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


Промежуточные таблицы истинности:
A∨D:
ADA∨D
000
011
101
111

(A∨D)∨C:
ADCA∨D(A∨D)∨C
00000
00101
01011
01111
10011
10111
11011
11111

¬C:
C¬C
01
10

(¬C)∨D:
CD¬C(¬C)∨D
0011
0111
1000
1101

¬A:
A¬A
01
10

B∨(¬A):
BA¬AB∨(¬A)
0011
0100
1011
1101

A∨C:
ACA∨C
000
011
101
111

((A∨D)∨C)∧((¬C)∨D):
ADCA∨D(A∨D)∨C¬C(¬C)∨D((A∨D)∨C)∧((¬C)∨D)
00000110
00101000
01011111
01111011
10011111
10111000
11011111
11111011

(B∨(¬A))∧(A∨C):
BAC¬AB∨(¬A)A∨C(B∨(¬A))∧(A∨C)
0001100
0011111
0100010
0110010
1001100
1011111
1100111
1110111

(((A∨D)∨C)∧((¬C)∨D))∨((B∨(¬A))∧(A∨C)):
ADCBA∨D(A∨D)∨C¬C(¬C)∨D((A∨D)∨C)∧((¬C)∨D)¬AB∨(¬A)A∨C(B∨(¬A))∧(A∨C)(((A∨D)∨C)∧((¬C)∨D))∨((B∨(¬A))∧(A∨C))
00000011011000
00010011011000
00100100011111
00110100011111
01001111111001
01011111111001
01101101111111
01111101111111
10001111100101
10011111101111
10101100000100
10111100001111
11001111100101
11011111101111
11101101100101
11111101101111

Y≡((((A∨D)∨C)∧((¬C)∨D))∨((B∨(¬A))∧(A∨C))):
YADCBA∨D(A∨D)∨C¬C(¬C)∨D((A∨D)∨C)∧((¬C)∨D)¬AB∨(¬A)A∨C(B∨(¬A))∧(A∨C)(((A∨D)∨C)∧((¬C)∨D))∨((B∨(¬A))∧(A∨C))Y≡((((A∨D)∨C)∧((¬C)∨D))∨((B∨(¬A))∧(A∨C)))
0000000110110001
0000100110110001
0001001000111110
0001101000111110
0010011111110010
0010111111110010
0011011011111110
0011111011111110
0100011111001010
0100111111011110
0101011000001001
0101111000011110
0110011111001010
0110111111011110
0111011011001010
0111111011011110
1000000110110000
1000100110110000
1001001000111111
1001101000111111
1010011111110011
1010111111110011
1011011011111111
1011111011111111
1100011111001011
1100111111011111
1101011000001000
1101111000011111
1110011111001011
1110111111011111
1111011011001011
1111111011011111

Общая таблица истинности:

