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


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

(A∧B)∧C:
ABCA∧B(A∧B)∧C
00000
00100
01000
01100
10000
10100
11010
11111

((A∧B)∧C)∧D:
ABCDA∧B(A∧B)∧C((A∧B)∧C)∧D
0000000
0001000
0010000
0011000
0100000
0101000
0110000
0111000
1000000
1001000
1010000
1011000
1100100
1101100
1110110
1111111

D∧A:
DAD∧A
000
010
100
111

(D∧A)∧D:
DAD∧A(D∧A)∧D
0000
0100
1000
1111

C∨((D∧A)∧D):
CDAD∧A(D∧A)∧DC∨((D∧A)∧D)
000000
001000
010000
011111
100001
101001
110001
111111

F∧(((A∧B)∧C)∧D):
FABCDA∧B(A∧B)∧C((A∧B)∧C)∧DF∧(((A∧B)∧C)∧D)
000000000
000010000
000100000
000110000
001000000
001010000
001100000
001110000
010000000
010010000
010100000
010110000
011001000
011011000
011101100
011111110
100000000
100010000
100100000
100110000
101000000
101010000
101100000
101110000
110000000
110010000
110100000
110110000
111001000
111011000
111101100
111111111

B∧(C∨((D∧A)∧D)):
BCDAD∧A(D∧A)∧DC∨((D∧A)∧D)B∧(C∨((D∧A)∧D))
00000000
00010000
00100000
00111110
01000010
01010010
01100010
01111110
10000000
10010000
10100000
10111111
11000011
11010011
11100011
11111111

A∨(B∧(C∨((D∧A)∧D))):
ABCDD∧A(D∧A)∧DC∨((D∧A)∧D)B∧(C∨((D∧A)∧D))A∨(B∧(C∨((D∧A)∧D)))
000000000
000100000
001000100
001100100
010000000
010100000
011000111
011100111
100000001
100111101
101000101
101111101
110000001
110111111
111000111
111111111

(F∧(((A∧B)∧C)∧D))≡(A∨(B∧(C∨((D∧A)∧D)))):
FABCDA∧B(A∧B)∧C((A∧B)∧C)∧DF∧(((A∧B)∧C)∧D)D∧A(D∧A)∧DC∨((D∧A)∧D)B∧(C∨((D∧A)∧D))A∨(B∧(C∨((D∧A)∧D)))(F∧(((A∧B)∧C)∧D))≡(A∨(B∧(C∨((D∧A)∧D))))
000000000000001
000010000000001
000100000001001
000110000001001
001000000000001
001010000000001
001100000001110
001110000001110
010000000000010
010010000111010
010100000001010
010110000111010
011001000000010
011011000111110
011101100001110
011111110111110
100000000000001
100010000000001
100100000001001
100110000001001
101000000000001
101010000000001
101100000001110
101110000001110
110000000000010
110010000111010
110100000001010
110110000111010
111001000000010
111011000111110
111101100001110
111111111111111

Общая таблица истинности:

FABCDA∧B(A∧B)∧C((A∧B)∧C)∧DD∧A(D∧A)∧DC∨((D∧A)∧D)F∧(((A∧B)∧C)∧D)B∧(C∨((D∧A)∧D))A∨(B∧(C∨((D∧A)∧D)))F∧(A∧B∧C∧D)≡A∨B∧(C∨D∧A∧D)
000000000000001
000010000000001
000100000010001
000110000010001
001000000000001
001010000000001
001100000010110
001110000010110
010000000000010
010010001110010
010100000010010
010110001110010
011001000000010
011011001110110
011101100010110
011111111110110
100000000000001
100010000000001
100100000010001
100110000010001
101000000000001
101010000000001
101100000010110
101110000010110
110000000000010
110010001110010
110100000010010
110110001110010
111001000000010
111011001110110
111101100010110
111111111111111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
FABCDF
000001
000011
000101
000111
001001
001011
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100001
100011
100101
100111
101001
101011
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111
Fсднф = ¬F∧¬A∧¬B∧¬C∧¬D ∨ ¬F∧¬A∧¬B∧¬C∧D ∨ ¬F∧¬A∧¬B∧C∧¬D ∨ ¬F∧¬A∧¬B∧C∧D ∨ ¬F∧¬A∧B∧¬C∧¬D ∨ ¬F∧¬A∧B∧¬C∧D ∨ F∧¬A∧¬B∧¬C∧¬D ∨ F∧¬A∧¬B∧¬C∧D ∨ F∧¬A∧¬B∧C∧¬D ∨ F∧¬A∧¬B∧C∧D ∨ F∧¬A∧B∧¬C∧¬D ∨ F∧¬A∧B∧¬C∧D ∨ F∧A∧B∧C∧D
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
FABCDF
000001
000011
000101
000111
001001
001011
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100001
100011
100101
100111
101001
101011
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111
Fскнф = (F∨A∨¬B∨¬C∨D) ∧ (F∨A∨¬B∨¬C∨¬D) ∧ (F∨¬A∨B∨C∨D) ∧ (F∨¬A∨B∨C∨¬D) ∧ (F∨¬A∨B∨¬C∨D) ∧ (F∨¬A∨B∨¬C∨¬D) ∧ (F∨¬A∨¬B∨C∨D) ∧ (F∨¬A∨¬B∨C∨¬D) ∧ (F∨¬A∨¬B∨¬C∨D) ∧ (F∨¬A∨¬B∨¬C∨¬D) ∧ (¬F∨A∨¬B∨¬C∨D) ∧ (¬F∨A∨¬B∨¬C∨¬D) ∧ (¬F∨¬A∨B∨C∨D) ∧ (¬F∨¬A∨B∨C∨¬D) ∧ (¬F∨¬A∨B∨¬C∨D) ∧ (¬F∨¬A∨B∨¬C∨¬D) ∧ (¬F∨¬A∨¬B∨C∨D) ∧ (¬F∨¬A∨¬B∨C∨¬D) ∧ (¬F∨¬A∨¬B∨¬C∨D)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
FABCDFж
000001
000011
000101
000111
001001
001011
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100001
100011
100101
100111
101001
101011
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111

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

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

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

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