YES
The TRS could be proven terminating. The proof took 1184 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (7ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (330ms).
| | – Problem 3 was processed with processor PolynomialLinearRange4iUR (313ms).
| | | – Problem 4 was processed with processor PolynomialOrderingProcessor (260ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(x, f(a, y)) | → | f#(a, x) | | f#(x, f(a, y)) | → | f#(a, h(f(a, x))) |
| f#(x, f(a, y)) | → | f#(a, f(f(a, h(f(a, x))), y)) | | f#(x, f(a, y)) | → | f#(f(a, h(f(a, x))), y) |
Rewrite Rules
| f(x, f(a, y)) | → | f(a, f(f(a, h(f(a, x))), y)) |
Original Signature
Termination of terms over the following signature is verified: f, a, h
Strategy
The following SCCs where found
| f#(x, f(a, y)) → f#(a, x) | f#(x, f(a, y)) → f#(f(a, h(f(a, x))), y) |
| f#(x, f(a, y)) → f#(a, f(f(a, h(f(a, x))), y)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(x, f(a, y)) | → | f#(a, x) | | f#(x, f(a, y)) | → | f#(f(a, h(f(a, x))), y) |
| f#(x, f(a, y)) | → | f#(a, f(f(a, h(f(a, x))), y)) |
Rewrite Rules
| f(x, f(a, y)) | → | f(a, f(f(a, h(f(a, x))), y)) |
Original Signature
Termination of terms over the following signature is verified: f, a, h
Strategy
Polynomial Interpretation
- a: 0
- f(x,y): y + 1
- f#(x,y): 2y + 2x
- h(x): 0
Improved Usable rules
| f(x, f(a, y)) | → | f(a, f(f(a, h(f(a, x))), y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(x, f(a, y)) | → | f#(a, x) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(x, f(a, y)) | → | f#(a, f(f(a, h(f(a, x))), y)) | | f#(x, f(a, y)) | → | f#(f(a, h(f(a, x))), y) |
Rewrite Rules
| f(x, f(a, y)) | → | f(a, f(f(a, h(f(a, x))), y)) |
Original Signature
Termination of terms over the following signature is verified: f, a, h
Strategy
Polynomial Interpretation
- a: 0
- f(x,y): 2y + 1
- f#(x,y): y
- h(x): 0
Improved Usable rules
| f(x, f(a, y)) | → | f(a, f(f(a, h(f(a, x))), y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(x, f(a, y)) | → | f#(f(a, h(f(a, x))), y) |
Problem 4: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| f#(x, f(a, y)) | → | f#(a, f(f(a, h(f(a, x))), y)) |
Rewrite Rules
| f(x, f(a, y)) | → | f(a, f(f(a, h(f(a, x))), y)) |
Original Signature
Termination of terms over the following signature is verified: f, a, h
Strategy
Polynomial Interpretation
- a: 2
- f(x,y): 4y + x - 1
- f#(x,y): 3y - 2
- h(x): -1
Improved Usable rules
| f(x, f(a, y)) | → | f(a, f(f(a, h(f(a, x))), y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(x, f(a, y)) | → | f#(a, f(f(a, h(f(a, x))), y)) |