YES
The TRS could be proven terminating. The proof took 192 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (5ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (166ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(f(x)) | → | f#(x) | | g#(s(0)) | → | f#(s(0)) |
| f#(s(x)) | → | f#(x) | | g#(s(0)) | → | g#(f(s(0))) |
Rewrite Rules
| f(f(x)) | → | f(x) | | f(s(x)) | → | f(x) |
| g(s(0)) | → | g(f(s(0))) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s
Strategy
The following SCCs where found
| f#(f(x)) → f#(x) | f#(s(x)) → f#(x) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| f#(f(x)) | → | f#(x) | | f#(s(x)) | → | f#(x) |
Rewrite Rules
| f(f(x)) | → | f(x) | | f(s(x)) | → | f(x) |
| g(s(0)) | → | g(f(s(0))) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(f(x)) | → | f#(x) | | f#(s(x)) | → | f#(x) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(f(x)) | → | f(x) | | f(s(x)) | → | f(x) |
| g(s(0)) | → | g(f(s(0))) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, s
Strategy
Polynomial Interpretation
- 0: 1
- f(x): 0
- g(x): 0
- g#(x): x + 1
- s(x): x + 1
Improved Usable rules
| f(s(x)) | → | f(x) | | f(f(x)) | → | f(x) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed: