Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96239)

The rewrite relation of the following TRS is considered.

0(1(1(2(x1)))) → 3(4(4(2(x1)))) (1)
2(1(0(2(5(5(x1)))))) → 4(0(5(2(5(x1))))) (2)
4(2(0(3(1(2(x1)))))) → 5(5(0(5(2(5(x1)))))) (3)
0(1(5(0(4(4(5(x1))))))) → 3(0(0(4(5(0(x1)))))) (4)
0(4(1(2(2(4(5(x1))))))) → 0(0(1(5(4(5(x1)))))) (5)
3(5(4(0(0(4(0(x1))))))) → 0(0(0(5(4(0(x1)))))) (6)
2(3(4(3(1(0(2(4(x1)))))))) → 0(4(5(3(1(4(x1)))))) (7)
2(5(2(4(4(3(1(0(0(x1))))))))) → 4(2(5(5(4(5(5(0(x1)))))))) (8)
5(4(1(5(5(5(3(3(0(x1))))))))) → 5(3(5(0(4(3(5(4(0(x1))))))))) (9)
4(0(5(1(2(1(3(3(0(5(3(x1))))))))))) → 1(2(1(4(4(3(2(5(0(2(3(x1))))))))))) (10)
4(5(2(5(0(0(2(2(0(4(1(x1))))))))))) → 4(4(4(3(3(1(3(5(5(1(1(x1))))))))))) (11)
1(2(3(1(4(1(3(2(0(5(1(4(x1)))))))))))) → 1(0(4(4(5(1(2(4(3(3(0(1(x1)))))))))))) (12)
1(1(5(1(4(3(4(1(1(5(1(2(5(x1))))))))))))) → 5(0(0(3(3(0(5(5(5(4(2(3(x1)))))))))))) (13)
1(4(0(0(1(3(4(3(3(0(3(0(4(5(x1)))))))))))))) → 5(5(3(3(5(0(4(5(4(1(5(0(0(x1))))))))))))) (14)
2(1(0(3(2(0(2(0(3(3(2(3(4(0(5(x1))))))))))))))) → 2(2(3(3(5(5(5(3(1(4(0(5(0(2(2(x1))))))))))))))) (15)
0(5(5(4(0(5(3(1(5(4(4(2(2(5(3(4(x1)))))))))))))))) → 0(4(2(1(2(0(2(2(4(3(3(2(2(4(4(3(3(4(x1)))))))))))))))))) (16)
2(4(4(5(0(1(5(0(3(5(1(1(5(2(0(4(4(x1))))))))))))))))) → 2(5(2(4(3(5(4(2(4(3(4(1(3(4(5(5(4(1(2(4(x1)))))))))))))))))))) (17)
1(2(2(2(5(0(0(3(3(5(2(4(5(1(1(0(4(3(x1)))))))))))))))))) → 1(1(3(0(0(0(3(1(5(1(0(2(3(0(0(x1))))))))))))))) (18)
0(2(1(4(1(2(5(0(0(1(3(1(5(5(5(5(1(3(4(x1))))))))))))))))))) → 3(5(5(1(3(2(2(2(5(2(5(5(5(2(0(4(1(1(4(4(x1)))))))))))))))))))) (19)
1(3(1(0(2(2(4(4(1(3(4(2(0(5(2(1(3(5(4(x1))))))))))))))))))) → 1(4(2(0(2(0(5(5(2(2(0(5(4(3(4(4(4(x1))))))))))))))))) (20)
1(3(2(0(0(1(1(5(5(3(1(3(0(2(3(4(5(5(5(x1))))))))))))))))))) → 3(0(0(5(2(5(2(2(1(4(2(2(0(0(2(5(5(3(0(2(x1)))))))))))))))))))) (21)
2(2(2(5(1(3(5(5(0(4(0(4(5(1(1(2(0(2(5(x1))))))))))))))))))) → 3(2(4(4(2(3(3(1(1(2(0(3(3(1(1(4(5(x1))))))))))))))))) (22)
2(5(3(4(5(1(1(5(5(1(2(1(3(3(3(5(4(5(1(x1))))))))))))))))))) → 5(4(5(4(5(1(1(0(4(2(5(4(1(2(0(1(3(1(x1)))))))))))))))))) (23)
3(2(1(3(3(5(5(5(4(2(4(4(5(1(5(1(3(0(4(x1))))))))))))))))))) → 0(4(4(3(0(1(1(3(0(4(2(2(3(3(0(4(0(1(x1)))))))))))))))))) (24)
2(5(5(2(2(0(1(2(0(3(3(5(1(3(2(2(0(2(3(4(3(x1))))))))))))))))))))) → 4(4(0(4(1(1(4(5(2(5(1(0(1(4(4(1(1(3(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 +
1
[51(x1)] = x1 +
0
[52(x1)] = x1 +
468
[53(x1)] = x1 +
7
[54(x1)] = x1 +
757
[55(x1)] = x1 +
223
[40(x1)] = x1 +
756
[41(x1)] = x1 +
0
[42(x1)] = x1 +
1
[43(x1)] = x1 +
1
[44(x1)] = x1 +
540
[45(x1)] = x1 +
756
[30(x1)] = x1 +
73
[31(x1)] = x1 +
757
[32(x1)] = x1 +
1
[33(x1)] = x1 +
756
[34(x1)] = x1 +
756
[35(x1)] = x1 +
0
[20(x1)] = x1 +
756
[21(x1)] = x1 +
433
[22(x1)] = x1 +
1
[23(x1)] = x1 +
36
[24(x1)] = x1 +
757
[25(x1)] = x1 +
756
[10(x1)] = x1 +
757
[11(x1)] = x1 +
0
[12(x1)] = x1 +
756
[13(x1)] = x1 +
756
[14(x1)] = x1 +
756
[15(x1)] = x1 +
756
[00(x1)] = x1 +
7
[01(x1)] = x1 +
0
[02(x1)] = x1 +
756
[03(x1)] = x1 +
37
[04(x1)] = x1 +
756
[05(x1)] = x1 +
0
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.