Таблица истинности для функции Z≡(¬X2∨¬X3)∧(¬X1∨X4):


Промежуточные таблицы истинности:
¬X2:
X2¬X2
01
10

¬X3:
X3¬X3
01
10

(¬X2)∨(¬X3):
X2X3¬X2¬X3(¬X2)∨(¬X3)
00111
01101
10011
11000

¬X1:
X1¬X1
01
10

(¬X1)∨X4:
X1X4¬X1(¬X1)∨X4
0011
0111
1000
1101

((¬X2)∨(¬X3))∧((¬X1)∨X4):
X2X3X1X4¬X2¬X3(¬X2)∨(¬X3)¬X1(¬X1)∨X4((¬X2)∨(¬X3))∧((¬X1)∨X4)
0000111111
0001111111
0010111000
0011111011
0100101111
0101101111
0110101000
0111101011
1000011111
1001011111
1010011000
1011011011
1100000110
1101000110
1110000000
1111000010

Z≡(((¬X2)∨(¬X3))∧((¬X1)∨X4)):
ZX2X3X1X4¬X2¬X3(¬X2)∨(¬X3)¬X1(¬X1)∨X4((¬X2)∨(¬X3))∧((¬X1)∨X4)Z≡(((¬X2)∨(¬X3))∧((¬X1)∨X4))
000001111110
000011111110
000101110001
000111110110
001001011110
001011011110
001101010001
001111010110
010000111110
010010111110
010100110001
010110110110
011000001101
011010001101
011100000001
011110000101
100001111111
100011111111
100101110000
100111110111
101001011111
101011011111
101101010000
101111010111
110000111111
110010111111
110100110000
110110110111
111000001100
111010001100
111100000000
111110000100

Общая таблица истинности:

ZX2X3X1X4¬X2¬X3(¬X2)∨(¬X3)¬X1(¬X1)∨X4((¬X2)∨(¬X3))∧((¬X1)∨X4)Z≡(¬X2∨¬X3)∧(¬X1∨X4)
000001111110
000011111110
000101110001
000111110110
001001011110
001011011110
001101010001
001111010110
010000111110
010010111110
010100110001
010110110110
011000001101
011010001101
011100000001
011110000101
100001111111
100011111111
100101110000
100111110111
101001011111
101011011111
101101010000
101111010111
110000111111
110010111111
110100110000
110110110111
111000001100
111010001100
111100000000
111110000100

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
ZX2X3X1X4F
000000
000010
000101
000110
001000
001010
001101
001110
010000
010010
010101
010110
011001
011011
011101
011111
100001
100011
100100
100111
101001
101011
101100
101111
110001
110011
110100
110111
111000
111010
111100
111110
Fсднф = ¬Z∧¬X2∧¬X3∧X1∧¬X4 ∨ ¬Z∧¬X2∧X3∧X1∧¬X4 ∨ ¬Z∧X2∧¬X3∧X1∧¬X4 ∨ ¬Z∧X2∧X3∧¬X1∧¬X4 ∨ ¬Z∧X2∧X3∧¬X1∧X4 ∨ ¬Z∧X2∧X3∧X1∧¬X4 ∨ ¬Z∧X2∧X3∧X1∧X4 ∨ Z∧¬X2∧¬X3∧¬X1∧¬X4 ∨ Z∧¬X2∧¬X3∧¬X1∧X4 ∨ Z∧¬X2∧¬X3∧X1∧X4 ∨ Z∧¬X2∧X3∧¬X1∧¬X4 ∨ Z∧¬X2∧X3∧¬X1∧X4 ∨ Z∧¬X2∧X3∧X1∧X4 ∨ Z∧X2∧¬X3∧¬X1∧¬X4 ∨ Z∧X2∧¬X3∧¬X1∧X4 ∨ Z∧X2∧¬X3∧X1∧X4
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
ZX2X3X1X4F
000000
000010
000101
000110
001000
001010
001101
001110
010000
010010
010101
010110
011001
011011
011101
011111
100001
100011
100100
100111
101001
101011
101100
101111
110001
110011
110100
110111
111000
111010
111100
111110
Fскнф = (Z∨X2∨X3∨X1∨X4) ∧ (Z∨X2∨X3∨X1∨¬X4) ∧ (Z∨X2∨X3∨¬X1∨¬X4) ∧ (Z∨X2∨¬X3∨X1∨X4) ∧ (Z∨X2∨¬X3∨X1∨¬X4) ∧ (Z∨X2∨¬X3∨¬X1∨¬X4) ∧ (Z∨¬X2∨X3∨X1∨X4) ∧ (Z∨¬X2∨X3∨X1∨¬X4) ∧ (Z∨¬X2∨X3∨¬X1∨¬X4) ∧ (¬Z∨X2∨X3∨¬X1∨X4) ∧ (¬Z∨X2∨¬X3∨¬X1∨X4) ∧ (¬Z∨¬X2∨X3∨¬X1∨X4) ∧ (¬Z∨¬X2∨¬X3∨X1∨X4) ∧ (¬Z∨¬X2∨¬X3∨X1∨¬X4) ∧ (¬Z∨¬X2∨¬X3∨¬X1∨X4) ∧ (¬Z∨¬X2∨¬X3∨¬X1∨¬X4)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
ZX2X3X1X4Fж
000000
000010
000101
000110
001000
001010
001101
001110
010000
010010
010101
010110
011001
011011
011101
011111
100001
100011
100100
100111
101001
101011
101100
101111
110001
110011
110100
110111
111000
111010
111100
111110

Построим полином Жегалкина:
Fж = C00000 ⊕ C10000∧Z ⊕ C01000∧X2 ⊕ C00100∧X3 ⊕ C00010∧X1 ⊕ C00001∧X4 ⊕ C11000∧Z∧X2 ⊕ C10100∧Z∧X3 ⊕ C10010∧Z∧X1 ⊕ C10001∧Z∧X4 ⊕ C01100∧X2∧X3 ⊕ C01010∧X2∧X1 ⊕ C01001∧X2∧X4 ⊕ C00110∧X3∧X1 ⊕ C00101∧X3∧X4 ⊕ C00011∧X1∧X4 ⊕ C11100∧Z∧X2∧X3 ⊕ C11010∧Z∧X2∧X1 ⊕ C11001∧Z∧X2∧X4 ⊕ C10110∧Z∧X3∧X1 ⊕ C10101∧Z∧X3∧X4 ⊕ C10011∧Z∧X1∧X4 ⊕ C01110∧X2∧X3∧X1 ⊕ C01101∧X2∧X3∧X4 ⊕ C01011∧X2∧X1∧X4 ⊕ C00111∧X3∧X1∧X4 ⊕ C11110∧Z∧X2∧X3∧X1 ⊕ C11101∧Z∧X2∧X3∧X4 ⊕ C11011∧Z∧X2∧X1∧X4 ⊕ C10111∧Z∧X3∧X1∧X4 ⊕ C01111∧X2∧X3∧X1∧X4 ⊕ C11111∧Z∧X2∧X3∧X1∧X4

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 = 1 => С10000 = 0 ⊕ 1 = 1
Fж(01000) = С00000 ⊕ С01000 = 0 => С01000 = 0 ⊕ 0 = 0
Fж(00100) = С00000 ⊕ С00100 = 0 => С00100 = 0 ⊕ 0 = 0
Fж(00010) = С00000 ⊕ С00010 = 1 => С00010 = 0 ⊕ 1 = 1
Fж(00001) = С00000 ⊕ С00001 = 0 => С00001 = 0 ⊕ 0 = 0
Fж(11000) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С11000 = 1 => С11000 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 1 => С10100 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 0 => С10010 = 0 ⊕ 1 ⊕ 1 ⊕ 0 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 1 => С10001 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 1 => С01100 = 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 1 => С01010 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 0 => С01001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 1 => С00110 = 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 0 => С00101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 0 => С00011 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 0 => С11100 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 1 => С11001 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 0 => С10110 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 1 => С10101 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 1 => С10011 = 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 1 => С01110 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 1 => С01101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(11110) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С11000 ⊕ С10100 ⊕ С10010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С11100 ⊕ С11010 ⊕ С10110 ⊕ С01110 ⊕ С11110 = 0 => С11110 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0
Fж(11101) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С11000 ⊕ С10100 ⊕ С10001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С11100 ⊕ С11001 ⊕ С10101 ⊕ С01101 ⊕ С11101 = 0 => С11101 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 1 => С11011 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 ⊕ 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 = 0 => С11111 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 = 0

Таким образом, полином Жегалкина будет равен:
Fж = Z ⊕ X1 ⊕ X2∧X3 ⊕ X1∧X4 ⊕ X2∧X3∧X1 ⊕ X2∧X3∧X1∧X4
Логическая схема, соответствующая полиному Жегалкина:

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