YADCBA∨D(A∨D)∨C¬C(¬C)∨D¬AB∨(¬A)A∨C((A∨D)∨C)∧((¬C)∨D)(B∨(¬A))∧(A∨C)(((A∨D)∨C)∧((¬C)∨D))∨((B∨(¬A))∧(A∨C))Y≡(A∨D∨C)∧(¬C∨D)∨(B∨¬A)∧(A∨C)
0000000111100001
0000100111100001
0001001001110110
0001101001110110
0010011111101010
0010111111101010
0011011011111110
0011111011111110
0100011110011010
0100111110111110
0101011000010001
0101111000110110
0110011110011010
0110111110111110
0111011010011010
0111111010111110
1000000111100000
1000100111100000
1001001001110111
1001101001110111
1010011111101011
1010111111101011
1011011011111111
1011111011111111
1100011110011011
1100111110111111
1101011000010000
1101111000110111
1110011110011011
1110111110111111
1111011010011011
1111111010111111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
YADCBF
000001
000011
000100
000110
001000
001010
001100
001110
010000
010010
010101
010110
011000
011010
011100
011110
100000
100010
100101
100111
101001
101011
101101
101111
110001
110011
110100
110111
111001
111011
111101
111111
Fсднф = ¬Y∧¬A∧¬D∧¬C∧¬B ∨ ¬Y∧¬A∧¬D∧¬C∧B ∨ ¬Y∧A∧¬D∧C∧¬B ∨ Y∧¬A∧¬D∧C∧¬B ∨ Y∧¬A∧¬D∧C∧B ∨ Y∧¬A∧D∧¬C∧¬B ∨ Y∧¬A∧D∧¬C∧B ∨ Y∧¬A∧D∧C∧¬B ∨ Y∧¬A∧D∧C∧B ∨ Y∧A∧¬D∧¬C∧¬B ∨ Y∧A∧¬D∧¬C∧B ∨ Y∧A∧¬D∧C∧B ∨ Y∧A∧D∧¬C∧¬B ∨ Y∧A∧D∧¬C∧B ∨ Y∧A∧D∧C∧¬B ∨ Y∧A∧D∧C∧B
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
YADCBF
000001
000011
000100
000110
001000
001010
001100
001110
010000
010010
010101
010110
011000
011010
011100
011110
100000
100010
100101
100111
101001
101011
101101
101111
110001
110011
110100
110111
111001
111011
111101
111111
Fскнф = (Y∨A∨D∨¬C∨B) ∧ (Y∨A∨D∨¬C∨¬B) ∧ (Y∨A∨¬D∨C∨B) ∧ (Y∨A∨¬D∨C∨¬B) ∧ (Y∨A∨¬D∨¬C∨B) ∧ (Y∨A∨¬D∨¬C∨¬B) ∧ (Y∨¬A∨D∨C∨B) ∧ (Y∨¬A∨D∨C∨¬B) ∧ (Y∨¬A∨D∨¬C∨¬B) ∧ (Y∨¬A∨¬D∨C∨B) ∧ (Y∨¬A∨¬D∨C∨¬B) ∧ (Y∨¬A∨¬D∨¬C∨B) ∧ (Y∨¬A∨¬D∨¬C∨¬B) ∧ (¬Y∨A∨D∨C∨B) ∧ (¬Y∨A∨D∨C∨¬B) ∧ (¬Y∨¬A∨D∨¬C∨B)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
YADCBFж
000001
000011
000100
000110
001000
001010
001100
001110
010000
010010
010101
010110
011000
011010
011100
011110
100000
100010
100101
100111
101001
101011
101101
101111
110001
110011
110100
110111
111001
111011
111101
111111

Построим полином Жегалкина:
Fж = C00000 ⊕ C10000∧Y ⊕ C01000∧A ⊕ C00100∧D ⊕ C00010∧C ⊕ C00001∧B ⊕ C11000∧Y∧A ⊕ C10100∧Y∧D ⊕ C10010∧Y∧C ⊕ C10001∧Y∧B ⊕ C01100∧A∧D ⊕ C01010∧A∧C ⊕ C01001∧A∧B ⊕ C00110∧D∧C ⊕ C00101∧D∧B ⊕ C00011∧C∧B ⊕ C11100∧Y∧A∧D ⊕ C11010∧Y∧A∧C ⊕ C11001∧Y∧A∧B ⊕ C10110∧Y∧D∧C ⊕ C10101∧Y∧D∧B ⊕ C10011∧Y∧C∧B ⊕ C01110∧A∧D∧C ⊕ C01101∧A∧D∧B ⊕ C01011∧A∧C∧B ⊕ C00111∧D∧C∧B ⊕ C11110∧Y∧A∧D∧C ⊕ C11101∧Y∧A∧D∧B ⊕ C11011∧Y∧A∧C∧B ⊕ C10111∧Y∧D∧C∧B ⊕ C01111∧A∧D∧C∧B ⊕ C11111∧Y∧A∧D∧C∧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 = 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 = 0 => С00010 = 1 ⊕ 0 = 1
Fж(00001) = С00000 ⊕ С00001 = 1 => С00001 = 1 ⊕ 1 = 0
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 1 => С11000 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 1 => С10100 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 1 => С10010 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 0 => С10001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 0 => С01100 = 1 ⊕ 1 ⊕ 1 ⊕ 0 = 1
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 1 => С01010 = 1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 0 => С01001 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 0 => С00110 = 1 ⊕ 1 ⊕ 1 ⊕ 0 = 1
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 0 => С00101 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 0 => С00011 = 1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 1 => С11100 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 1 => С11001 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 1 => С10110 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 1 => С10101 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 1 => С10011 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 0 => С01110 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 0 => С01101 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 1
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(10111) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С10100 ⊕ С10010 ⊕ С10001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С10110 ⊕ С10101 ⊕ С10011 ⊕ С00111 ⊕ С10111 = 1 => С10111 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 0 => С01111 = 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 = 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 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0

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

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