YES
The TRS could be proven terminating. The proof took 1248 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (17ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (524ms).
| | – Problem 3 was processed with processor PolynomialLinearRange4iUR (499ms).
| | | – Problem 4 was processed with processor PolynomialLinearRange4iUR (157ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(f(x)) | → | f#(g(f(x), x)) | | f#(f(x)) | → | f#(x) |
| f#(f(x)) | → | f#(h(f(x), f(x))) | | f#(f(x)) | → | h#(f(x), f(x)) |
| f#(f(x)) | → | g#(f(x), x) | | h#(x, x) | → | g#(x, 0) |
Rewrite Rules
| f(f(x)) | → | f(g(f(x), x)) | | f(f(x)) | → | f(h(f(x), f(x))) |
| g(x, y) | → | y | | h(x, x) | → | g(x, 0) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, h
Strategy
The following SCCs where found
| f#(f(x)) → f#(g(f(x), x)) | f#(f(x)) → f#(x) |
| f#(f(x)) → f#(h(f(x), f(x))) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(f(x)) | → | f#(g(f(x), x)) | | f#(f(x)) | → | f#(x) |
| f#(f(x)) | → | f#(h(f(x), f(x))) |
Rewrite Rules
| f(f(x)) | → | f(g(f(x), x)) | | f(f(x)) | → | f(h(f(x), f(x))) |
| g(x, y) | → | y | | h(x, x) | → | g(x, 0) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, h
Strategy
Polynomial Interpretation
- 0: 0
- f(x): 2x + 1
- f#(x): x
- g(x,y): 2y + 1
- h(x,y): 1
Improved Usable rules
| h(x, x) | → | g(x, 0) | | g(x, y) | → | y |
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#(g(f(x), x)) | | f#(f(x)) | → | f#(h(f(x), f(x))) |
Rewrite Rules
| f(f(x)) | → | f(g(f(x), x)) | | f(f(x)) | → | f(h(f(x), f(x))) |
| g(x, y) | → | y | | h(x, x) | → | g(x, 0) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, h
Strategy
Polynomial Interpretation
- 0: 0
- f(x): 2x + 1
- f#(x): x + 1
- g(x,y): y
- h(x,y): x
Improved Usable rules
| f(f(x)) | → | f(h(f(x), f(x))) | | f(f(x)) | → | f(g(f(x), x)) |
| h(x, x) | → | g(x, 0) | | g(x, y) | → | y |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(f(x)) | → | f#(g(f(x), x)) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(f(x)) | → | f#(h(f(x), f(x))) |
Rewrite Rules
| f(f(x)) | → | f(g(f(x), x)) | | f(f(x)) | → | f(h(f(x), f(x))) |
| g(x, y) | → | y | | h(x, x) | → | g(x, 0) |
Original Signature
Termination of terms over the following signature is verified: f, g, 0, h
Strategy
Polynomial Interpretation
- 0: 0
- f(x): x + 1
- f#(x): 2x + 1
- g(x,y): 2y
- h(x,y): 0
Improved Usable rules
| h(x, x) | → | g(x, 0) | | g(x, y) | → | y |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(f(x)) | → | f#(h(f(x), f(x))) |