Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/86452)

The rewrite relation of the following TRS is considered.

0(1(2(x1))) → 0(3(0(x1))) (1)
4(4(3(3(x1)))) → 5(4(4(x1))) (2)
3(0(0(3(5(x1))))) → 5(5(2(5(x1)))) (3)
5(4(1(4(1(2(x1)))))) → 4(4(1(4(0(0(x1)))))) (4)
3(2(4(2(4(1(1(3(x1)))))))) → 5(4(1(1(5(5(3(x1))))))) (5)
3(3(2(0(2(1(5(3(x1)))))))) → 2(0(1(3(1(0(3(3(x1)))))))) (6)
3(2(5(5(2(5(1(4(3(x1))))))))) → 4(4(0(0(0(2(3(0(5(x1))))))))) (7)
3(3(4(0(0(2(4(0(0(x1))))))))) → 0(2(0(1(5(0(0(2(x1)))))))) (8)
5(1(2(0(0(4(5(1(2(x1))))))))) → 5(2(5(5(3(1(2(0(x1)))))))) (9)
0(0(4(2(2(1(0(3(1(5(x1)))))))))) → 0(0(5(5(1(4(4(2(5(x1))))))))) (10)
3(1(1(2(4(5(5(4(3(1(2(x1))))))))))) → 5(0(4(1(1(4(0(1(0(3(2(x1))))))))))) (11)
1(4(0(4(0(1(3(4(3(3(4(0(x1)))))))))))) → 1(5(2(2(5(4(3(3(5(4(0(x1))))))))))) (12)
2(0(5(5(5(2(2(0(2(1(5(2(x1)))))))))))) → 2(2(0(2(4(5(0(5(0(3(0(4(x1)))))))))))) (13)
2(3(5(0(0(2(3(0(2(1(0(1(x1)))))))))))) → 2(2(1(5(4(4(1(2(4(0(3(x1))))))))))) (14)
0(2(2(0(2(0(5(1(4(2(3(4(3(x1))))))))))))) → 0(2(5(4(0(4(1(1(4(5(4(0(0(x1))))))))))))) (15)
0(3(2(2(3(2(0(2(1(1(4(1(3(x1))))))))))))) → 0(0(0(2(0(3(5(1(5(1(2(3(5(x1))))))))))))) (16)
2(0(2(1(4(1(5(5(5(5(5(0(3(x1))))))))))))) → 2(2(0(3(5(2(2(2(3(1(2(0(5(3(x1)))))))))))))) (17)
3(4(0(5(3(1(4(2(0(3(3(4(1(x1))))))))))))) → 2(4(3(5(5(1(5(4(2(2(4(3(x1)))))))))))) (18)
3(4(3(5(1(0(5(4(4(4(2(1(2(x1))))))))))))) → 2(3(3(1(5(2(2(1(3(3(2(2(3(x1))))))))))))) (19)
3(5(0(5(4(2(2(4(2(3(5(2(5(x1))))))))))))) → 3(3(2(3(5(1(4(1(0(0(5(0(5(x1))))))))))))) (20)
4(3(4(4(2(4(4(1(3(5(5(1(0(x1))))))))))))) → 4(5(1(1(5(3(5(3(2(1(3(2(2(x1))))))))))))) (21)
1(1(4(0(0(4(3(1(4(1(1(0(3(1(5(x1))))))))))))))) → 1(1(3(0(0(3(0(4(5(5(2(1(0(0(4(x1))))))))))))))) (22)
2(5(4(5(1(1(1(1(3(2(1(5(0(1(2(x1))))))))))))))) → 5(2(2(3(1(0(2(0(5(5(0(1(2(5(2(x1))))))))))))))) (23)
3(2(3(1(1(2(2(5(5(2(5(3(5(4(1(2(x1)))))))))))))))) → 2(0(5(3(3(0(4(5(3(5(5(0(3(0(0(0(0(x1))))))))))))))))) (24)
1(1(1(3(0(0(2(0(5(3(1(3(5(5(4(1(3(x1))))))))))))))))) → 1(2(1(5(3(4(3(2(0(0(5(4(0(5(3(5(2(x1))))))))))))))))) (25)
4(5(5(4(4(2(0(5(1(3(2(1(3(0(0(4(5(x1))))))))))))))))) → 4(5(2(5(0(5(1(5(0(0(3(4(5(0(2(1(0(x1))))))))))))))))) (26)
0(3(1(2(0(4(4(5(3(0(4(4(3(4(0(5(0(3(x1)))))))))))))))))) → 4(4(3(4(0(2(2(2(3(0(1(1(0(5(3(1(0(x1))))))))))))))))) (27)
3(2(5(4(1(0(2(2(0(5(3(2(5(3(1(0(1(0(x1)))))))))))))))))) → 2(0(0(3(3(1(4(3(5(1(5(0(5(0(1(0(5(0(x1)))))))))))))))))) (28)
4(1(3(3(3(3(3(5(3(5(2(3(2(2(5(4(2(1(3(x1))))))))))))))))))) → 4(1(2(2(0(0(5(0(0(0(2(1(2(4(0(0(4(0(0(0(x1)))))))))))))))))))) (29)
1(2(2(1(3(0(5(5(2(1(5(5(3(4(0(0(0(0(0(0(3(x1))))))))))))))))))))) → 1(1(5(3(4(4(5(3(1(2(0(1(2(5(0(1(2(2(0(0(x1)))))))))))))))))))) (30)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{5(☐), 4(☐), 3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 180 ruless (increase limit for explicit display).

1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,...,5}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 6):

[5(x1)] = 6x1 + 0
[4(x1)] = 6x1 + 1
[3(x1)] = 6x1 + 2
[2(x1)] = 6x1 + 3
[1(x1)] = 6x1 + 4
[0(x1)] = 6x1 + 5

We obtain the labeled TRS

There are 1080 ruless (increase limit for explicit display).

1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[50(x1)] = x1 +
36855/59
[51(x1)] = x1 +
56511/59
[52(x1)] = x1 +
45402/59
[53(x1)] = x1 +
1163/59
[54(x1)] = x1 +
62050/59
[55(x1)] = x1 +
5
[40(x1)] = x1 +
5
[41(x1)] = x1 +
0
[42(x1)] = x1 +
61755/59
[43(x1)] = x1 +
61755/59
[44(x1)] = x1 +
50094/59
[45(x1)] = x1 +
0
[30(x1)] = x1 +
62994/59
[31(x1)] = x1 +
61755/59
[32(x1)] = x1 +
61814/59
[33(x1)] = x1 +
50458/59
[34(x1)] = x1 +
52095/59
[35(x1)] = x1 +
5539/59
[20(x1)] = x1 +
61755/59
[21(x1)] = x1 +
11887/59
[22(x1)] = x1 +
61755/59
[23(x1)] = x1 +
6
[24(x1)] = x1 +
62109/59
[25(x1)] = x1 +
10083/59
[10(x1)] = x1 +
6
[11(x1)] = x1 +
57201/59
[12(x1)] = x1 +
62050/59
[13(x1)] = x1 +
62109/59
[14(x1)] = x1 +
0
[15(x1)] = x1 +
35259/59
[00(x1)] = x1 +
0
[01(x1)] = x1 +
36037/59
[02(x1)] = x1 +
32361/59
[03(x1)] = x1 +
24445/59
[04(x1)] = x1 +
1
[05(x1)] = x1 +
5
all of the following rules can be deleted.

There are 1080 ruless (increase limit for explicit display).

1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.