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


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

A∧B:
ABA∧B
000
010
100
111

(B∧A)∧(A∧B):
BAB∧AA∧B(B∧A)∧(A∧B)
00000
01000
10000
11111

((B∧A)∧(A∧B))∧X:
BAXB∧AA∧B(B∧A)∧(A∧B)((B∧A)∧(A∧B))∧X
0000000
0010000
0100000
0110000
1000000
1010000
1101110
1111111

(((B∧A)∧(A∧B))∧X)∧O:
BAXOB∧AA∧B(B∧A)∧(A∧B)((B∧A)∧(A∧B))∧X(((B∧A)∧(A∧B))∧X)∧O
000000000
000100000
001000000
001100000
010000000
010100000
011000000
011100000
100000000
100100000
101000000
101100000
110011100
110111100
111011110
111111111

((((B∧A)∧(A∧B))∧X)∧O)∧R:
BAXORB∧AA∧B(B∧A)∧(A∧B)((B∧A)∧(A∧B))∧X(((B∧A)∧(A∧B))∧X)∧O((((B∧A)∧(A∧B))∧X)∧O)∧R
00000000000
00001000000
00010000000
00011000000
00100000000
00101000000
00110000000
00111000000
01000000000
01001000000
01010000000
01011000000
01100000000
01101000000
01110000000
01111000000
10000000000
10001000000
10010000000
10011000000
10100000000
10101000000
10110000000
10111000000
11000111000
11001111000
11010111000
11011111000
11100111100
11101111100
11110111110
11111111111

(((((B∧A)∧(A∧B))∧X)∧O)∧R)∧B:
BAXORB∧AA∧B(B∧A)∧(A∧B)((B∧A)∧(A∧B))∧X(((B∧A)∧(A∧B))∧X)∧O((((B∧A)∧(A∧B))∧X)∧O)∧R(((((B∧A)∧(A∧B))∧X)∧O)∧R)∧B
000000000000
000010000000
000100000000
000110000000
001000000000
001010000000
001100000000
001110000000
010000000000
010010000000
010100000000
010110000000
011000000000
011010000000
011100000000
011110000000
100000000000
100010000000
100100000000
100110000000
101000000000
101010000000
101100000000
101110000000
110001110000
110011110000
110101110000
110111110000
111001111000
111011111000
111101111100
111111111111

Общая таблица истинности:

BAXORB∧AA∧B(B∧A)∧(A∧B)((B∧A)∧(A∧B))∧X(((B∧A)∧(A∧B))∧X)∧O((((B∧A)∧(A∧B))∧X)∧O)∧R(B∧A)∧(A∧B)∧X∧O∧R∧B
000000000000
000010000000
000100000000
000110000000
001000000000
001010000000
001100000000
001110000000
010000000000
010010000000
010100000000
010110000000
011000000000
011010000000
011100000000
011110000000
100000000000
100010000000
100100000000
100110000000
101000000000
101010000000
101100000000
101110000000
110001110000
110011110000
110101110000
110111110000
111001111000
111011111000
111101111100
111111111111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
BAXORF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111
Fсднф = B∧A∧X∧O∧R
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
BAXORF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111
Fскнф = (B∨A∨X∨O∨R) ∧ (B∨A∨X∨O∨¬R) ∧ (B∨A∨X∨¬O∨R) ∧ (B∨A∨X∨¬O∨¬R) ∧ (B∨A∨¬X∨O∨R) ∧ (B∨A∨¬X∨O∨¬R) ∧ (B∨A∨¬X∨¬O∨R) ∧ (B∨A∨¬X∨¬O∨¬R) ∧ (B∨¬A∨X∨O∨R) ∧ (B∨¬A∨X∨O∨¬R) ∧ (B∨¬A∨X∨¬O∨R) ∧ (B∨¬A∨X∨¬O∨¬R) ∧ (B∨¬A∨¬X∨O∨R) ∧ (B∨¬A∨¬X∨O∨¬R) ∧ (B∨¬A∨¬X∨¬O∨R) ∧ (B∨¬A∨¬X∨¬O∨¬R) ∧ (¬B∨A∨X∨O∨R) ∧ (¬B∨A∨X∨O∨¬R) ∧ (¬B∨A∨X∨¬O∨R) ∧ (¬B∨A∨X∨¬O∨¬R) ∧ (¬B∨A∨¬X∨O∨R) ∧ (¬B∨A∨¬X∨O∨¬R) ∧ (¬B∨A∨¬X∨¬O∨R) ∧ (¬B∨A∨¬X∨¬O∨¬R) ∧ (¬B∨¬A∨X∨O∨R) ∧ (¬B∨¬A∨X∨O∨¬R) ∧ (¬B∨¬A∨X∨¬O∨R) ∧ (¬B∨¬A∨X∨¬O∨¬R) ∧ (¬B∨¬A∨¬X∨O∨R) ∧ (¬B∨¬A∨¬X∨O∨¬R) ∧ (¬B∨¬A∨¬X∨¬O∨R)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
BAXORFж
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111

Построим полином Жегалкина:
Fж = C00000 ⊕ C10000∧B ⊕ C01000∧A ⊕ C00100∧X ⊕ C00010∧O ⊕ C00001∧R ⊕ C11000∧B∧A ⊕ C10100∧B∧X ⊕ C10010∧B∧O ⊕ C10001∧B∧R ⊕ C01100∧A∧X ⊕ C01010∧A∧O ⊕ C01001∧A∧R ⊕ C00110∧X∧O ⊕ C00101∧X∧R ⊕ C00011∧O∧R ⊕ C11100∧B∧A∧X ⊕ C11010∧B∧A∧O ⊕ C11001∧B∧A∧R ⊕ C10110∧B∧X∧O ⊕ C10101∧B∧X∧R ⊕ C10011∧B∧O∧R ⊕ C01110∧A∧X∧O ⊕ C01101∧A∧X∧R ⊕ C01011∧A∧O∧R ⊕ C00111∧X∧O∧R ⊕ C11110∧B∧A∧X∧O ⊕ C11101∧B∧A∧X∧R ⊕ C11011∧B∧A∧O∧R ⊕ C10111∧B∧X∧O∧R ⊕ C01111∧A∧X∧O∧R ⊕ C11111∧B∧A∧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 = 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 = 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 = 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 ⊕ 0 ⊕ 0 ⊕ 0 = 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 = 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 ⊕ 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 ⊕ 0 ⊕ 0 ⊕ 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 = 0 => С10111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
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 = 1 => С11111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1

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

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