YES
The TRS could be proven terminating. The proof took 668 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (9ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (174ms).
| | – Problem 4 was processed with processor PolynomialLinearRange4iUR (98ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (93ms).
| | – Problem 5 was processed with processor PolynomialLinearRange4iUR (137ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| g#(0) | → | f#(0) | | f#(f(x)) | → | f#(x) |
| f#(1) | → | g#(1) | | g#(0) | → | g#(f(0)) |
| g#(g(x)) | → | g#(x) | | f#(1) | → | f#(g(1)) |
Rewrite Rules
| f(1) | → | f(g(1)) | | f(f(x)) | → | f(x) |
| g(0) | → | g(f(0)) | | g(g(x)) | → | g(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0
Strategy
The following SCCs where found
| g#(0) → g#(f(0)) | g#(g(x)) → g#(x) |
| f#(f(x)) → f#(x) | f#(1) → f#(g(1)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| g#(0) | → | g#(f(0)) | | g#(g(x)) | → | g#(x) |
Rewrite Rules
| f(1) | → | f(g(1)) | | f(f(x)) | → | f(x) |
| g(0) | → | g(f(0)) | | g(g(x)) | → | g(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0
Strategy
Polynomial Interpretation
- 0: 0
- 1: 1
- f(x): 0
- g(x): x + 1
- g#(x): x + 1
Improved Usable rules
| f(1) | → | f(g(1)) | | f(f(x)) | → | f(x) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(1) | → | f(g(1)) | | f(f(x)) | → | f(x) |
| g(0) | → | g(f(0)) | | g(g(x)) | → | g(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0
Strategy
Polynomial Interpretation
- 0: 1
- 1: 3
- f(x): 0
- g(x): 0
- g#(x): x + 1
Improved Usable rules
| f(1) | → | f(g(1)) | | f(f(x)) | → | f(x) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(f(x)) | → | f#(x) | | f#(1) | → | f#(g(1)) |
Rewrite Rules
| f(1) | → | f(g(1)) | | f(f(x)) | → | f(x) |
| g(0) | → | g(f(0)) | | g(g(x)) | → | g(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0
Strategy
Polynomial Interpretation
- 0: 0
- 1: 0
- f(x): 2x + 1
- f#(x): 2x + 1
- g(x): 0
Improved Usable rules
| g(g(x)) | → | g(x) | | g(0) | → | g(f(0)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(1) | → | f(g(1)) | | f(f(x)) | → | f(x) |
| g(0) | → | g(f(0)) | | g(g(x)) | → | g(x) |
Original Signature
Termination of terms over the following signature is verified: f, g, 1, 0
Strategy
Polynomial Interpretation
- 0: 3
- 1: 2
- f(x): 0
- f#(x): x + 1
- g(x): 0
Improved Usable rules
| g(g(x)) | → | g(x) | | g(0) | → | g(f(0)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed: