Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/25192)
The rewrite relation of the following TRS is considered.
|
0(x1) |
→ |
1(x1) |
(1) |
|
4(5(4(5(x1)))) |
→ |
4(4(5(5(x1)))) |
(2) |
|
5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) |
→ |
2(x1) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [0(x1)] |
= |
1 · x1 + 1 |
| [1(x1)] |
= |
1 · x1
|
| [4(x1)] |
= |
1 · x1 + 1 |
| [5(x1)] |
= |
1 · x1
|
| [2(x1)] |
= |
1 · x1 + 5 |
all of the following rules can be deleted.
|
0(x1) |
→ |
1(x1) |
(1) |
|
5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) |
→ |
2(x1) |
(3) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
4#(5(4(5(x1)))) |
→ |
4#(4(5(5(x1)))) |
(4) |
|
4#(5(4(5(x1)))) |
→ |
4#(5(5(x1))) |
(5) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.