Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96104)

The rewrite relation of the following TRS is considered.

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

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 150 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 900 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 +
5
[51(x1)] = x1 +
121534/71
[52(x1)] = x1 +
93
[53(x1)] = x1 +
1
[54(x1)] = x1 +
0
[55(x1)] = x1 +
31
[40(x1)] = x1 +
121534/71
[41(x1)] = x1 +
8987/142
[42(x1)] = x1 +
121179/71
[43(x1)] = x1 +
121179/71
[44(x1)] = x1 +
15345/71
[45(x1)] = x1 +
121179/71
[30(x1)] = x1 +
123380/71
[31(x1)] = x1 +
139593/142
[32(x1)] = x1 +
84537/71
[33(x1)] = x1 +
39897/71
[34(x1)] = x1 +
121179/71
[35(x1)] = x1 +
134757/142
[20(x1)] = x1 +
5
[21(x1)] = x1 +
121534/71
[22(x1)] = x1 +
105183/71
[23(x1)] = x1 +
121179/71
[24(x1)] = x1 +
1
[25(x1)] = x1 +
0
[10(x1)] = x1 +
121321/142
[11(x1)] = x1 +
121534/71
[12(x1)] = x1 +
121179/71
[13(x1)] = x1 +
91933/142
[14(x1)] = x1 +
214365/142
[15(x1)] = x1 +
121179/71
[00(x1)] = x1 +
0
[01(x1)] = x1 +
121179/71
[02(x1)] = x1 +
121179/71
[03(x1)] = x1 +
0
[04(x1)] = x1 +
166519/142
[05(x1)] = x1 +
121250/71
all of the following rules can be deleted.

There are 900 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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