Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/68322)

The rewrite relation of the following TRS is considered.

0(1(0(0(x1)))) → 1(1(2(x1))) (1)
2(3(3(4(x1)))) → 4(1(4(x1))) (2)
4(2(4(0(x1)))) → 0(0(1(1(x1)))) (3)
2(1(0(1(0(x1))))) → 1(3(4(0(x1)))) (4)
5(3(3(4(5(x1))))) → 5(4(1(5(x1)))) (5)
0(3(4(5(3(0(x1)))))) → 0(3(5(1(0(2(x1)))))) (6)
2(4(1(1(3(5(x1)))))) → 2(0(4(4(5(x1))))) (7)
0(3(1(2(1(1(0(x1))))))) → 1(1(3(4(2(0(x1)))))) (8)
0(3(5(4(5(3(0(x1))))))) → 0(3(4(3(5(0(4(x1))))))) (9)
2(1(1(3(5(5(3(x1))))))) → 1(3(5(4(0(5(x1)))))) (10)
1(2(3(3(3(5(1(1(x1)))))))) → 1(5(0(1(5(0(1(x1))))))) (11)
2(0(2(2(4(3(1(0(x1)))))))) → 2(1(1(5(4(2(2(x1))))))) (12)
3(0(5(0(5(0(3(3(x1)))))))) → 5(5(1(0(2(4(1(x1))))))) (13)
3(3(2(4(2(3(4(3(3(x1))))))))) → 5(4(4(5(0(4(3(x1))))))) (14)
4(2(0(2(5(2(2(3(1(0(x1)))))))))) → 5(0(5(1(2(5(0(5(x1)))))))) (15)
4(3(5(5(2(5(3(3(0(1(x1)))))))))) → 4(2(1(4(0(2(3(3(4(2(x1)))))))))) (16)
1(3(1(1(0(0(2(0(3(4(0(x1))))))))))) → 1(2(0(0(2(3(4(3(2(1(x1)))))))))) (17)
2(1(2(1(3(2(4(4(0(0(3(x1))))))))))) → 5(1(0(5(3(1(4(3(0(3(x1)))))))))) (18)
4(0(0(1(0(4(3(3(0(2(2(x1))))))))))) → 5(3(5(3(4(2(4(4(2(x1))))))))) (19)
1(3(1(1(1(3(0(2(0(0(3(0(x1)))))))))))) → 1(3(1(4(5(5(4(3(0(5(x1)))))))))) (20)
3(3(5(5(5(5(2(4(1(0(1(5(x1)))))))))))) → 5(0(5(1(1(1(3(0(1(2(0(5(x1)))))))))))) (21)
5(2(1(4(2(2(1(2(4(1(2(0(x1)))))))))))) → 5(2(5(2(2(4(1(3(4(0(4(x1))))))))))) (22)
3(5(1(4(4(1(2(2(5(2(3(4(4(x1))))))))))))) → 0(0(3(2(3(5(2(4(0(1(0(0(4(5(x1)))))))))))))) (23)
4(0(3(4(0(4(3(0(2(3(0(5(2(x1))))))))))))) → 2(3(4(5(1(3(5(0(5(5(2(2(x1)))))))))))) (24)
0(5(3(3(0(1(4(5(4(4(1(1(3(0(x1)))))))))))))) → 0(1(5(0(3(5(2(4(0(5(3(1(3(2(x1)))))))))))))) (25)
2(3(0(4(3(3(1(1(0(2(1(4(5(5(x1)))))))))))))) → 2(5(0(2(2(2(0(3(5(3(0(4(5(x1))))))))))))) (26)
3(5(0(2(5(3(2(2(1(0(3(0(0(2(x1)))))))))))))) → 3(5(3(2(1(2(3(4(3(1(1(0(5(x1))))))))))))) (27)
4(1(0(5(1(2(5(5(3(4(2(1(3(0(0(x1))))))))))))))) → 1(5(4(5(0(3(5(1(2(4(3(3(0(2(x1)))))))))))))) (28)
3(0(0(4(1(4(0(1(4(4(0(4(4(5(5(0(x1)))))))))))))))) → 2(3(0(4(4(1(1(1(0(2(0(0(1(4(0(5(1(x1))))))))))))))))) (29)
3(1(0(2(5(0(0(4(1(3(0(1(3(4(2(3(x1)))))))))))))))) → 5(3(0(3(1(3(4(5(2(1(2(4(3(4(3(x1))))))))))))))) (30)
2(0(2(3(5(1(1(5(4(4(3(4(2(2(2(4(5(x1))))))))))))))))) → 2(2(5(3(0(3(5(4(3(4(0(1(5(4(5(5(x1)))))))))))))))) (31)
2(5(4(3(0(1(2(2(5(2(3(4(3(3(5(0(2(x1))))))))))))))))) → 4(4(3(3(4(0(2(1(3(3(5(5(0(3(3(1(4(x1))))))))))))))))) (32)
5(3(3(3(5(0(2(1(0(0(4(5(1(3(2(0(0(x1))))))))))))))))) → 5(4(3(3(0(3(1(2(5(2(4(3(1(1(4(1(x1)))))))))))))))) (33)
5(4(3(4(3(4(5(2(2(0(5(0(4(3(4(0(2(2(x1)))))))))))))))))) → 5(1(2(5(1(1(3(3(0(4(3(2(5(2(1(2(4(2(x1)))))))))))))))))) (34)
1(0(3(1(1(2(2(0(3(3(2(0(0(1(1(5(3(5(2(x1))))))))))))))))))) → 0(1(3(1(1(4(5(0(1(2(1(4(2(2(2(4(4(x1))))))))))))))))) (35)
2(4(5(2(3(2(5(1(4(1(3(5(0(0(1(1(3(1(4(x1))))))))))))))))))) → 5(2(4(4(4(4(5(4(5(3(4(1(4(5(5(0(x1)))))))))))))))) (36)
4(0(0(1(2(4(0(2(4(1(2(1(0(4(3(1(1(0(2(x1))))))))))))))))))) → 4(0(3(1(4(1(0(3(1(0(4(5(3(0(0(1(4(4(x1)))))))))))))))))) (37)
1(2(4(1(0(2(1(5(0(0(5(1(4(0(4(5(3(1(2(5(0(x1))))))))))))))))))))) → 1(0(2(3(5(4(1(1(1(1(1(5(5(4(1(4(4(4(1(2(0(x1))))))))))))))))))))) (38)
2(4(4(3(3(3(5(2(3(0(2(5(1(3(3(0(4(3(4(0(1(x1))))))))))))))))))))) → 5(1(1(4(1(4(2(5(2(3(0(1(3(1(1(5(5(1(3(3(x1)))))))))))))))))))) (39)
5(3(4(2(5(2(5(1(1(5(3(5(1(5(0(4(1(0(2(2(0(x1))))))))))))))))))))) → 5(0(3(5(4(5(5(1(2(2(1(4(0(0(1(2(5(1(5(4(x1)))))))))))))))))))) (40)

