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