Промежуточные таблицы истинности:¬D:
A∧B:
C∧(¬D):
| C | D | ¬D | C∧(¬D) |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
(A∧B)∨(C∧(¬D)):
| A | B | C | D | A∧B | ¬D | C∧(¬D) | (A∧B)∨(C∧(¬D)) |
| 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 |
F≡((A∧B)∨(C∧(¬D))):
| F | A | B | C | D | A∧B | ¬D | C∧(¬D) | (A∧B)∨(C∧(¬D)) | F≡((A∧B)∨(C∧(¬D))) |
| 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
| 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
| 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
| 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
Общая таблица истинности:
| F | A | B | C | D | ¬D | A∧B | C∧(¬D) | (A∧B)∨(C∧(¬D)) | F≡A∧B∨C∧¬D |
| 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
| 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
| 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |
| 0 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 |
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
| F | A | B | C | D | F |
| 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 0 | 1 | 1 | 1 |
| 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 |
| 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 |
F
сднф = ¬F∧¬A∧¬B∧¬C∧¬D ∨ ¬F∧¬A∧¬B∧¬C∧D ∨ ¬F∧¬A∧¬B∧C∧D ∨ ¬F∧¬A∧B∧¬C∧¬D ∨ ¬F∧¬A∧B∧¬C∧D ∨ ¬F∧¬A∧B∧C∧D ∨ ¬F∧A∧¬B∧¬C∧¬D ∨ ¬F∧A∧¬B∧¬C∧D ∨ ¬F∧A∧¬B∧C∧D ∨ F∧¬A∧¬B∧C∧¬D ∨ F∧¬A∧B∧C∧¬D ∨ F∧A∧¬B∧C∧¬D ∨ F∧A∧B∧¬C∧¬D ∨ F∧A∧B∧¬C∧D ∨ F∧A∧B∧C∧¬D ∨ F∧A∧B∧C∧D
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
| F | A | B | C | D | F |
| 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 0 | 1 | 1 | 1 |
| 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 |
| 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 |
| 1 | 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 |
F
скнф = (F∨A∨B∨¬C∨D) ∧ (F∨A∨¬B∨¬C∨D) ∧ (F∨¬A∨B∨¬C∨D) ∧ (F∨¬A∨¬B∨C∨D) ∧ (F∨¬A∨¬B∨C∨¬D) ∧ (F∨¬A∨¬B∨¬C∨D) ∧ (F∨¬A∨¬B∨¬C∨¬D) ∧ (¬F∨A∨B∨C∨D) ∧ (¬F∨A∨B∨C∨¬D) ∧ (¬F∨A∨B∨¬C∨¬D) ∧ (¬F∨A∨¬B∨C∨D) ∧ (¬F∨A∨¬B∨C∨¬D) ∧ (¬F∨A∨¬B∨¬C∨¬D) ∧ (¬F∨¬A∨B∨C∨D) ∧ (¬F∨¬A∨B∨C∨¬D) ∧ (¬F∨¬A∨B∨¬C∨¬D)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
| F | A | B | C | D | Fж |
| 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 0 | 1 | 1 | 1 |
| 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 | 1 |
| 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 1 | 1 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 | 0 | 1 |
| 1 | 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 1 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 |
| 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 | 0 | 1 |
| 1 | 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 1 | 0 | 0 | 1 |
| 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∧F ⊕ C
01000∧A ⊕ C
00100∧B ⊕ C
00010∧C ⊕ C
00001∧D ⊕ C
11000∧F∧A ⊕ C
10100∧F∧B ⊕ C
10010∧F∧C ⊕ C
10001∧F∧D ⊕ C
01100∧A∧B ⊕ C
01010∧A∧C ⊕ C
01001∧A∧D ⊕ C
00110∧B∧C ⊕ C
00101∧B∧D ⊕ C
00011∧C∧D ⊕ C
11100∧F∧A∧B ⊕ C
11010∧F∧A∧C ⊕ C
11001∧F∧A∧D ⊕ C
10110∧F∧B∧C ⊕ C
10101∧F∧B∧D ⊕ C
10011∧F∧C∧D ⊕ C
01110∧A∧B∧C ⊕ C
01101∧A∧B∧D ⊕ C
01011∧A∧C∧D ⊕ C
00111∧B∧C∧D ⊕ C
11110∧F∧A∧B∧C ⊕ C
11101∧F∧A∧B∧D ⊕ C
11011∧F∧A∧C∧D ⊕ C
10111∧F∧B∧C∧D ⊕ C
01111∧A∧B∧C∧D ⊕ C
11111∧F∧A∧B∧C∧D
Так как F
ж(00000) = 1, то С
00000 = 1.
