YES
The TRS could be proven terminating. The proof took 275 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (35ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (124ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| p#(s(0)) | → | 0# | | f#(s(0)) | → | f#(p(s(0))) |
| activate#(n__f(X)) | → | f#(activate(X)) | | activate#(n__s(X)) | → | activate#(X) |
| f#(0) | → | 0# | | f#(s(0)) | → | s#(0) |
| f#(s(0)) | → | p#(s(0)) | | f#(s(0)) | → | 0# |
| activate#(n__f(X)) | → | activate#(X) | | activate#(n__0) | → | 0# |
| activate#(n__s(X)) | → | s#(activate(X)) |
Rewrite Rules
| f(0) | → | cons(0, n__f(n__s(n__0))) | | f(s(0)) | → | f(p(s(0))) |
| p(s(0)) | → | 0 | | f(X) | → | n__f(X) |
| s(X) | → | n__s(X) | | 0 | → | n__0 |
| activate(n__f(X)) | → | f(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__0) | → | 0 | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__s, f, activate, n__0, 0, s, p, n__f, cons
Strategy
The following SCCs where found
| activate#(n__s(X)) → activate#(X) | activate#(n__f(X)) → activate#(X) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(0) | → | cons(0, n__f(n__s(n__0))) | | f(s(0)) | → | f(p(s(0))) |
| p(s(0)) | → | 0 | | f(X) | → | n__f(X) |
| s(X) | → | n__s(X) | | 0 | → | n__0 |
| activate(n__f(X)) | → | f(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__0) | → | 0 | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__s, f, activate, n__0, 0, s, p, n__f, cons
Strategy
Polynomial Interpretation
- 0: 0
- activate(x): 0
- cons(x,y): 0
- f(x): 0
- f#(x): x
- n__0: 0
- n__f(x): 0
- n__s(x): 3
- p(x): 1
- s(x): 3x + 2
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| activate#(n__s(X)) | → | activate#(X) | | activate#(n__f(X)) | → | activate#(X) |
Rewrite Rules
| f(0) | → | cons(0, n__f(n__s(n__0))) | | f(s(0)) | → | f(p(s(0))) |
| p(s(0)) | → | 0 | | f(X) | → | n__f(X) |
| s(X) | → | n__s(X) | | 0 | → | n__0 |
| activate(n__f(X)) | → | f(activate(X)) | | activate(n__s(X)) | → | s(activate(X)) |
| activate(n__0) | → | 0 | | activate(X) | → | X |
Original Signature
Termination of terms over the following signature is verified: n__s, f, activate, n__0, 0, s, p, n__f, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| activate#(n__s(X)) | → | activate#(X) | | activate#(n__f(X)) | → | activate#(X) |