Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/86526)

The rewrite relation of the following TRS is considered.

0(1(2(0(x1)))) → 0(1(0(x1))) (1)
2(3(4(4(x1)))) → 0(3(4(x1))) (2)
2(2(4(3(2(5(2(x1))))))) → 0(3(0(0(0(x1))))) (3)
3(2(2(0(4(1(3(0(x1)))))))) → 3(4(2(0(1(0(5(0(x1)))))))) (4)
2(2(2(3(0(1(2(2(5(x1))))))))) → 0(1(1(1(3(5(2(5(x1)))))))) (5)
3(0(4(3(1(1(0(1(1(x1))))))))) → 3(4(1(2(3(3(4(3(x1)))))))) (6)
2(4(2(0(3(5(0(3(2(4(x1)))))))))) → 0(0(0(1(0(1(3(2(x1)))))))) (7)
4(2(3(3(1(0(5(2(5(3(x1)))))))))) → 5(4(4(0(2(0(0(4(3(x1))))))))) (8)
4(3(2(0(1(5(0(3(0(5(x1)))))))))) → 4(4(1(2(5(5(2(4(2(1(x1)))))))))) (9)
0(3(3(1(3(0(3(4(5(0(5(x1))))))))))) → 0(3(4(1(5(2(0(2(1(3(x1)))))))))) (10)
5(3(4(0(4(4(0(5(2(2(0(x1))))))))))) → 5(4(5(3(1(0(0(4(5(2(x1)))))))))) (11)
2(1(5(1(2(4(2(2(2(1(5(1(x1)))))))))))) → 2(1(3(2(5(1(2(0(2(3(4(x1))))))))))) (12)
2(2(5(5(1(2(1(5(5(5(3(0(2(x1))))))))))))) → 4(4(4(1(1(4(1(5(5(1(2(x1))))))))))) (13)
3(3(5(1(5(4(2(3(2(1(0(0(2(0(x1)))))))))))))) → 4(3(0(3(5(1(1(3(1(5(4(3(1(x1))))))))))))) (14)
2(3(3(0(3(3(3(1(4(2(0(5(4(1(2(2(x1)))))))))))))))) → 3(5(1(0(3(0(3(3(4(5(4(2(2(5(1(x1))))))))))))))) (15)
3(5(4(3(2(3(0(3(5(2(4(4(1(0(5(4(x1)))))))))))))))) → 3(5(4(5(0(5(1(0(2(3(3(4(5(2(2(2(4(x1))))))))))))))))) (16)
4(1(1(4(5(2(4(1(4(2(5(3(5(2(0(0(x1)))))))))))))))) → 4(3(0(2(3(3(5(4(4(0(0(0(5(4(2(3(x1)))))))))))))))) (17)
4(4(1(2(0(2(0(5(4(3(0(0(1(2(2(2(x1)))))))))))))))) → 4(5(4(2(4(1(0(4(0(1(1(2(4(4(4(x1))))))))))))))) (18)
0(2(1(1(5(2(5(0(3(5(5(0(3(5(4(2(2(x1))))))))))))))))) → 4(2(2(5(1(5(1(5(4(4(4(4(0(2(5(3(0(x1))))))))))))))))) (19)
2(5(5(1(0(0(2(1(2(3(5(1(1(1(1(2(2(4(x1)))))))))))))))))) → 4(3(2(1(5(4(0(3(2(3(5(5(2(4(1(4(x1)))))))))))))))) (20)
2(5(5(3(4(2(0(0(2(0(4(0(4(4(2(5(4(0(x1)))))))))))))))))) → 5(1(1(4(0(5(5(2(5(0(1(2(4(2(4(4(x1)))))))))))))))) (21)
3(0(1(1(5(1(2(1(0(3(2(3(2(0(3(4(3(4(0(x1))))))))))))))))))) → 1(5(3(0(5(4(4(0(0(4(1(3(4(3(1(4(1(x1))))))))))))))))) (22)
3(3(5(4(4(1(5(5(2(1(5(4(4(5(1(0(5(3(1(x1))))))))))))))))))) → 1(0(0(0(1(0(4(5(4(0(2(3(5(3(5(4(4(0(3(x1))))))))))))))))))) (23)
3(5(2(2(4(2(0(3(4(4(3(5(4(4(1(4(5(2(4(x1))))))))))))))))))) → 4(2(1(2(0(3(2(3(0(3(2(5(1(4(1(2(3(4(x1)))))))))))))))))) (24)
4(0(4(4(5(2(2(0(4(4(0(3(5(3(5(5(0(5(2(x1))))))))))))))))))) → 5(2(3(3(2(2(0(4(4(1(5(5(3(4(2(3(5(2(x1)))))))))))))))))) (25)
4(2(2(4(3(4(4(3(1(3(2(2(2(0(1(2(5(4(1(x1))))))))))))))))))) → 5(4(4(5(4(2(0(2(3(2(3(4(2(1(2(1(4(1(x1)))))))))))))))))) (26)
3(3(4(5(1(3(1(3(4(1(4(0(2(2(0(4(2(4(2(2(x1)))))))))))))))))))) → 1(0(1(4(3(3(4(4(4(3(5(0(1(2(0(2(3(3(x1)))))))))))))))))) (27)
3(5(2(0(2(0(4(5(2(1(4(2(5(0(5(2(5(3(2(5(x1)))))))))))))))))))) → 1(2(5(4(5(3(3(0(0(5(5(4(4(0(3(0(3(1(x1)))))))))))))))))) (28)
2(3(1(5(2(3(0(3(3(1(3(2(1(3(1(4(3(5(1(5(4(x1))))))))))))))))))))) → 3(1(1(4(5(4(5(3(4(2(4(5(3(5(4(1(3(4(0(4(x1)))))))))))))))))))) (29)
2(4(1(3(0(5(2(2(2(4(4(5(2(0(3(4(5(1(5(5(0(x1))))))))))))))))))))) → 4(3(5(4(2(4(5(1(0(2(3(5(5(0(4(4(5(3(3(5(1(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 +
170392/145
[51(x1)] = x1 +
0
[52(x1)] = x1 +
11
[53(x1)] = x1 +
201887/145
[54(x1)] = x1 +
0
[55(x1)] = x1 +
201742/145
[40(x1)] = x1 +
38076/145
[41(x1)] = x1 +
41907/145
[42(x1)] = x1 +
64334/145
[43(x1)] = x1 +
201742/145
[44(x1)] = x1 +
0
[45(x1)] = x1 +
0
[30(x1)] = x1 +
183094/145
[31(x1)] = x1 +
0
[32(x1)] = x1 +
45334/145
[33(x1)] = x1 +
201742/145
[34(x1)] = x1 +
46929/145
[35(x1)] = x1 +
37848/29
[20(x1)] = x1 +
201742/145
[21(x1)] = x1 +
201742/145
[22(x1)] = x1 +
32025/29
[23(x1)] = x1 +
201887/145
[24(x1)] = x1 +
201742/145
[25(x1)] = x1 +
201742/145
[10(x1)] = x1 +
118028/145
[11(x1)] = x1 +
0
[12(x1)] = x1 +
117686/145
[13(x1)] = x1 +
1
[14(x1)] = x1 +
159448/145
[15(x1)] = x1 +
0
[00(x1)] = x1 +
202322/145
[01(x1)] = x1 +
201742/145
[02(x1)] = x1 +
201742/145
[03(x1)] = x1 +
154622/145
[04(x1)] = x1 +
4
[05(x1)] = x1 +
4
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.