Таблица истинности для функции A∧V∧C∧V∧D∧V∧¬B:


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

A∧V:
AVA∧V
000
010
100
111

(A∧V)∧C:
AVCA∧V(A∧V)∧C
00000
00100
01000
01100
10000
10100
11010
11111

((A∧V)∧C)∧V:
AVCA∧V(A∧V)∧C((A∧V)∧C)∧V
000000
001000
010000
011000
100000
101000
110100
111111

(((A∧V)∧C)∧V)∧D:
AVCDA∧V(A∧V)∧C((A∧V)∧C)∧V(((A∧V)∧C)∧V)∧D
00000000
00010000
00100000
00110000
01000000
01010000
01100000
01110000
10000000
10010000
10100000
10110000
11001000
11011000
11101110
11111111

((((A∧V)∧C)∧V)∧D)∧V:
AVCDA∧V(A∧V)∧C((A∧V)∧C)∧V(((A∧V)∧C)∧V)∧D((((A∧V)∧C)∧V)∧D)∧V
000000000
000100000
001000000
001100000
010000000
010100000
011000000
011100000
100000000
100100000
101000000
101100000
110010000
110110000
111011100
111111111

(((((A∧V)∧C)∧V)∧D)∧V)∧(¬B):
AVCDBA∧V(A∧V)∧C((A∧V)∧C)∧V(((A∧V)∧C)∧V)∧D((((A∧V)∧C)∧V)∧D)∧V¬B(((((A∧V)∧C)∧V)∧D)∧V)∧(¬B)
000000000010
000010000000
000100000010
000110000000
001000000010
001010000000
001100000010
001110000000
010000000010
010010000000
010100000010
010110000000
011000000010
011010000000
011100000010
011110000000
100000000010
100010000000
100100000010
100110000000
101000000010
101010000000
101100000010
101110000000
110001000010
110011000000
110101000010
110111000000
111001110010
111011110000
111101111111
111111111100

Общая таблица истинности:

AVCDB¬BA∧V(A∧V)∧C((A∧V)∧C)∧V(((A∧V)∧C)∧V)∧D((((A∧V)∧C)∧V)∧D)∧VA∧V∧C∧V∧D∧V∧¬B
000001000000
000010000000
000101000000
000110000000
001001000000
001010000000
001101000000
001110000000
010001000000
010010000000
010101000000
010110000000
011001000000
011010000000
011101000000
011110000000
100001000000
100010000000
100101000000
100110000000
101001000000
101010000000
101101000000
101110000000
110001100000
110010100000
110101100000
110110100000
111001111000
111010111000
111101111111
111110111110

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
AVCDBF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111101
111110
Fсднф = A∧V∧C∧D∧¬B
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
AVCDBF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111101
111110
Fскнф = (A∨V∨C∨D∨B) ∧ (A∨V∨C∨D∨¬B) ∧ (A∨V∨C∨¬D∨B) ∧ (A∨V∨C∨¬D∨¬B) ∧ (A∨V∨¬C∨D∨B) ∧ (A∨V∨¬C∨D∨¬B) ∧ (A∨V∨¬C∨¬D∨B) ∧ (A∨V∨¬C∨¬D∨¬B) ∧ (A∨¬V∨C∨D∨B) ∧ (A∨¬V∨C∨D∨¬B) ∧ (A∨¬V∨C∨¬D∨B) ∧ (A∨¬V∨C∨¬D∨¬B) ∧ (A∨¬V∨¬C∨D∨B) ∧ (A∨¬V∨¬C∨D∨¬B) ∧ (A∨¬V∨¬C∨¬D∨B) ∧ (A∨¬V∨¬C∨¬D∨¬B) ∧ (¬A∨V∨C∨D∨B) ∧ (¬A∨V∨C∨D∨¬B) ∧ (¬A∨V∨C∨¬D∨B) ∧ (¬A∨V∨C∨¬D∨¬B) ∧ (¬A∨V∨¬C∨D∨B) ∧ (¬A∨V∨¬C∨D∨¬B) ∧ (¬A∨V∨¬C∨¬D∨B) ∧ (¬A∨V∨¬C∨¬D∨¬B) ∧ (¬A∨¬V∨C∨D∨B) ∧ (¬A∨¬V∨C∨D∨¬B) ∧ (¬A∨¬V∨C∨¬D∨B) ∧ (¬A∨¬V∨C∨¬D∨¬B) ∧ (¬A∨¬V∨¬C∨D∨B) ∧ (¬A∨¬V∨¬C∨D∨¬B) ∧ (¬A∨¬V∨¬C∨¬D∨¬B)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
AVCDBFж
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111101
111110

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

Так как Fж(00000) = 0, то С00000 = 0.

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

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

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