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 240 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 1440 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 +
14416157/1658
[51(x1)] = x1 +
2952191/829
[52(x1)] = x1 +
29151577/1658
[53(x1)] = x1 +
29161525/1658
[54(x1)] = x1 +
48
[55(x1)] = x1 +
36
[40(x1)] = x1 +
4385855/829
[41(x1)] = x1 +
29163183/1658
[42(x1)] = x1 +
11743617/829
[43(x1)] = x1 +
11783409/829
[44(x1)] = x1 +
22999877/1658
[45(x1)] = x1 +
7585947/829
[30(x1)] = x1 +
27759841/1658
[31(x1)] = x1 +
20259193/1658
[32(x1)] = x1 +
29232819/1658
[33(x1)] = x1 +
3047519/829
[34(x1)] = x1 +
25952611/1658
[35(x1)] = x1 +
29181421/1658
[20(x1)] = x1 +
28772879/1658
[21(x1)] = x1 +
6567772/829
[22(x1)] = x1 +
29231161/1658
[23(x1)] = x1 +
29231161/1658
[24(x1)] = x1 +
29151577/1658
[25(x1)] = x1 +
9933009/1658
[10(x1)] = x1 +
258
[11(x1)] = x1 +
3178517/829
[12(x1)] = x1 +
14269235/1658
[13(x1)] = x1 +
9966789/829
[14(x1)] = x1 +
2766019/829
[15(x1)] = x1 +
14269235/1658
[00(x1)] = x1 +
12750457/1658
[01(x1)] = x1 +
29151577/1658
[02(x1)] = x1 +
11743617/829
[03(x1)] = x1 +
29231161/1658
[04(x1)] = x1 +
22951795/1658
[05(x1)] = x1 +
11744446/829
all of the following rules can be deleted.

There are 1440 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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