YES
The TRS could be proven terminating. The proof took 171 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (4ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (99ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(s(0)) | → | f#(p(s(0))) | | activate#(n__f(X)) | → | f#(X) |
| f#(s(0)) | → | p#(s(0)) |
Rewrite Rules
| f(0) | → | cons(0, n__f(s(0))) | | f(s(0)) | → | f(p(s(0))) |
| p(s(0)) | → | 0 | | f(X) | → | n__f(X) |
| activate(n__f(X)) | → | f(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, 0, s, p, n__f, cons
Strategy
The following SCCs where found
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(0) | → | cons(0, n__f(s(0))) | | f(s(0)) | → | f(p(s(0))) |
| p(s(0)) | → | 0 | | f(X) | → | n__f(X) |
| activate(n__f(X)) | → | f(X) | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: f, activate, 0, s, p, n__f, cons
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- cons(x,y): 0
- f(x): 0
- f#(x): 2x
- n__f(x): 0
- p(x): 0
- s(x): 1
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed: