Промежуточные таблицы истинности:¬A:
¬B:
(¬A)∧B:
A | B | ¬A | (¬A)∧B |
0 | 0 | 1 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 |
1 | 1 | 0 | 0 |
((¬A)∧B)∧C:
A | B | C | ¬A | (¬A)∧B | ((¬A)∧B)∧C |
0 | 0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 0 | 0 |
(¬B)∧D:
B | D | ¬B | (¬B)∧D |
0 | 0 | 1 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 |
1 | 1 | 0 | 0 |
(((¬A)∧B)∧C)→((¬B)∧D):
A | B | C | D | ¬A | (¬A)∧B | ((¬A)∧B)∧C | ¬B | (¬B)∧D | (((¬A)∧B)∧C)→((¬B)∧D) |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 |
0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 |
0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
¬C:
(¬C)∧D:
C | D | ¬C | (¬C)∧D |
0 | 0 | 1 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 |
1 | 1 | 0 | 0 |
((¬C)∧D)→D:
C | D | ¬C | (¬C)∧D | ((¬C)∧D)→D |
0 | 0 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 1 |
((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D):
A | B | C | D | ¬A | (¬A)∧B | ((¬A)∧B)∧C | ¬B | (¬B)∧D | (((¬A)∧B)∧C)→((¬B)∧D) | ¬C | (¬C)∧D | ((¬C)∧D)→D | ((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D) |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 |
0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 |
(((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D))∨B:
A | B | C | D | ¬A | (¬A)∧B | ((¬A)∧B)∧C | ¬B | (¬B)∧D | (((¬A)∧B)∧C)→((¬B)∧D) | ¬C | (¬C)∧D | ((¬C)∧D)→D | ((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D) | (((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D))∨B |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
A⊕(¬B):
A | B | ¬B | A⊕(¬B) |
0 | 0 | 1 | 1 |
0 | 1 | 0 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 |
B∧C:
(A⊕(¬B))→(B∧C):
A | B | C | ¬B | A⊕(¬B) | B∧C | (A⊕(¬B))→(B∧C) |
0 | 0 | 0 | 1 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 0 | 0 |
0 | 1 | 0 | 0 | 0 | 0 | 1 |
0 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 1 |
1 | 1 | 0 | 0 | 1 | 0 | 0 |
1 | 1 | 1 | 0 | 1 | 1 | 1 |
((((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D))∨B)→((A⊕(¬B))→(B∧C)):
A | B | C | D | ¬A | (¬A)∧B | ((¬A)∧B)∧C | ¬B | (¬B)∧D | (((¬A)∧B)∧C)→((¬B)∧D) | ¬C | (¬C)∧D | ((¬C)∧D)→D | ((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D) | (((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D))∨B | ¬B | A⊕(¬B) | B∧C | (A⊕(¬B))→(B∧C) | ((((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D))∨B)→((A⊕(¬B))→(B∧C)) |
0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |
Общая таблица истинности:
A | B | C | D | ¬A | ¬B | (¬A)∧B | ((¬A)∧B)∧C | (¬B)∧D | (((¬A)∧B)∧C)→((¬B)∧D) | ¬C | (¬C)∧D | ((¬C)∧D)→D | ((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D) | (((((¬A)∧B)∧C)→((¬B)∧D))∧(((¬C)∧D)→D))∨B | A⊕(¬B) | B∧C | (A⊕(¬B))→(B∧C) | (((¬A∧B∧C→¬B∧D)∧(¬C∧D→D))∨B)→((A⊕¬B)→B∧C) |
0 | 0 | 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
0 | 1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 |
0 | 1 | 1 | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 |
1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |
1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Логическая схема:
Совершенная дизъюнктивная нормальная форма (СДНФ):
По таблице истинности:
