Таблица истинности для функции F≡(A∧B)∧U∧(C∧B):


Промежуточные таблицы истинности:
A∧B:
ABA∧B
000
010
100
111

C∧B:
CBC∧B
000
010
100
111

(A∧B)∧U:
ABUA∧B(A∧B)∧U
00000
00100
01000
01100
10000
10100
11010
11111

((A∧B)∧U)∧(C∧B):
ABUCA∧B(A∧B)∧UC∧B((A∧B)∧U)∧(C∧B)
00000000
00010000
00100000
00110000
01000000
01010010
01100000
01110010
10000000
10010000
10100000
10110000
11001000
11011010
11101100
11111111

F≡(((A∧B)∧U)∧(C∧B)):
FABUCA∧B(A∧B)∧UC∧B((A∧B)∧U)∧(C∧B)F≡(((A∧B)∧U)∧(C∧B))
0000000001
0000100001
0001000001
0001100001
0010000001
0010100101
0011000001
0011100101
0100000001
0100100001
0101000001
0101100001
0110010001
0110110101
0111011001
0111111110
1000000000
1000100000
1001000000
1001100000
1010000000
1010100100
1011000000
1011100100
1100000000
1100100000
1101000000
1101100000
1110010000
1110110100
1111011000
1111111111

Общая таблица истинности:

FABUCA∧BC∧B(A∧B)∧U((A∧B)∧U)∧(C∧B)F≡(A∧B)∧U∧(C∧B)
0000000001
0000100001
0001000001
0001100001
0010000001
0010101001
0011000001
0011101001
0100000001
0100100001
0101000001
0101100001
0110010001
0110111001
0111010101
0111111110
1000000000
1000100000
1001000000
1001100000
1010000000
1010101000
1011000000
1011101000
1100000000
1100100000
1101000000
1101100000
1110010000
1110111000
1111010100
1111111111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FABUCF
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011011
011101
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111
Fсднф = ¬F∧¬A∧¬B∧¬U∧¬C ∨ ¬F∧¬A∧¬B∧¬U∧C ∨ ¬F∧¬A∧¬B∧U∧¬C ∨ ¬F∧¬A∧¬B∧U∧C ∨ ¬F∧¬A∧B∧¬U∧¬C ∨ ¬F∧¬A∧B∧¬U∧C ∨ ¬F∧¬A∧B∧U∧¬C ∨ ¬F∧¬A∧B∧U∧C ∨ ¬F∧A∧¬B∧¬U∧¬C ∨ ¬F∧A∧¬B∧¬U∧C ∨ ¬F∧A∧¬B∧U∧¬C ∨ ¬F∧A∧¬B∧U∧C ∨ ¬F∧A∧B∧¬U∧¬C ∨ ¬F∧A∧B∧¬U∧C ∨ ¬F∧A∧B∧U∧¬C ∨ F∧A∧B∧U∧C
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
FABUCF
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011011
011101
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111
Fскнф = (F∨¬A∨¬B∨¬U∨¬C) ∧ (¬F∨A∨B∨U∨C) ∧ (¬F∨A∨B∨U∨¬C) ∧ (¬F∨A∨B∨¬U∨C) ∧ (¬F∨A∨B∨¬U∨¬C) ∧ (¬F∨A∨¬B∨U∨C) ∧ (¬F∨A∨¬B∨U∨¬C) ∧ (¬F∨A∨¬B∨¬U∨C) ∧ (¬F∨A∨¬B∨¬U∨¬C) ∧ (¬F∨¬A∨B∨U∨C) ∧ (¬F∨¬A∨B∨U∨¬C) ∧ (¬F∨¬A∨B∨¬U∨C) ∧ (¬F∨¬A∨B∨¬U∨¬C) ∧ (¬F∨¬A∨¬B∨U∨C) ∧ (¬F∨¬A∨¬B∨U∨¬C) ∧ (¬F∨¬A∨¬B∨¬U∨C)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
FABUCFж
000001
000011
000101
000111
001001
001011
001101
001111
010001
010011
010101
010111
011001
011011
011101
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111

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

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

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

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