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