A | B | C | D | F |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 |
F
сднф = ¬A∧B∧¬C∧¬D ∨ ¬A∧B∧¬C∧D ∨ ¬A∧B∧C∧¬D ∨ ¬A∧B∧C∧D ∨ A∧¬B∧¬C∧¬D ∨ A∧¬B∧¬C∧D ∨ A∧¬B∧C∧¬D ∨ A∧¬B∧C∧D ∨ A∧B∧C∧¬D ∨ A∧B∧C∧D
Логическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
По таблице истинности:
A | B | C | D | F |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 |
F
скнф = (A∨B∨C∨D) ∧ (A∨B∨C∨¬D) ∧ (A∨B∨¬C∨D) ∧ (A∨B∨¬C∨¬D) ∧ (¬A∨¬B∨C∨D) ∧ (¬A∨¬B∨C∨¬D)
Логическая cхема:
Построение полинома Жегалкина:
По таблице истинности функции
A | B | C | D | Fж |
0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
0 | 0 | 1 | 0 | 0 |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 0 | 0 | 1 |
0 | 1 | 0 | 1 | 1 |
0 | 1 | 1 | 0 | 1 |
0 | 1 | 1 | 1 | 1 |
1 | 0 | 0 | 0 | 1 |
1 | 0 | 0 | 1 | 1 |
1 | 0 | 1 | 0 | 1 |
1 | 0 | 1 | 1 | 1 |
1 | 1 | 0 | 0 | 0 |
1 | 1 | 0 | 1 | 0 |
1 | 1 | 1 | 0 | 1 |
1 | 1 | 1 | 1 | 1 |
Построим полином Жегалкина:
F
ж = C
0000 ⊕ C
1000∧A ⊕ C
0100∧B ⊕ C
0010∧C ⊕ C
0001∧D ⊕ C
1100∧A∧B ⊕ C
1010∧A∧C ⊕ C
1001∧A∧D ⊕ C
0110∧B∧C ⊕ C
0101∧B∧D ⊕ C
0011∧C∧D ⊕ C
1110∧A∧B∧C ⊕ C
1101∧A∧B∧D ⊕ C
1011∧A∧C∧D ⊕ C
0111∧B∧C∧D ⊕ C
1111∧A∧B∧C∧D
Так как F
ж(0000) = 0, то С
0000 = 0.
Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
F
ж(1000) = С
0000 ⊕ С
1000 = 1 => С
1000 = 0 ⊕ 1 = 1
F
ж(0100) = С
0000 ⊕ С
0100 = 1 => С
0100 = 0 ⊕ 1 = 1
F
ж(0010) = С
0000 ⊕ С
0010 = 0 => С
0010 = 0 ⊕ 0 = 0
F
ж(0001) = С
0000 ⊕ С
0001 = 0 => С
0001 = 0 ⊕ 0 = 0
F
ж(1100) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
1100 = 0 => С
1100 = 0 ⊕ 1 ⊕ 1 ⊕ 0 = 0
F
ж(1010) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
1010 = 1 => С
1010 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F
ж(1001) = С
0000 ⊕ С
1000 ⊕ С
0001 ⊕ С
1001 = 1 => С
1001 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F
ж(0110) = С
0000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0110 = 1 => С
0110 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F
ж(0101) = С
0000 ⊕ С
0100 ⊕ С
0001 ⊕ С
0101 = 1 => С
0101 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
F
ж(0011) = С
0000 ⊕ С
0010 ⊕ С
0001 ⊕ С
0011 = 0 => С
0011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(1110) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0010 ⊕ С
1100 ⊕ С
1010 ⊕ С
0110 ⊕ С
1110 = 1 => С
1110 = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
F
ж(1101) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0001 ⊕ С
1100 ⊕ С
1001 ⊕ С
0101 ⊕ С
1101 = 0 => С
1101 = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
F
ж(1011) = С
0000 ⊕ С
1000 ⊕ С
0010 ⊕ С
0001 ⊕ С
1010 ⊕ С
1001 ⊕ С
0011 ⊕ С
1011 = 1 => С
1011 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(0111) = С
0000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0001 ⊕ С
0110 ⊕ С
0101 ⊕ С
0011 ⊕ С
0111 = 1 => С
0111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
F
ж(1111) = С
0000 ⊕ С
1000 ⊕ С
0100 ⊕ С
0010 ⊕ С
0001 ⊕ С
1100 ⊕ С
1010 ⊕ С
1001 ⊕ С
0110 ⊕ С
0101 ⊕ С
0011 ⊕ С
1110 ⊕ С
1101 ⊕ С
1011 ⊕ С
0111 ⊕ С
1111 = 1 => С
1111 = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Таким образом, полином Жегалкина будет равен:
F
ж = A ⊕ B ⊕ A∧B∧C
Логическая схема, соответствующая полиному Жегалкина: