Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/188261)
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 +
|
[51(x1)] |
= |
x1 +
|
[52(x1)] |
= |
x1 +
|
[53(x1)] |
= |
x1 +
|
[54(x1)] |
= |
x1 +
|
[55(x1)] |
= |
x1 +
|
[40(x1)] |
= |
x1 +
|
[41(x1)] |
= |
x1 +
|
[42(x1)] |
= |
x1 +
|
[43(x1)] |
= |
x1 +
|
[44(x1)] |
= |
x1 +
|
[45(x1)] |
= |
x1 +
|
[30(x1)] |
= |
x1 +
|
[31(x1)] |
= |
x1 +
|
[32(x1)] |
= |
x1 +
|
[33(x1)] |
= |
x1 +
|
[34(x1)] |
= |
x1 +
|
[35(x1)] |
= |
x1 +
|
[20(x1)] |
= |
x1 +
|
[21(x1)] |
= |
x1 +
|
[22(x1)] |
= |
x1 +
|
[23(x1)] |
= |
x1 +
|
[24(x1)] |
= |
x1 +
|
[25(x1)] |
= |
x1 +
|
[10(x1)] |
= |
x1 +
|
[11(x1)] |
= |
x1 +
|
[12(x1)] |
= |
x1 +
|
[13(x1)] |
= |
x1 +
|
[14(x1)] |
= |
x1 +
|
[15(x1)] |
= |
x1 +
|
[00(x1)] |
= |
x1 +
|
[01(x1)] |
= |
x1 +
|
[02(x1)] |
= |
x1 +
|
[03(x1)] |
= |
x1 +
|
[04(x1)] |
= |
x1 +
|
[05(x1)] |
= |
x1 +
|
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.