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 = 1 ⊕ 0 = 1
F
ж(01000) = С
00000 ⊕ С
01000 = 1 => С
01000 = 1 ⊕ 1 = 0
F
ж(00100) = С
00000 ⊕ С
00100 = 1 => С
00100 = 1 ⊕ 1 = 0
F
ж(00010) = С
00000 ⊕ С
00010 = 0 => С
00010 = 1 ⊕ 0 = 1
F
ж(00001) = С
00000 ⊕ С
00001 = 1 => С
00001 = 1 ⊕ 1 = 0
F
ж(11000) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
11000 = 0 => С
11000 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F
ж(10100) = С
00000 ⊕ С
10000 ⊕ С
00100 ⊕ С
10100 = 0 => С
10100 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F
ж(10010) = С
00000 ⊕ С
10000 ⊕ С
00010 ⊕ С
10010 = 1 => С
10010 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
F
ж(10001) = С
00000 ⊕ С
10000 ⊕ С
00001 ⊕ С
10001 = 0 => С
10001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
F
ж(01100) = С
00000 ⊕ С
01000 ⊕ С
00100 ⊕ С
01100 = 0 => С
01100 = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
F
ж(01010) = С
00000 ⊕ С
01000 ⊕ С
00010 ⊕ С
01010 = 0 => С
01010 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(01001) = С
00000 ⊕ С
01000 ⊕ С
00001 ⊕ С
01001 = 1 => С
01001 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(00110) = С
00000 ⊕ С
00100 ⊕ С
00010 ⊕ С
00110 = 0 => С
00110 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(00101) = С
00000 ⊕ С
00100 ⊕ С
00001 ⊕ С
00101 = 1 => С
00101 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(00011) = С
00000 ⊕ С
00010 ⊕ С
00001 ⊕ С
00011 = 1 => С
00011 = 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
F
ж(11100) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
00100 ⊕ С
11000 ⊕ С
10100 ⊕ С
01100 ⊕ С
11100 = 1 => С
11100 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(11010) = С
00000 ⊕ С
10000 ⊕ С
01000 ⊕ С
00010 ⊕ С
11000 ⊕ С
10010 ⊕ С
01010 ⊕ С
11010 = 1 => С
11010 = 1 ⊕ 1 ⊕ 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 = 0 => С
11001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(10110) = С
00000 ⊕ С
10000 ⊕ С
00100 ⊕ С
00010 ⊕ С
10100 ⊕ С
10010 ⊕ С
00110 ⊕ С
10110 = 1 => С
10110 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(10101) = С
00000 ⊕ С
10000 ⊕ С
00100 ⊕ С
00001 ⊕ С
10100 ⊕ С
10001 ⊕ С
00101 ⊕ С
10101 = 0 => С
10101 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(10011) = С
00000 ⊕ С
10000 ⊕ С
00010 ⊕ С
00001 ⊕ С
10010 ⊕ С
10001 ⊕ С
00011 ⊕ С
10011 = 0 => С
10011 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
F
ж(01110) = С
00000 ⊕ С
01000 ⊕ С
00100 ⊕ С
00010 ⊕ С
01100 ⊕ С
01010 ⊕ С
00110 ⊕ С
01110 = 0 => С
01110 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
F
ж(01101) = С
00000 ⊕ С
01000 ⊕ С
00100 ⊕ С
00001 ⊕ С
01100 ⊕ С
01001 ⊕ С
00101 ⊕ С
01101 = 0 => С
01101 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(01011) = С
00000 ⊕ С
01000 ⊕ С
00010 ⊕ С
00001 ⊕ С
01010 ⊕ С
01001 ⊕ С
00011 ⊕ С
01011 = 1 => С
01011 = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
F
ж(00111) = С
00000 ⊕ С
00100 ⊕ С
00010 ⊕ С
00001 ⊕ С
00110 ⊕ С
00101 ⊕ С
00011 ⊕ С
00111 = 1 => С
00111 = 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 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 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 0 => С
11011 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 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 = 0 => С
10111 = 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 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 = 0 => С
01111 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 1
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 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Таким образом, полином Жегалкина будет равен:
F
ж = 1 ⊕ F ⊕ C ⊕ A∧B ⊕ C∧D ⊕ A∧B∧C ⊕ A∧B∧C∧D
Логическая схема, соответствующая полиному Жегалкина:
