YES
The TRS could be proven terminating. The proof took 246 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (3ms).
| – Problem 2 was processed with processor PolynomialOrderingProcessor (93ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| -#(x, s(y)) | → | p#(s(y)) | | -#(x, s(y)) | → | -#(x, p(s(y))) |
Rewrite Rules
| -(0, y) | → | 0 | | -(x, 0) | → | x |
| -(x, s(y)) | → | if(greater(x, s(y)), s(-(x, p(s(y)))), 0) | | p(0) | → | 0 |
| p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: 0, greater, s, if, p, -
Strategy
The following SCCs where found
| -#(x, s(y)) → -#(x, p(s(y))) |
Problem 2: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| -#(x, s(y)) | → | -#(x, p(s(y))) |
Rewrite Rules
| -(0, y) | → | 0 | | -(x, 0) | → | x |
| -(x, s(y)) | → | if(greater(x, s(y)), s(-(x, p(s(y)))), 0) | | p(0) | → | 0 |
| p(s(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: 0, greater, s, if, p, -
Strategy
Polynomial Interpretation
- -(x,y): -2
- -#(x,y): 2y + x - 1
- 0: -1
- greater(x,y): -2
- if(x,y,z): -2
- p(x): x - 1
- s(x): x + 2
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| -#(x, s(y)) | → | -#(x, p(s(y))) |