Задание 2. Логика предикатов
Согласно варианту (см. табл. 2):
преобразовать формулу к виду ПНФ, затем КНФ и ССФ,
сформировать множество дизъюнктов К,
выполнить унификацию дизъюнктов множества К,
доказать истинность заключения:
а) методом дедукции,
b) методом резолюции.
Таблица 2
Вариант |
Формула |
1 |
x(A(x)B(y))y(A(x)(B(y)C(z)))z(C(z)) |
2 |
x(B(x)z(A(z)))y(A(z)C(y))(C(y)B(x)) |
3 |
x(A(x)y(B(y)))y(A(x)C(z)B(y)) |
4 |
x(A(x)z(C(z)))y(C(z)B(y))(A(x)B(y)) |
5 |
x(A(x)y(B(y)C(z)))z(A(x)B(y)C(z)) |
6 |
x(A(x)B(z))y(C(y)A(x))z(C(y)B(z)) |
7 |
x((A(x)B(y))y((C(y)A(x)))(C(y)y(B(y))) |
8 |
x(A(x)B(y))y(A(x)(B(y)C(z)))(A(x)z(C(z))) |
9 |
x(A(x)B(y)A(x)y(B(y)C(z)))z(C(z)) |
10 |
x(A(x)z(B(y)C(z)))y(B(y)(A(x)C(z))) |
11 |
(x(A(x))z(B(z)))z((B(x)C(z))C(z)) |
12 |
(x(A(x))z(B(z)))(B(x)A(x)) |
13 |
(x(A(x))y(B(y)))y(C(y)A(x)C(y)B(x)) |
14 |
x(A(x)y(B(y)))(B(y)A(x)) |
15 |
x(A(x)y(B(y)))(B(y)A(x)) |
16 |
x(A(x)B(x))y(B(x)C(y))z(C(y)D(z)) |
17 |
x(A(x)B(x))z(C(z)A(x))y(C(z)B(y)) |
18 |
x(B(x)y(A(y)))y(B(y)(A(x)C(z)))z(C(z))) |
19 |
(x(A(x)B(y)))(A(x)y(B(y)C(z)))z(C(z)) |
20 |
x(A(x)B(x))(y(C(y)A(x))z(C(z)B(x))) |
21 |
x(B(x)A(y))(B(x)y(A(y)C(z)))z(C(z)) |
22 |
(x(A(x)B(z))y(C(y)A(x)))z(C(y)B(z)) |
23 |
(x(B(x))y(A(y)))(A(y)y(C(y)))(A(x)C(y)) |
24 |
((x(A(x))x(B(x)))y(A(x)C(y)))(B(x)C(y)) |
25 |
x(A(x)y(B(y)))(A(x)y(B(y)))B(y) |
26 |
x(A(x)y(B(y)))(A(x)B(x))B(x) |
27 |
(x(B(x))y(A(y)))(B(x)A(y))A(z) |
28 |
(x(B(x))z(C(z)))(A(y)B(x)A(y)C(z)) |
29 |
x(A(x)B(y))yz((C(z)A(x))(C(z)B(y))) |
30 |
(x(A(x))z(C(z)))y(C(z)B(y))(A(x)B(y)) |
31 |
(x(A(x))y(B(y)))y(C(y)x(D(x)))(A(x)C(y)) |
32 |
x(A(x)B(y))(A(x)y(B(y)C(z)))z(C(z)) |
33 |
(x(A(x)B(z))y(C(y)A(x)))z(C(y)B(z)) |
34 |
x(A(x)B(y))z(C(z)A(x))y(C(z)B(y)) |
35 |
(x(A(x)z(B(z)))y(A(x)C(y)))(B(z)C(y)) |
36 |
x(B(x)y(A(y)))y(B(x)(A(y)C(z))) |
37 |
x(A(x)B(x))y((C(y)A(x))(C(y)B(x))) |