Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/188004)

The rewrite relation of the following TRS is considered.

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

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 1080 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 6480 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 +
0
[51(x1)] = x1 +
0
[52(x1)] = x1 +
364/3
[53(x1)] = x1 +
364/3
[54(x1)] = x1 +
367/3
[55(x1)] = x1 +
238/3
[40(x1)] = x1 +
1
[41(x1)] = x1 +
0
[42(x1)] = x1 +
0
[43(x1)] = x1 +
0
[44(x1)] = x1 +
367/3
[45(x1)] = x1 +
0
[30(x1)] = x1 +
308/3
[31(x1)] = x1 +
1
[32(x1)] = x1 +
0
[33(x1)] = x1 +
364/3
[34(x1)] = x1 +
123
[35(x1)] = x1 +
31/3
[20(x1)] = x1 +
364/3
[21(x1)] = x1 +
364/3
[22(x1)] = x1 +
364/3
[23(x1)] = x1 +
140/3
[24(x1)] = x1 +
364/3
[25(x1)] = x1 +
5/3
[10(x1)] = x1 +
367/3
[11(x1)] = x1 +
123
[12(x1)] = x1 +
123
[13(x1)] = x1 +
367/3
[14(x1)] = x1 +
123
[15(x1)] = x1 +
364/3
[00(x1)] = x1 +
0
[01(x1)] = x1 +
350/3
[02(x1)] = x1 +
0
[03(x1)] = x1 +
53
[04(x1)] = x1 +
367/3
[05(x1)] = x1 +
0
all of the following rules can be deleted.

There are 6480 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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