YES
The TRS could be proven terminating. The proof took 740 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (4ms).
| – Problem 2 was processed with processor PolynomialOrderingProcessor (443ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(s(x)) | → | p#(s(x)) | | f#(s(x)) | → | f#(f(p(s(x)))) |
| f#(s(x)) | → | f#(p(s(x))) |
Rewrite Rules
| f(s(x)) | → | s(f(f(p(s(x))))) | | f(0) | → | 0 |
| p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, p
Strategy
The following SCCs where found
| f#(s(x)) → f#(f(p(s(x)))) | f#(s(x)) → f#(p(s(x))) |
Problem 2: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| f#(s(x)) | → | f#(f(p(s(x)))) | | f#(s(x)) | → | f#(p(s(x))) |
Rewrite Rules
| f(s(x)) | → | s(f(f(p(s(x))))) | | f(0) | → | 0 |
| p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, p
Strategy
Polynomial Interpretation
- 0: 1
- f(x): x
- f#(x): 2x - 2
- p(x): x - 2
- s(x): x + 2
Improved Usable rules
| f(s(x)) | → | s(f(f(p(s(x))))) | | p(s(x)) | → | x |
| f(0) | → | 0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(s(x)) | → | f#(p(s(x))) | | f#(s(x)) | → | f#(f(p(s(x)))) |