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