Для функции ¬X∨Y⊕¬Z|P→Y≡P↓X∧ZC:


Промежуточные таблицы истинности:
¬X:
X¬X
01
10

¬Z:
Z¬Z
01
10

(¬Z)|P:
ZP¬Z(¬Z)|P
0011
0110
1001
1101

P↓X:
PXP↓X
001
010
100
110

(P↓X)∧ZC:
PXZCP↓X(P↓X)∧ZC
00010
00111
01000
01100
10000
10100
11000
11100

(¬X)∨Y:
XY¬X(¬X)∨Y
0011
0111
1000
1101

((¬X)∨Y)⊕((¬Z)|P):
XYZP¬X(¬X)∨Y¬Z(¬Z)|P((¬X)∨Y)⊕((¬Z)|P)
000011110
000111101
001011010
001111010
010011110
010111101
011011010
011111010
100000111
100100100
101000011
101100011
110001110
110101101
111001010
111101010

(((¬X)∨Y)⊕((¬Z)|P))→Y:
XYZP¬X(¬X)∨Y¬Z(¬Z)|P((¬X)∨Y)⊕((¬Z)|P)(((¬X)∨Y)⊕((¬Z)|P))→Y
0000111101
0001111010
0010110101
0011110101
0100111101
0101111011
0110110101
0111110101
1000001110
1001001001
1010000110
1011000110
1100011101
1101011011
1110010101
1111010101

((((¬X)∨Y)⊕((¬Z)|P))→Y)≡((P↓X)∧ZC):
XYZPZC¬X(¬X)∨Y¬Z(¬Z)|P((¬X)∨Y)⊕((¬Z)|P)(((¬X)∨Y)⊕((¬Z)|P))→YP↓X(P↓X)∧ZC((((¬X)∨Y)⊕((¬Z)|P))→Y)≡((P↓X)∧ZC)
00000111101100
00001111101111
00010111010001
00011111010001
00100110101100
00101110101111
00110110101000
00111110101000
01000111101100
01001111101111
01010111011000
01011111011000
01100110101100
01101110101111
01110110101000
01111110101000
10000001110001
10001001110001
10010001001000
10011001001000
10100000110001
10101000110001
10110000110001
10111000110001
11000011101000
11001011101000
11010011011000
11011011011000
11100010101000
11101010101000
11110010101000
11111010101000

Общая таблица истинности:

XYZPZC¬X¬Z(¬Z)|PP↓X(P↓X)∧ZC(¬X)∨Y((¬X)∨Y)⊕((¬Z)|P)(((¬X)∨Y)⊕((¬Z)|P))→Y¬X∨Y⊕¬Z|P→Y≡P↓X∧ZC
00000111101010
00001111111011
00010110001101
00011110001101
00100101101010
00101101111011
00110101001010
00111101001010
01000111101010
01001111111011
01010110001110
01011110001110
01100101101010
01101101111011
01110101001010
01111101001010
10000011000101
10001011000101
10010010000010
10011010000010
10100001000101
10101001000101
10110001000101
10111001000101
11000011001010
11001011001010
11010010001110
11011010001110
11100001001010
11101001001010
11110001001010
11111001001010

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
XYZPZCF
000000
000011
000101
000111
001000
001011
001100
001110
010000
010011
010100
010110
011000
011011
011100
011110
100001
100011
100100
100110
101001
101011
101101
101111
110000
110010
110100
110110
111000
111010
111100
111110
Fсднф = ¬X∧¬Y∧¬Z∧¬P∧ZC ∨ ¬X∧¬Y∧¬Z∧P∧¬ZC ∨ ¬X∧¬Y∧¬Z∧P∧ZC ∨ ¬X∧¬Y∧Z∧¬P∧ZC ∨ ¬X∧Y∧¬Z∧¬P∧ZC ∨ ¬X∧Y∧Z∧¬P∧ZC ∨ X∧¬Y∧¬Z∧¬P∧¬ZC ∨ X∧¬Y∧¬Z∧¬P∧ZC ∨ X∧¬Y∧Z∧¬P∧¬ZC ∨ X∧¬Y∧Z∧¬P∧ZC ∨ X∧¬Y∧Z∧P∧¬ZC ∨ X∧¬Y∧Z∧P∧ZC
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
XYZPZCF
000000
000011
000101
000111
001000
001011
001100
001110
010000
010011
010100
010110
011000
011011
011100
011110
100001
100011
100100
100110
101001
101011
101101
101111
110000
110010
110100
110110
111000
111010
111100
111110
Fскнф = (X∨Y∨Z∨P∨ZC) ∧ (X∨Y∨¬Z∨P∨ZC) ∧ (X∨Y∨¬Z∨¬P∨ZC) ∧ (X∨Y∨¬Z∨¬P∨¬ZC) ∧ (X∨¬Y∨Z∨P∨ZC) ∧ (X∨¬Y∨Z∨¬P∨ZC) ∧ (X∨¬Y∨Z∨¬P∨¬ZC) ∧ (X∨¬Y∨¬Z∨P∨ZC) ∧ (X∨¬Y∨¬Z∨¬P∨ZC) ∧ (X∨¬Y∨¬Z∨¬P∨¬ZC) ∧ (¬X∨Y∨Z∨¬P∨ZC) ∧ (¬X∨Y∨Z∨¬P∨¬ZC) ∧ (¬X∨¬Y∨Z∨P∨ZC) ∧ (¬X∨¬Y∨Z∨P∨¬ZC) ∧ (¬X∨¬Y∨Z∨¬P∨ZC) ∧ (¬X∨¬Y∨Z∨¬P∨¬ZC) ∧ (¬X∨¬Y∨¬Z∨P∨ZC) ∧ (¬X∨¬Y∨¬Z∨P∨¬ZC) ∧ (¬X∨¬Y∨¬Z∨¬P∨ZC) ∧ (¬X∨¬Y∨¬Z∨¬P∨¬ZC)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
XYZPZCFж
000000
000011
000101
000111
001000
001011
001100
001110
010000
010011
010100
010110
011000
011011
011100
011110
100001
100011
100100
100110
101001
101011
101101
101111
110000
110010
110100
110110
111000
111010
111100
111110

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

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

Таким образом, полином Жегалкина будет равен:
Fж = X ⊕ P ⊕ ZC ⊕ X∧Y ⊕ X∧ZC ⊕ Y∧P ⊕ Z∧P ⊕ P∧ZC ⊕ X∧P∧ZC ⊕ Y∧Z∧P
Логическая схема, соответствующая полиному Жегалкина:

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