Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/187254)
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 Rule Removal
Using the
matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [5(x1)] |
= |
x1 +
|
| [4(x1)] |
= |
x1 +
|
| [3(x1)] |
= |
x1 +
|
| [2(x1)] |
= |
x1 +
|
| [1(x1)] |
= |
x1 +
|
| [0(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
There are 179 ruless (increase limit for explicit display).
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
0#(2(4(3(4(1(3(5(5(5(5(4(1(4(0(4(x1)))))))))))))))) |
→ |
0#(4(4(1(5(5(5(4(1(3(2(0(4(3(5(4(x1)))))))))))))))) |
(181) |
|
0#(2(4(3(4(1(3(5(5(5(5(4(1(4(0(4(x1)))))))))))))))) |
→ |
0#(4(3(5(4(x1))))) |
(182) |
1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [5(x1)] |
= |
x1 +
|
| [4(x1)] |
= |
x1 +
|
| [3(x1)] |
= |
x1 +
|
| [2(x1)] |
= |
x1 +
|
| [1(x1)] |
= |
x1 +
|
| [0(x1)] |
= |
x1 +
|
| [0#(x1)] |
= |
x1 +
|
together with the usable
rule
|
0(2(4(3(4(1(3(5(5(5(5(4(1(4(0(4(x1)))))))))))))))) |
→ |
0(4(4(1(5(5(5(4(1(3(2(0(4(3(5(4(x1)))))))))))))))) |
(82) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
|
0#(2(4(3(4(1(3(5(5(5(5(4(1(4(0(4(x1)))))))))))))))) |
→ |
0#(4(3(5(4(x1))))) |
(182) |
and
no rules
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.