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