Для функции ¬(A∨V)≡(¬B→C∧B)→D∧¬C:


Промежуточные таблицы истинности:
A∨V:
AVA∨V
000
011
101
111

¬B:
B¬B
01
10

C∧B:
CBC∧B
000
010
100
111

(¬B)→(C∧B):
BC¬BC∧B(¬B)→(C∧B)
00100
01100
10001
11011

¬(A∨V):
AVA∨V¬(A∨V)
0001
0110
1010
1110

¬C:
C¬C
01
10

D∧(¬C):
DC¬CD∧(¬C)
0010
0100
1011
1100

((¬B)→(C∧B))→(D∧(¬C)):
BCD¬BC∧B(¬B)→(C∧B)¬CD∧(¬C)((¬B)→(C∧B))→(D∧(¬C))
000100101
001100111
010100001
011100001
100001100
101001111
110011000
111011000

(¬(A∨V))≡(((¬B)→(C∧B))→(D∧(¬C))):
AVBCDA∨V¬(A∨V)¬BC∧B(¬B)→(C∧B)¬CD∧(¬C)((¬B)→(C∧B))→(D∧(¬C))(¬(A∨V))≡(((¬B)→(C∧B))→(D∧(¬C)))
00000011001011
00001011001111
00010011000011
00011011000011
00100010011000
00101010011111
00110010110000
00111010110000
01000101001010
01001101001110
01010101000010
01011101000010
01100100011001
01101100011110
01110100110001
01111100110001
10000101001010
10001101001110
10010101000010
10011101000010
10100100011001
10101100011110
10110100110001
10111100110001
11000101001010
11001101001110
11010101000010
11011101000010
11100100011001
11101100011110
11110100110001
11111100110001

Общая таблица истинности:

AVBCDA∨V¬BC∧B(¬B)→(C∧B)¬(A∨V)¬CD∧(¬C)((¬B)→(C∧B))→(D∧(¬C))¬(A∨V)≡(¬B→C∧B)→D∧¬C
00000010011011
00001010011111
00010010010011
00011010010011
00100000111000
00101000111111
00110001110000
00111001110000
01000110001010
01001110001110
01010110000010
01011110000010
01100100101001
01101100101110
01110101100001
01111101100001
10000110001010
10001110001110
10010110000010
10011110000010
10100100101001
10101100101110
10110101100001
10111101100001
11000110001010
11001110001110
11010110000010
11011110000010
11100100101001
11101100101110
11110101100001
11111101100001

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
AVBCDF
000001
000011
000101
000111
001000
001011
001100
001110
010000
010010
010100
010110
011001
011010
011101
011111
100000
100010
100100
100110
101001
101010
101101
101111
110000
110010
110100
110110
111001
111010
111101
111111
Fсднф = ¬A∧¬V∧¬B∧¬C∧¬D ∨ ¬A∧¬V∧¬B∧¬C∧D ∨ ¬A∧¬V∧¬B∧C∧¬D ∨ ¬A∧¬V∧¬B∧C∧D ∨ ¬A∧¬V∧B∧¬C∧D ∨ ¬A∧V∧B∧¬C∧¬D ∨ ¬A∧V∧B∧C∧¬D ∨ ¬A∧V∧B∧C∧D ∨ A∧¬V∧B∧¬C∧¬D ∨ A∧¬V∧B∧C∧¬D ∨ A∧¬V∧B∧C∧D ∨ A∧V∧B∧¬C∧¬D ∨ A∧V∧B∧C∧¬D ∨ A∧V∧B∧C∧D
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
AVBCDF
000001
000011
000101
000111
001000
001011
001100
001110
010000
010010
010100
010110
011001
011010
011101
011111
100000
100010
100100
100110
101001
101010
101101
101111
110000
110010
110100
110110
111001
111010
111101
111111
Fскнф = (A∨V∨¬B∨C∨D) ∧ (A∨V∨¬B∨¬C∨D) ∧ (A∨V∨¬B∨¬C∨¬D) ∧ (A∨¬V∨B∨C∨D) ∧ (A∨¬V∨B∨C∨¬D) ∧ (A∨¬V∨B∨¬C∨D) ∧ (A∨¬V∨B∨¬C∨¬D) ∧ (A∨¬V∨¬B∨C∨¬D) ∧ (¬A∨V∨B∨C∨D) ∧ (¬A∨V∨B∨C∨¬D) ∧ (¬A∨V∨B∨¬C∨D) ∧ (¬A∨V∨B∨¬C∨¬D) ∧ (¬A∨V∨¬B∨C∨¬D) ∧ (¬A∨¬V∨B∨C∨D) ∧ (¬A∨¬V∨B∨C∨¬D) ∧ (¬A∨¬V∨B∨¬C∨D) ∧ (¬A∨¬V∨B∨¬C∨¬D) ∧ (¬A∨¬V∨¬B∨C∨¬D)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
AVBCDFж
000001
000011
000101
000111
001000
001011
001100
001110
010000
010010
010100
010110
011001
011010
011101
011111
100000
100010
100100
100110
101001
101010
101101
101111
110000
110010
110100
110110
111001
111010
111101
111111

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

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 = 0 => С01000 = 1 ⊕ 0 = 1
Fж(00100) = С00000 ⊕ С00100 = 0 => С00100 = 1 ⊕ 0 = 1
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 ⊕ 1 ⊕ 0 = 1
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 1 => С10100 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 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 ⊕ 1 ⊕ 1 ⊕ 1 = 0
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 0 => С01010 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 0 => С01001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 0 => С00110 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 1 => С00101 = 1 ⊕ 1 ⊕ 0 ⊕ 1 = 1
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 1 => С00011 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 1 => С11100 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 1 => С10110 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 0 => С10101 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 0 => С01101 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 1 => С11110 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 0 => С11101 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 1 => С10111 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(01111) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С01100 ⊕ С01010 ⊕ С01001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С01110 ⊕ С01101 ⊕ С01011 ⊕ С00111 ⊕ С01111 = 1 => С01111 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 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 = 1 => С11111 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0

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

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