Certification Problem
Input (TPDB TRS_Standard/Endrullis_06/direct)
The rewrite relation of the following TRS is considered.
|
h(x,c(y,z)) |
→ |
h(c(s(y),x),z) |
(1) |
|
h(c(s(x),c(s(0),y)),z) |
→ |
h(y,c(s(0),c(x,z))) |
(2) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
h#(x,c(y,z)) |
→ |
h#(c(s(y),x),z) |
(3) |
|
h#(c(s(x),c(s(0),y)),z) |
→ |
h#(y,c(s(0),c(x,z))) |
(4) |
1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [c(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [s(x1)] |
= |
1 · x1
|
| [0] |
= |
0 |
| [h#(x1, x2)] |
= |
1 · x1 + 1 · x2
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the rationals with delta = 1/4
| [h#(x1, x2)] |
= |
0 + 1 · x1 + 1/2 · x2
|
| [c(x1, x2)] |
= |
0 + 1/4 · x1 + 1 · x2
|
| [s(x1)] |
= |
0 + 1/2 · x1
|
| [0] |
= |
4 |
the
pair
|
h#(c(s(x),c(s(0),y)),z) |
→ |
h#(y,c(s(0),c(x,z))) |
(4) |
could be deleted.
1.1.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
h#(x,c(y,z)) |
→ |
h#(c(s(y),x),z) |
(3) |
|
| 2 |
> |
2 |
As there is no critical graph in the transitive closure, there are no infinite chains.