Для функции F≡A∧V∧¬B∧C:


Промежуточные таблицы истинности:
¬B:
B¬B
01
10

A∧V:
AVA∧V
000
010
100
111

(A∧V)∧(¬B):
AVBA∧V¬B(A∧V)∧(¬B)
000010
001000
010010
011000
100010
101000
110111
111100

((A∧V)∧(¬B))∧C:
AVBCA∧V¬B(A∧V)∧(¬B)((A∧V)∧(¬B))∧C
00000100
00010100
00100000
00110000
01000100
01010100
01100000
01110000
10000100
10010100
10100000
10110000
11001110
11011111
11101000
11111000

F≡(((A∧V)∧(¬B))∧C):
FAVBCA∧V¬B(A∧V)∧(¬B)((A∧V)∧(¬B))∧CF≡(((A∧V)∧(¬B))∧C)
0000001001
0000101001
0001000001
0001100001
0010001001
0010101001
0011000001
0011100001
0100001001
0100101001
0101000001
0101100001
0110011101
0110111110
0111010001
0111110001
1000001000
1000101000
1001000000
1001100000
1010001000
1010101000
1011000000
1011100000
1100001000
1100101000
1101000000
1101100000
1110011100
1110111111
1111010000
1111110000

Общая таблица истинности:

FAVBC¬BA∧V(A∧V)∧(¬B)((A∧V)∧(¬B))∧CF≡A∧V∧¬B∧C
0000010001
0000110001
0001000001
0001100001
0010010001
0010110001
0011000001
0011100001
0100010001
0100110001
0101000001
0101100001
0110011101
0110111110
0111001001
0111101001
1000010000
1000110000
1001000000
1001100000
1010010000
1010110000
1011000000
1011100000
1100010000
1100110000
1101000000
1101100000
1110011100
1110111111
1111001000
1111101000

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FAVBCF
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011010
011101
011111
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111011
111100
111110
Fсднф = ¬F∧¬A∧¬V∧¬B∧¬C ∨ ¬F∧¬A∧¬V∧¬B∧C ∨ ¬F∧¬A∧¬V∧B∧¬C ∨ ¬F∧¬A∧¬V∧B∧C ∨ ¬F∧¬A∧V∧¬B∧¬C ∨ ¬F∧¬A∧V∧¬B∧C ∨ ¬F∧¬A∧V∧B∧¬C ∨ ¬F∧¬A∧V∧B∧C ∨ ¬F∧A∧¬V∧¬B∧¬C ∨ ¬F∧A∧¬V∧¬B∧C ∨ ¬F∧A∧¬V∧B∧¬C ∨ ¬F∧A∧¬V∧B∧C ∨ ¬F∧A∧V∧¬B∧¬C ∨ ¬F∧A∧V∧B∧¬C ∨ ¬F∧A∧V∧B∧C ∨ F∧A∧V∧¬B∧C
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
FAVBCF
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011010
011101
011111
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111011
111100
111110
Fскнф = (F∨¬A∨¬V∨B∨¬C) ∧ (¬F∨A∨V∨B∨C) ∧ (¬F∨A∨V∨B∨¬C) ∧ (¬F∨A∨V∨¬B∨C) ∧ (¬F∨A∨V∨¬B∨¬C) ∧ (¬F∨A∨¬V∨B∨C) ∧ (¬F∨A∨¬V∨B∨¬C) ∧ (¬F∨A∨¬V∨¬B∨C) ∧ (¬F∨A∨¬V∨¬B∨¬C) ∧ (¬F∨¬A∨V∨B∨C) ∧ (¬F∨¬A∨V∨B∨¬C) ∧ (¬F∨¬A∨V∨¬B∨C) ∧ (¬F∨¬A∨V∨¬B∨¬C) ∧ (¬F∨¬A∨¬V∨B∨C) ∧ (¬F∨¬A∨¬V∨¬B∨C) ∧ (¬F∨¬A∨¬V∨¬B∨¬C)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
FAVBCFж
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011010
011101
011111
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111011
111100
111110

Построим полином Жегалкина:
Fж = C00000 ⊕ C10000∧F ⊕ C01000∧A ⊕ C00100∧V ⊕ C00010∧B ⊕ C00001∧C ⊕ C11000∧F∧A ⊕ C10100∧F∧V ⊕ C10010∧F∧B ⊕ C10001∧F∧C ⊕ C01100∧A∧V ⊕ C01010∧A∧B ⊕ C01001∧A∧C ⊕ C00110∧V∧B ⊕ C00101∧V∧C ⊕ C00011∧B∧C ⊕ C11100∧F∧A∧V ⊕ C11010∧F∧A∧B ⊕ C11001∧F∧A∧C ⊕ C10110∧F∧V∧B ⊕ C10101∧F∧V∧C ⊕ C10011∧F∧B∧C ⊕ C01110∧A∧V∧B ⊕ C01101∧A∧V∧C ⊕ C01011∧A∧B∧C ⊕ C00111∧V∧B∧C ⊕ C11110∧F∧A∧V∧B ⊕ C11101∧F∧A∧V∧C ⊕ C11011∧F∧A∧B∧C ⊕ C10111∧F∧V∧B∧C ⊕ C01111∧A∧V∧B∧C ⊕ C11111∧F∧A∧V∧B∧C

Так как Fж(00000) = 1, то С00000 = 1.

Далее подставляем все остальные наборы в порядке возрастания числа единиц, подставляя вновь полученные значения в следующие формулы:
Fж(10000) = С00000 ⊕ С10000 = 0 => С10000 = 1 ⊕ 0 = 1
Fж(01000) = С00000 ⊕ С01000 = 1 => С01000 = 1 ⊕ 1 = 0
Fж(00100) = С00000 ⊕ С00100 = 1 => С00100 = 1 ⊕ 1 = 0
Fж(00010) = С00000 ⊕ С00010 = 1 => С00010 = 1 ⊕ 1 = 0
Fж(00001) = С00000 ⊕ С00001 = 1 => С00001 = 1 ⊕ 1 = 0
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 0 => С11000 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 0 => С10100 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 0 => С10010 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 0 => С10001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 1 => С01100 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 1 => С01010 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 1 => С01001 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 1 => С00110 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 1 => С00101 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 1 => С00011 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 0 => С11100 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 0 => С10110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 0 => С10101 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 1 => С01110 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 0 => С01101 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 1
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 1 => С01011 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 1 => С00111 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 0 => С11110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 1 => С11101 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 0 => С11011 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 0 => С10111 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 1 => С01111 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(11111) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С11100 ⊕ С11010 ⊕ С11001 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С11110 ⊕ С11101 ⊕ С11011 ⊕ С10111 ⊕ С01111 ⊕ С11111 = 0 => С11111 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0

Таким образом, полином Жегалкина будет равен:
Fж = 1 ⊕ F ⊕ A∧V∧C ⊕ A∧V∧B∧C
Логическая схема, соответствующая полиному Жегалкина:

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