
Задание 6
Найти формулы ПНФ и ССФ, выполнить унификацию атомов дизъюнктов.
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Формула |
1 |
x(A(x) B(y))y(B(y) A(x)) |
2 |
x( A(x)y( C(y)))x((D(x)A(x)) |
3 |
x(A(x)z(B(z)))y( A(x) C(y)C(y) B(x)) |
4 |
x(A(x)y(B(y)))x( A(x) B(y)) |
5 |
x(A(x)B(y)) y(A(x)(B(y)C(z))z(A(x)C(z)) |
6 |
x(A(x)y(B(y)C(z)))z(A(x) B(y)C(z)) |
7 |
x(A(x)B(z)) y(C(y)A(x))z(C(y)B(z)) |
8 |
x(A(x)B(y))y((C(y)A(x))(C(y)y(B(y))) |
9 |
x(B(x))y(A(y)B(x)) |
10 |
x(A(x)B(x))(y(C(y)A(x))z(C(z)B(x))) |
11 |
x(A(x)B(x))y((C(y)A(x))(C(y)B(x))) |
12 |
x(A(x) B(y))y(B(y) A(x)) |
13 |
x(A(x)B(z))y((C(y)A(x))z(C(y)B(z))) |
14 |
x(A(x)B(y)) z(C(z)A(x))y(C(z)B(y)) |
15 |
(x( A(x)y( C(y)))(C(x)A(x)) |
16 |
x(A(x)B(x)) y(B(x)C(y)) z(C(y)D(z))) |
17 |
x( A(x)y( B(y)))(B(x)A(x)) |
18 |
x( A(x)x(B(x)))(B(x)A(x)) |
19 |
(x( A(x)y(B(y))))( B(x)A(x)) |
20 |
x(A(x)B(y)) y(A(x)(B(y)C(z)))z(A(x)C(z)) |
21 |
x(A(x)z(B(y)C(z)))y(B(y)(A(x)C(z))) |
22 |
(x(A(x))x(B(x)))z((B(x)C(z))(A(x)C(z))) |
23 |
(x( A(x))x( B(x)))( B(x)A(x)) |
24 |
(x(A(x)))(x(B(x)))y(C(y) A(x)C(y) B(x)) |
25 |
x( A(x)y(B(y)))( B(y)A(x)) |
26 |
(x(B(x))x(A(x))) y((A(x)C(y))( C(y) B(x))) |
27 |
x( A(x)y(B(y)))(B(y)A(x)) |
28 |
x( A(x)y( B(y)))(B(y)A(x)) |
29 |
x(A(x)B(x)) y(B(x)C(y) z(C(y)D(z))) |
30 |
(x(A(x)B(x)) z(C(z)A(x)))y(C(z)B(y)) |
Задание 7
Доказать, что функция примитивно рекурсивна
Вариант |
Функция |
Вариант |
Функция |
1 |
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11 |
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2 |
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12 |
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3 |
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13 |
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4 |
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14 |
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5 |
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15 |
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6 |
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16 |
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7 |
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17 |
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8 |
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18 |
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9 |
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19 |
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10 |
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20 |
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21 |
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23 |
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22 |
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24 |
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