному Жегалкина:

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