Таблица истинности для функции X∧Y∧Z∧T∧F∧(X∧Y∧Z∧T):


Промежуточные таблицы истинности:
X∧Y:
XYX∧Y
000
010
100
111

(X∧Y)∧Z:
XYZX∧Y(X∧Y)∧Z
00000
00100
01000
01100
10000
10100
11010
11111

((X∧Y)∧Z)∧T:
XYZTX∧Y(X∧Y)∧Z((X∧Y)∧Z)∧T
0000000
0001000
0010000
0011000
0100000
0101000
0110000
0111000
1000000
1001000
1010000
1011000
1100100
1101100
1110110
1111111

(((X∧Y)∧Z)∧T)∧F:
XYZTFX∧Y(X∧Y)∧Z((X∧Y)∧Z)∧T(((X∧Y)∧Z)∧T)∧F
000000000
000010000
000100000
000110000
001000000
001010000
001100000
001110000
010000000
010010000
010100000
010110000
011000000
011010000
011100000
011110000
100000000
100010000
100100000
100110000
101000000
101010000
101100000
101110000
110001000
110011000
110101000
110111000
111001100
111011100
111101110
111111111

((((X∧Y)∧Z)∧T)∧F)∧(((X∧Y)∧Z)∧T):
XYZTFX∧Y(X∧Y)∧Z((X∧Y)∧Z)∧T(((X∧Y)∧Z)∧T)∧FX∧Y(X∧Y)∧Z((X∧Y)∧Z)∧T((((X∧Y)∧Z)∧T)∧F)∧(((X∧Y)∧Z)∧T)
0000000000000
0000100000000
0001000000000
0001100000000
0010000000000
0010100000000
0011000000000
0011100000000
0100000000000
0100100000000
0101000000000
0101100000000
0110000000000
0110100000000
0111000000000
0111100000000
1000000000000
1000100000000
1001000000000
1001100000000
1010000000000
1010100000000
1011000000000
1011100000000
1100010001000
1100110001000
1101010001000
1101110001000
1110011001100
1110111001100
1111011101110
1111111111111

Общая таблица истинности:

XYZTFX∧Y(X∧Y)∧Z((X∧Y)∧Z)∧T(((X∧Y)∧Z)∧T)∧FX∧Y∧Z∧T∧F∧(X∧Y∧Z∧T)
0000000000
0000100000
0001000000
0001100000
0010000000
0010100000
0011000000
0011100000
0100000000
0100100000
0101000000
0101100000
0110000000
0110100000
0111000000
0111100000
1000000000
1000100000
1001000000
1001100000
1010000000
1010100000
1011000000
1011100000
1100010000
1100110000
1101010000
1101110000
1110011000
1110111000
1111011100
1111111111

Логическая схема:

Совершенная дизъюнктивная нормальная форма (СДНФ):

По таблице истинности:
XYZTFF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111
Fсднф = X∧Y∧Z∧T∧F
Логическая cхема:

Совершенная конъюнктивная нормальная форма (СКНФ):

По таблице истинности:
XYZTFF
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111
Fскнф = (X∨Y∨Z∨T∨F) ∧ (X∨Y∨Z∨T∨¬F) ∧ (X∨Y∨Z∨¬T∨F) ∧ (X∨Y∨Z∨¬T∨¬F) ∧ (X∨Y∨¬Z∨T∨F) ∧ (X∨Y∨¬Z∨T∨¬F) ∧ (X∨Y∨¬Z∨¬T∨F) ∧ (X∨Y∨¬Z∨¬T∨¬F) ∧ (X∨¬Y∨Z∨T∨F) ∧ (X∨¬Y∨Z∨T∨¬F) ∧ (X∨¬Y∨Z∨¬T∨F) ∧ (X∨¬Y∨Z∨¬T∨¬F) ∧ (X∨¬Y∨¬Z∨T∨F) ∧ (X∨¬Y∨¬Z∨T∨¬F) ∧ (X∨¬Y∨¬Z∨¬T∨F) ∧ (X∨¬Y∨¬Z∨¬T∨¬F) ∧ (¬X∨Y∨Z∨T∨F) ∧ (¬X∨Y∨Z∨T∨¬F) ∧ (¬X∨Y∨Z∨¬T∨F) ∧ (¬X∨Y∨Z∨¬T∨¬F) ∧ (¬X∨Y∨¬Z∨T∨F) ∧ (¬X∨Y∨¬Z∨T∨¬F) ∧ (¬X∨Y∨¬Z∨¬T∨F) ∧ (¬X∨Y∨¬Z∨¬T∨¬F) ∧ (¬X∨¬Y∨Z∨T∨F) ∧ (¬X∨¬Y∨Z∨T∨¬F) ∧ (¬X∨¬Y∨Z∨¬T∨F) ∧ (¬X∨¬Y∨Z∨¬T∨¬F) ∧ (¬X∨¬Y∨¬Z∨T∨F) ∧ (¬X∨¬Y∨¬Z∨T∨¬F) ∧ (¬X∨¬Y∨¬Z∨¬T∨F)
Логическая cхема:

Построение полинома Жегалкина:

По таблице истинности функции
XYZTFFж
000000
000010
000100
000110
001000
001010
001100
001110
010000
010010
010100
010110
011000
011010
011100
011110
100000
100010
100100
100110
101000
101010
101100
101110
110000
110010
110100
110110
111000
111010
111100
111111

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

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 = 0 => С10000 = 0 ⊕ 0 = 0
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 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10100) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С10100 = 0 => С10100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10010) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С10010 = 0 => С10010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10001) = С00000 ⊕ С10000 ⊕ С00001 ⊕ С10001 = 0 => С10001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 = 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 = 0 => С11100 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11010) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00010 ⊕ С11000 ⊕ С10010 ⊕ С01010 ⊕ С11010 = 0 => С11010 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(11001) = С00000 ⊕ С10000 ⊕ С01000 ⊕ С00001 ⊕ С11000 ⊕ С10001 ⊕ С01001 ⊕ С11001 = 0 => С11001 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10110) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00010 ⊕ С10100 ⊕ С10010 ⊕ С00110 ⊕ С10110 = 0 => С10110 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10101) = С00000 ⊕ С10000 ⊕ С00100 ⊕ С00001 ⊕ С10100 ⊕ С10001 ⊕ С00101 ⊕ С10101 = 0 => С10101 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 0
Fж(10011) = С00000 ⊕ С10000 ⊕ С00010 ⊕ С00001 ⊕ С10010 ⊕ С10001 ⊕ С00011 ⊕ С10011 = 0 => С10011 = 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0 = 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 = 0 => С11110 = 0 ⊕ 0 ⊕ 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 = 0 ⊕ 0 ⊕ 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 = 0 ⊕ 0 ⊕ 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 = 0 ⊕ 0 ⊕ 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 = 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 ⊕ 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 ⊕ 0 ⊕ 0 ⊕ 1 = 1

Таким образом, полином Жегалкина будет равен:
Fж = X∧Y∧Z∧T∧F
Логическая схема, соответствующая полиному Жегалкина:

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