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


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

C∧V:
CVC∧V
000
010
100
111

(C∧V)∧D:
CVDC∧V(C∧V)∧D
00000
00100
01000
01100
10000
10100
11010
11111

A∨((C∧V)∧D):
ACVDC∧V(C∧V)∧DA∨((C∧V)∧D)
0000000
0001000
0010000
0011000
0100000
0101000
0110100
0111111
1000001
1001001
1010001
1011001
1100001
1101001
1110101
1111111

(A∨((C∧V)∧D))∨(¬B):
ACVDBC∧V(C∧V)∧DA∨((C∧V)∧D)¬B(A∨((C∧V)∧D))∨(¬B)
0000000011
0000100000
0001000011
0001100000
0010000011
0010100000
0011000011
0011100000
0100000011
0100100000
0101000011
0101100000
0110010011
0110110000
0111011111
0111111101
1000000111
1000100101
1001000111
1001100101
1010000111
1010100101
1011000111
1011100101
1100000111
1100100101
1101000111
1101100101
1110010111
1110110101
1111011111
1111111101

Общая таблица истинности:

ACVDB¬BC∧V(C∧V)∧DA∨((C∧V)∧D)A∨C∧V∧D∨(¬B)
0000010001
0000100000
0001010001
0001100000
0010010001
0010100000
0011010001
0011100000
0100010001
0100100000
0101010001
0101100000
0110011001
0110101000
0111011111
0111101111
1000010011
1000100011
1001010011
1001100011
1010010011
1010100011
1011010011
1011100011
1100010011
1100100011
1101010011
1101100011
1110011011
1110101011
1111011111
1111101111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ACVDBF
000001
000010
000101
000110
001001
001010
001101
001110
010001
010010
010101
010110
011001
011010
011101
011111
100001
100011
100101
100111
101001
101011
101101
101111
110001
110011
110101
110111
111001
111011
111101
111111
Fсднф = ¬A∧¬C∧¬V∧¬D∧¬B ∨ ¬A∧¬C∧¬V∧D∧¬B ∨ ¬A∧¬C∧V∧¬D∧¬B ∨ ¬A∧¬C∧V∧D∧¬B ∨ ¬A∧C∧¬V∧¬D∧¬B ∨ ¬A∧C∧¬V∧D∧¬B ∨ ¬A∧C∧V∧¬D∧¬B ∨ ¬A∧C∧V∧D∧¬B ∨ ¬A∧C∧V∧D∧B ∨ A∧¬C∧¬V∧¬D∧¬B ∨ A∧¬C∧¬V∧¬D∧B ∨ A∧¬C∧¬V∧D∧¬B ∨ A∧¬C∧¬V∧D∧B ∨ A∧¬C∧V∧¬D∧¬B ∨ A∧¬C∧V∧¬D∧B ∨ A∧¬C∧V∧D∧¬B ∨ A∧¬C∧V∧D∧B ∨ A∧C∧¬V∧¬D∧¬B ∨ A∧C∧¬V∧¬D∧B ∨ A∧C∧¬V∧D∧¬B ∨ A∧C∧¬V∧D∧B ∨ A∧C∧V∧¬D∧¬B ∨ A∧C∧V∧¬D∧B ∨ A∧C∧V∧D∧¬B ∨ A∧C∧V∧D∧B
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ACVDBF
000001
000010
000101
000110
001001
001010
001101
001110
010001
010010
010101
010110
011001
011010
011101
011111
100001
100011
100101
100111
101001
101011
101101
101111
110001
110011
110101
110111
111001
111011
111101
111111
Fскнф = (A∨C∨V∨D∨¬B) ∧ (A∨C∨V∨¬D∨¬B) ∧ (A∨C∨¬V∨D∨¬B) ∧ (A∨C∨¬V∨¬D∨¬B) ∧ (A∨¬C∨V∨D∨¬B) ∧ (A∨¬C∨V∨¬D∨¬B) ∧ (A∨¬C∨¬V∨D∨¬B)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ACVDBFж
000001
000010
000101
000110
001001
001010
001101
001110
010001
010010
010101
010110
011001
011010
011101
011111
100001
100011
100101
100111
101001
101011
101101
101111
110001
110011
110101
110111
111001
111011
111101
111111

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

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 = 1 => С10000 = 1 ⊕ 1 = 0
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 = 0 => С00001 = 1 ⊕ 0 = 1
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 1 => С11000 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 1 => С10100 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 1 => С10010 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 1 => С10001 = 1 ⊕ 0 ⊕ 1 ⊕ 1 = 1
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 = 0 => С01001 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 1 => С00110 = 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 0 => С00101 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 0 => С00011 = 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 1 => С11100 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 1 => С11010 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 1 => С11001 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 1 => С10110 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 1 => С10101 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 1 => С10011 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 = 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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 1 => С11101 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 1 => С11011 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 1 => С10111 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 1 => С11111 = 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 1

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

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