YES
The TRS could be proven terminating. The proof took 127 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (10ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (79ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(f(X)) | → | f#(g(f(g(f(X))))) | | f#(f(X)) | → | f#(g(f(X))) |
| f#(g(f(X))) | → | f#(g(X)) | | f#(f(X)) | → | f#(X) |
Rewrite Rules
| f(f(X)) | → | f(g(f(g(f(X))))) | | f(g(f(X))) | → | f(g(X)) |
Original Signature
Termination of terms over the following signature is verified: f, g
Strategy
The following SCCs where found
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(f(X)) | → | f(g(f(g(f(X))))) | | f(g(f(X))) | → | f(g(X)) |
Original Signature
Termination of terms over the following signature is verified: f, g
Strategy
Polynomial Interpretation
- f(x): 2x + 1
- f#(x): 2x
- g(x): x
There are no 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(g(f(g(f(X))))) | | f(g(f(X))) | → | f(g(X)) |
Original Signature
Termination of terms over the following signature is verified: f, g
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: