Общая таблица истинности:
A | B | C | ABC | AB | ¬ABC | ¬C | ¬(¬ABC) | A∨B | (A∨B)∨(¬C) | (¬(¬ABC))∨AB | A∨B∨¬C≡¬(¬ABC)∨AB |
0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
A | B | C | ABC | AB | F |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 1 | 1 |
0 | 0 | 0 | 1 | 0 | 1 |
0 | 0 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 1 |
0 | 1 | 0 | 1 | 0 | 1 |
0 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 |
1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 1 |
F
сднф = ¬A∧¬B∧¬C∧¬ABC∧AB ∨ ¬A∧¬B∧¬C∧ABC∧¬AB ∨ ¬A∧¬B∧¬C∧ABC∧AB ∨ ¬A∧¬B∧C∧¬ABC∧¬AB ∨ ¬A∧B∧¬C∧¬ABC∧AB ∨ ¬A∧B∧¬C∧ABC∧¬AB ∨ ¬A∧B∧¬C∧ABC∧AB ∨ ¬A∧B∧C∧¬ABC∧AB ∨ ¬A∧B∧C∧ABC∧¬AB ∨ ¬A∧B∧C∧ABC∧AB ∨ A∧¬B∧¬C∧¬ABC∧AB ∨ A∧¬B∧¬C∧ABC∧¬AB ∨ A∧¬B∧¬C∧ABC∧AB ∨ A∧¬B∧C∧¬ABC∧AB ∨ A∧¬B∧C∧ABC∧¬AB ∨ A∧¬B∧C∧ABC∧AB ∨ A∧B∧¬C∧¬ABC∧AB ∨ A∧B∧¬C∧ABC∧¬AB ∨ A∧B∧¬C∧ABC∧AB ∨ A∧B∧C∧¬ABC∧AB ∨ A∧B∧C∧ABC∧¬AB ∨ A∧B∧C∧ABC∧AB
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
A | B | C | ABC | AB | F |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 1 | 1 |
0 | 0 | 0 | 1 | 0 | 1 |
0 | 0 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 1 |
0 | 1 | 0 | 1 | 0 | 1 |
0 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 |
1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 1 |
F
скнф = (A∨B∨C∨ABC∨AB) ∧ (A∨B∨¬C∨ABC∨¬AB) ∧ (A∨B∨¬C∨¬ABC∨AB) ∧ (A∨B∨¬C∨¬ABC∨¬AB) ∧ (A∨¬B∨C∨ABC∨AB) ∧ (A∨¬B∨¬C∨ABC∨AB) ∧ (¬A∨B∨C∨ABC∨AB) ∧ (¬A∨B∨¬C∨ABC∨AB) ∧ (¬A∨¬B∨C∨ABC∨AB) ∧ (¬A∨¬B∨¬C∨ABC∨AB)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
A | B | C | ABC | AB | Fж |
0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 1 | 1 |
0 | 0 | 0 | 1 | 0 | 1 |
0 | 0 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 0 | 1 |
0 | 0 | 1 | 0 | 1 | 0 |
0 | 0 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 1 |
0 | 1 | 0 | 1 | 0 | 1 |
0 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 1 |
1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 | 1 |
Построим полином Жегалкина:
F
ж = C
00000 ⊕ C
10000∧A ⊕ C
01000∧B ⊕ C
00100∧C ⊕ C
00010∧ABC ⊕ C
00001∧AB ⊕ C
11000∧A∧B ⊕ C
10100∧A∧C ⊕ C
10010∧A∧ABC ⊕ C
10001∧A∧AB ⊕ C
01100∧B∧C ⊕ C
01010∧B∧ABC ⊕ C
01001∧B∧AB ⊕ C
00110∧C∧ABC ⊕ C
00101∧C∧AB ⊕ C
00011∧ABC∧AB ⊕ C
11100∧A∧B∧C ⊕ C
11010∧A∧B∧ABC ⊕ C
11001∧A∧B∧AB ⊕ C
10110∧A∧C∧ABC ⊕ C
10101∧A∧C∧AB ⊕ C
10011∧A∧ABC∧AB ⊕ C
01110∧B∧C∧ABC ⊕ C
01101∧B∧C∧AB ⊕ C
01011∧B∧ABC∧AB ⊕ C
00111∧C∧ABC∧AB ⊕ C
11110∧A∧B∧C∧ABC ⊕ C
11101∧A∧B∧C∧AB ⊕ C
11011∧A∧B∧ABC∧AB ⊕ C
10111∧A∧C∧ABC∧AB ⊕ C
01111∧B∧C∧ABC∧AB ⊕ C
11111∧A∧B∧C∧ABC∧AB
Так как F
ж(00000) = 0, то С
00000 = 0.
Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
F
ж(10000) = С
00000 ⊕ С
10000 = 0 => С
10000 = 0 ⊕ 0 = 0
F
ж(01000) = С
00000 ⊕ С
01000 = 0 => С
01000 = 0 ⊕ 0 = 0
F
ж(00100) = С
00000 ⊕ С
00100 = 1 => С
00100 = 0 ⊕ 1 = 1
F
ж(00010) = С
00000 ⊕ С
00010 = 1 => С
00010 = 0 ⊕ 1 = 1
F
ж(00001) = С
00000 ⊕ С
00001 = 1 => С
00001 = 0 ⊕ 1 = 1
F
ж(11000) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
11000 = 0 => С
11000 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(10100) = С
00000 ⊕ С
10000 ⊕ С
00100 ⊕ С
10100 = 0 => С
10100 = 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
F
ж(10010) = С
00000 ⊕ С
10000 ⊕ С
00010 ⊕ С
10010 = 1 => С
10010 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(10001) = С
00000 ⊕ С
10000 ⊕ С
00001 ⊕ С
10001 = 1 => С
10001 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(01100) = С
00000 ⊕ С
01000 ⊕ С
00100 ⊕ С
01100 = 0 => С
01100 = 0 ⊕ 0 ⊕ 1 ⊕ 0 = 1
F
ж(01010) = С
00000 ⊕ С
01000 ⊕ С
00010 ⊕ С
01010 = 1 => С
01010 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(01001) = С
00000 ⊕ С
01000 ⊕ С
00001 ⊕ С
01001 = 1 => С
01001 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(00110) = С
00000 ⊕ С
00100 ⊕ С
00010 ⊕ С
00110 = 0 => С
00110 = 0 ⊕ 1 ⊕ 1 ⊕ 0 = 0
F
ж(00101) = С
00000 ⊕ С
00100 ⊕ С
00001 ⊕ С
00101 = 0 => С
00101 = 0 ⊕ 1 ⊕ 1 ⊕ 0 = 0
F
ж(00011) = С
00000 ⊕ С
00010 ⊕ С
00001 ⊕ С
00011 = 1 => С
00011 = 0 ⊕ 1 ⊕ 1 ⊕ 1 = 1
F
ж(11100) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
00100 ⊕ С
11000 ⊕ С
10100 ⊕ С
01100 ⊕ С
11100 = 0 => С
11100 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 = 1
F
ж(11010) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
00010 ⊕ С
11000 ⊕ С
10010 ⊕ С
01010 ⊕ С
11010 = 1 => С
11010 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(11001) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
00001 ⊕ С
11000 ⊕ С
10001 ⊕ С
01001 ⊕ С
11001 = 1 => С
11001 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(10110) = С
00000 ⊕ С
10000 ⊕ С
00100 ⊕ С
00010 ⊕ С
10100 ⊕ С
10010 ⊕ С
00110 ⊕ С
10110 = 1 => С
10110 = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(10101) = С
00000 ⊕ С
10000 ⊕ С
00100 ⊕ С
00001 ⊕ С
10100 ⊕ С
10001 ⊕ С
00101 ⊕ С
10101 = 1 => С
10101 = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(10011) = С
00000 ⊕ С
10000 ⊕ С
00010 ⊕ С
00001 ⊕ С
10010 ⊕ С
10001 ⊕ С
00011 ⊕ С
10011 = 1 => С
10011 = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(01110) = С
00000 ⊕ С
01000 ⊕ С
00100 ⊕ С
00010 ⊕ С
01100 ⊕ С
01010 ⊕ С
00110 ⊕ С
01110 = 1 => С
01110 = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(01101) = С
00000 ⊕ С
01000 ⊕ С
00100 ⊕ С
00001 ⊕ С
01100 ⊕ С
01001 ⊕ С
00101 ⊕ С
01101 = 1 => С
01101 = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(01011) = С
00000 ⊕ С
01000 ⊕ С
00010 ⊕ С
00001 ⊕ С
01010 ⊕ С
01001 ⊕ С
00011 ⊕ С
01011 = 1 => С
01011 = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(00111) = С
00000 ⊕ С
00100 ⊕ С
00010 ⊕ С
00001 ⊕ С
00110 ⊕ С
00101 ⊕ С
00011 ⊕ С
00111 = 0 => С
00111 = 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(11110) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
00100 ⊕ С
00010 ⊕ С
11000 ⊕ С
10100 ⊕ С
10010 ⊕ С
01100 ⊕ С
01010 ⊕ С
00110 ⊕ С
11100 ⊕ С
11010 ⊕ С
10110 ⊕ С
01110 ⊕ С
11110 = 1 => С
11110 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(11101) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
00100 ⊕ С
00001 ⊕ С
11000 ⊕ С
10100 ⊕ С
10001 ⊕ С
01100 ⊕ С
01001 ⊕ С
00101 ⊕ С
11100 ⊕ С
11001 ⊕ С
10101 ⊕ С
01101 ⊕ С
11101 = 1 => С
11101 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(11011) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
00010 ⊕ С
00001 ⊕ С
11000 ⊕ С
10010 ⊕ С
10001 ⊕ С
01010 ⊕ С
01001 ⊕ С
00011 ⊕ С
11010 ⊕ С
11001 ⊕ С
10011 ⊕ С
01011 ⊕ С
11011 = 1 => С
11011 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(10111) = С
00000 ⊕ С
10000 ⊕ С
00100 ⊕ С
00010 ⊕ С
00001 ⊕ С
10100 ⊕ С
10010 ⊕ С
10001 ⊕ С
00110 ⊕ С
00101 ⊕ С
00011 ⊕ С
10110 ⊕ С
10101 ⊕ С
10011 ⊕ С
00111 ⊕ С
10111 = 1 => С
10111 = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(01111) = С
00000 ⊕ С
01000 ⊕ С
00100 ⊕ С
00010 ⊕ С
00001 ⊕ С
01100 ⊕ С
01010 ⊕ С
01001 ⊕ С
00110 ⊕ С
00101 ⊕ С
00011 ⊕ С
01110 ⊕ С
01101 ⊕ С
01011 ⊕ С
00111 ⊕ С
01111 = 1 => С
01111 = 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(11111) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
00100 ⊕ С
00010 ⊕ С
00001 ⊕ С
11000 ⊕ С
10100 ⊕ С
10010 ⊕ С
10001 ⊕ С
01100 ⊕ С
01010 ⊕ С
01001 ⊕ С
00110 ⊕ С
00101 ⊕ С
00011 ⊕ С
11100 ⊕ С
11010 ⊕ С
11001 ⊕ С
10110 ⊕ С
10101 ⊕ С
10011 ⊕ С
01110 ⊕ С
01101 ⊕ С
01011 ⊕ С
00111 ⊕ С
11110 ⊕ С
11101 ⊕ С
11011 ⊕ С
10111 ⊕ С
01111 ⊕ С
11111 = 1 => С
11111 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Таким образом, полином Жегалкина будет равен:
F
ж = C ⊕ ABC ⊕ AB ⊕ A∧C ⊕ B∧C ⊕ ABC∧AB ⊕ A∧B∧C
Логическая схема, соответствующая полиному Жегалкина: