Certification Problem
Input (TPDB TRS_Standard/AG01/#3.26)
The rewrite relation of the following TRS is considered.
|
f(x) |
→ |
s(x) |
(1) |
|
f(s(s(x))) |
→ |
s(f(f(x))) |
(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.
|
f#(s(s(x))) |
→ |
f#(f(x)) |
(3) |
|
f#(s(s(x))) |
→ |
f#(x) |
(4) |
1.1 Switch to Innermost Termination
The TRS does not have overlaps with the pairs and is locally confluent:
20
Hence, it suffices to show innermost termination in the following.
1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [f#(x1)] |
= |
2 + x1
|
| [f(x1)] |
= |
2 + x1
|
| [s(x1)] |
= |
2 + x1
|
the
pairs
|
f#(s(s(x))) |
→ |
f#(f(x)) |
(3) |
|
f#(s(s(x))) |
→ |
f#(x) |
(4) |
could be deleted.
1.1.1.1 P is empty
There are no pairs anymore.