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


Промежуточные таблицы истинности:
A∧V:
AVA∧V
000
010
100
111

(A∧V)∧B:
AVBA∧V(A∧V)∧B
00000
00100
01000
01100
10000
10100
11010
11111

((A∧V)∧B)→C:
AVBCA∧V(A∧V)∧B((A∧V)∧B)→C
0000001
0001001
0010001
0011001
0100001
0101001
0110001
0111001
1000001
1001001
1010001
1011001
1100101
1101101
1110110
1111111

F≡(((A∧V)∧B)→C):
FAVBCA∧V(A∧V)∧B((A∧V)∧B)→CF≡(((A∧V)∧B)→C)
000000010
000010010
000100010
000110010
001000010
001010010
001100010
001110010
010000010
010010010
010100010
010110010
011001010
011011010
011101101
011111110
100000011
100010011
100100011
100110011
101000011
101010011
101100011
101110011
110000011
110010011
110100011
110110011
111001011
111011011
111101100
111111111

Общая таблица истинности:

FAVBCA∧V(A∧V)∧B((A∧V)∧B)→CF≡(A∧V∧B)→C
000000010
000010010
000100010
000110010
001000010
001010010
001100010
001110010
010000010
010010010
010100010
010110010
011001010
011011010
011101101
011111110
100000011
100010011
100100011
100110011
101000011
101010011
101100011
101110011
110000011
110010011
110100011
110110011
111001011
111011011
111101100
111111111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FAVBCF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011101
011110
100001
100011
100101
100111
101001
101011
101101
101111
110001
110011
110101
110111
111001
111011
111100
111111
Fсднф = ¬F∧A∧V∧B∧¬C ∨ F∧¬A∧¬V∧¬B∧¬C ∨ F∧¬A∧¬V∧¬B∧C ∨ F∧¬A∧¬V∧B∧¬C ∨ F∧¬A∧¬V∧B∧C ∨ F∧¬A∧V∧¬B∧¬C ∨ F∧¬A∧V∧¬B∧C ∨ F∧¬A∧V∧B∧¬C ∨ F∧¬A∧V∧B∧C ∨ F∧A∧¬V∧¬B∧¬C ∨ F∧A∧¬V∧¬B∧C ∨ F∧A∧¬V∧B∧¬C ∨ F∧A∧¬V∧B∧C ∨ F∧A∧V∧¬B∧¬C ∨ F∧A∧V∧¬B∧C ∨ F∧A∧V∧B∧C
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
FAVBCF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011101
011110
100001
100011
100101
100111
101001
101011
101101
101111
110001
110011
110101
110111
111001
111011
111100
111111
Fскнф = (F∨A∨V∨B∨C) ∧ (F∨A∨V∨B∨¬C) ∧ (F∨A∨V∨¬B∨C) ∧ (F∨A∨V∨¬B∨¬C) ∧ (F∨A∨¬V∨B∨C) ∧ (F∨A∨¬V∨B∨¬C) ∧ (F∨A∨¬V∨¬B∨C) ∧ (F∨A∨¬V∨¬B∨¬C) ∧ (F∨¬A∨V∨B∨C) ∧ (F∨¬A∨V∨B∨¬C) ∧ (F∨¬A∨V∨¬B∨C) ∧ (F∨¬A∨V∨¬B∨¬C) ∧ (F∨¬A∨¬V∨B∨C) ∧ (F∨¬A∨¬V∨B∨¬C) ∧ (F∨¬A∨¬V∨¬B∨¬C) ∧ (¬F∨¬A∨¬V∨¬B∨C)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
FAVBCFж
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011101
011110
100001
100011
100101
100111
101001
101011
101101
101111
110001
110011
110101
110111
111001
111011
111100
111111

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

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 = 0 => С00010 = 0 ⊕ 0 = 0
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 = 1 => С10010 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 1 => С10001 = 0 ⊕ 1 ⊕ 0 ⊕ 1 = 0
Fж(01100) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С01100 = 0 => С01100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01010) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С01010 = 0 => С01010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01001) = С00000 ⊕ С01000 ⊕ С00001 ⊕ С01001 = 0 => С01001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00110) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00110 = 0 => С00110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00101) = С00000 ⊕ С00100 ⊕ С00001 ⊕ С00101 = 0 => С00101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00011) = С00000 ⊕ С00010 ⊕ С00001 ⊕ С00011 = 0 => С00011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11100) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00100 ⊕ С11000 ⊕ С10100 ⊕ С01100 ⊕ С11100 = 1 => С11100 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 1 => С11010 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 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 = 1 => С10110 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 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 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(01110) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00010 ⊕ С01100 ⊕ С01010 ⊕ С00110 ⊕ С01110 = 1 => С01110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(01101) = С00000 ⊕ С01000 ⊕ С00100 ⊕ С00001 ⊕ С01100 ⊕ С01001 ⊕ С00101 ⊕ С01101 = 0 => С01101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(01011) = С00000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С01011 = 0 => С01011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(00111) = С00000 ⊕ С00100 ⊕ С00010 ⊕ С00001 ⊕ С00110 ⊕ С00101 ⊕ С00011 ⊕ С00111 = 0 => С00111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 1 => С11101 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 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 = 0 => С01111 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 1 = 0

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

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