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


Промежуточные таблицы истинности:
A→B:
ABA→B
001
011
100
111

A∨C:
ACA∨C
000
011
101
111

A→V:
AVA→V
001
011
100
111

(A∨C)↓(A→V):
ACVA∨CA→V(A∨C)↓(A→V)
000010
001010
010110
011110
100100
101110
110100
111110

F∧(A→B):
FABA→BF∧(A→B)
00010
00110
01000
01110
10011
10111
11000
11111

(F∧(A→B))∨((A∨C)↓(A→V)):
FABCVA→BF∧(A→B)A∨CA→V(A∨C)↓(A→V)(F∧(A→B))∨((A∨C)↓(A→V))
00000100100
00001100100
00010101100
00011101100
00100100100
00101100100
00110101100
00111101100
01000001000
01001001100
01010001000
01011001100
01100101000
01101101100
01110101000
01111101100
10000110101
10001110101
10010111101
10011111101
10100110101
10101110101
10110111101
10111111101
11000001000
11001001100
11010001000
11011001100
11100111001
11101111101
11110111001
11111111101

Общая таблица истинности:

FABCVA→BA∨CA→V(A∨C)↓(A→V)F∧(A→B)F∧(A→B)∨(A∨C)↓(A→V)
00000101000
00001101000
00010111000
00011111000
00100101000
00101101000
00110111000
00111111000
01000010000
01001011000
01010010000
01011011000
01100110000
01101111000
01110110000
01111111000
10000101011
10001101011
10010111011
10011111011
10100101011
10101101011
10110111011
10111111011
11000010000
11001011000
11010010000
11011011000
11100110011
11101111011
11110110011
11111111011

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FABCVF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100001
100011
100101
100111
101001
101011
101101
101111
110000
110010
110100
110110
111001
111011
111101
111111
Fсднф = F∧¬A∧¬B∧¬C∧¬V ∨ F∧¬A∧¬B∧¬C∧V ∨ F∧¬A∧¬B∧C∧¬V ∨ F∧¬A∧¬B∧C∧V ∨ F∧¬A∧B∧¬C∧¬V ∨ F∧¬A∧B∧¬C∧V ∨ F∧¬A∧B∧C∧¬V ∨ F∧¬A∧B∧C∧V ∨ F∧A∧B∧¬C∧¬V ∨ F∧A∧B∧¬C∧V ∨ F∧A∧B∧C∧¬V ∨ F∧A∧B∧C∧V
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
FABCVF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100001
100011
100101
100111
101001
101011
101101
101111
110000
110010
110100
110110
111001
111011
111101
111111
Fскнф = (F∨A∨B∨C∨V) ∧ (F∨A∨B∨C∨¬V) ∧ (F∨A∨B∨¬C∨V) ∧ (F∨A∨B∨¬C∨¬V) ∧ (F∨A∨¬B∨C∨V) ∧ (F∨A∨¬B∨C∨¬V) ∧ (F∨A∨¬B∨¬C∨V) ∧ (F∨A∨¬B∨¬C∨¬V) ∧ (F∨¬A∨B∨C∨V) ∧ (F∨¬A∨B∨C∨¬V) ∧ (F∨¬A∨B∨¬C∨V) ∧ (F∨¬A∨B∨¬C∨¬V) ∧ (F∨¬A∨¬B∨C∨V) ∧ (F∨¬A∨¬B∨C∨¬V) ∧ (F∨¬A∨¬B∨¬C∨V) ∧ (F∨¬A∨¬B∨¬C∨¬V) ∧ (¬F∨¬A∨B∨C∨V) ∧ (¬F∨¬A∨B∨C∨¬V) ∧ (¬F∨¬A∨B∨¬C∨V) ∧ (¬F∨¬A∨B∨¬C∨¬V)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
FABCVFж
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100001
100011
100101
100111
101001
101011
101101
101111
110000
110010
110100
110110
111001
111011
111101
111111

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

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 = 0 => С11000 = 0 ⊕ 1 ⊕ 0 ⊕ 0 = 1
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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 = 1
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 = 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 = 0 => С01110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
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 = 1 => С11110 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0
Fж(11011) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С00001 ⊕ С11000 ⊕ С10010 ⊕ С10001 ⊕ С01010 ⊕ С01001 ⊕ С00011 ⊕ С11010 ⊕ С11001 ⊕ С10011 ⊕ С01011 ⊕ С11011 = 0 => С11011 = 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 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 = 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 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
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 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 1 = 0

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

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