Таблица истинности для функции (A⊕B)∧(A∧V∧B)≡A∧B:
Промежуточная таблица истинности: A⊕B
| A | B | A⊕B |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Промежуточная таблица истинности: A∧V
| A | V | A∧V |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
Промежуточная таблица истинности: (A∧V)∧B
| A | V | B | A∧V | (A∧V)∧B |
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 1 |
Промежуточная таблица истинности: (A⊕B)∧((A∧V)∧B)
| A | B | V | A⊕B | A∧V | (A∧V)∧B | (A⊕B)∧((A∧V)∧B) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 | 0 | 0 |
| 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 0 | 1 | 1 | 0 |
Промежуточная таблица истинности: A∧B
| A | B | A∧B |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
Промежуточная таблица истинности: ((A⊕B)∧((A∧V)∧B))≡(A∧B)
| A | B | V | A⊕B | A∧V | (A∧V)∧B | (A⊕B)∧((A∧V)∧B) | A∧B | ((A⊕B)∧((A∧V)∧B))≡(A∧B) |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 |
Общая таблица истинности:
| A | B | V | A⊕B | A∧V | (A∧V)∧B | (A⊕B)∧((A∧V)∧B) | A∧B | (A⊕B)∧(A∧V∧B)≡A∧B |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
| 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 0 | 1 | 1 | 0 | 1 | 0 |
Совершенная дизъюнктивная нормальная форма (СДНФ):
Fсднф = ¬A∧¬B∧¬V ∨ ¬A∧¬B∧V ∨ ¬A∧B∧¬V ∨ ¬A∧B∧V ∨ A∧¬B∧¬V ∨ A∧¬B∧VЛогическая cхема:
Совершенная конъюнктивная нормальная форма (СКНФ):
Fскнф = (¬A∨¬B∨V) ∧ (¬A∨¬B∨¬V)Логическая cхема:
Построение полинома Жегалкина:
Fж = C000 ⊕ C100∧A ⊕ C010∧B ⊕ C001∧V ⊕ C110∧A∧B ⊕ C101∧A∧V ⊕ C011∧B∧V ⊕ C111∧A∧B∧VТак как Fж(000) = 1, то С000 = 1.
Fж(100) = С000 ⊕ С100 = 1 => С100 = 1 ⊕ 1 = 0
Fж(010) = С000 ⊕ С010 = 1 => С010 = 1 ⊕ 1 = 0
Fж(001) = С000 ⊕ С001 = 1 => С001 = 1 ⊕ 1 = 0
Fж(110) = С000 ⊕ С100 ⊕ С010 ⊕ С110 = 0 => С110 = 1 ⊕ 0 ⊕ 0 ⊕ 0 = 1
Fж(101) = С000 ⊕ С100 ⊕ С001 ⊕ С101 = 1 => С101 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(011) = С000 ⊕ С010 ⊕ С001 ⊕ С011 = 1 => С011 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(111) = С000 ⊕ С100 ⊕ С010 ⊕ С001 ⊕ С110 ⊕ С101 ⊕ С011 ⊕ С111 = 0 => С111 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Таким образом, полином Жегалкина будет равен:
Fж = 1 ⊕ A∧B
Логическая схема, соответствующая полиному Жегалкина: