YES
The TRS could be proven terminating. The proof took 145 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (3ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (125ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| g#(0) | → | f#(0) | | f#(f(x)) | → | f#(x) |
| g#(0) | → | g#(f(0)) |
Rewrite Rules
| f(f(x)) | → | f(x) | | g(0) | → | g(f(0)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0
Strategy
The following SCCs where found
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(f(x)) | → | f(x) | | g(0) | → | g(f(0)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0
Strategy
Polynomial Interpretation
- 0: 2
- f(x): 0
- g(x): 0
- g#(x): x + 1
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
Rewrite Rules
| f(f(x)) | → | f(x) | | g(0) | → | g(f(0)) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: