YES
The TRS could be proven terminating. The proof took 1936 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (54ms).
| – Problem 2 was processed with processor PolynomialOrderingProcessor (450ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (594ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(x, f(s(s(y)), nil)) | → | f#(y, f(s(0), nil)) | | f#(x, f(s(s(y)), f(z, w))) | → | f#(s(x), f(y, f(s(z), w))) |
| L#(f(s(s(y)), f(z, w))) | → | f#(y, f(s(z), w)) | | f#(x, f(s(s(y)), nil)) | → | f#(s(x), f(y, f(s(0), nil))) |
| L#(f(s(s(y)), f(z, w))) | → | f#(s(0), f(y, f(s(z), w))) | | f#(x, f(s(s(y)), nil)) | → | f#(s(0), nil) |
| L#(f(s(s(y)), f(z, w))) | → | L#(f(s(0), f(y, f(s(z), w)))) | | L#(f(s(s(y)), f(z, w))) | → | f#(s(z), w) |
| f#(x, f(s(s(y)), f(z, w))) | → | f#(y, f(s(z), w)) | | f#(x, f(s(s(y)), f(z, w))) | → | f#(s(z), w) |
Rewrite Rules
| f(x, f(s(s(y)), f(z, w))) | → | f(s(x), f(y, f(s(z), w))) | | L(f(s(s(y)), f(z, w))) | → | L(f(s(0), f(y, f(s(z), w)))) |
| f(x, f(s(s(y)), nil)) | → | f(s(x), f(y, f(s(0), nil))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, L, nil
Strategy
The following SCCs where found
| f#(x, f(s(s(y)), nil)) → f#(y, f(s(0), nil)) | f#(x, f(s(s(y)), f(z, w))) → f#(s(x), f(y, f(s(z), w))) |
| f#(x, f(s(s(y)), nil)) → f#(s(x), f(y, f(s(0), nil))) | f#(x, f(s(s(y)), f(z, w))) → f#(y, f(s(z), w)) |
| f#(x, f(s(s(y)), f(z, w))) → f#(s(z), w) |
| L#(f(s(s(y)), f(z, w))) → L#(f(s(0), f(y, f(s(z), w)))) |
Problem 2: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| L#(f(s(s(y)), f(z, w))) | → | L#(f(s(0), f(y, f(s(z), w)))) |
Rewrite Rules
| f(x, f(s(s(y)), f(z, w))) | → | f(s(x), f(y, f(s(z), w))) | | L(f(s(s(y)), f(z, w))) | → | L(f(s(0), f(y, f(s(z), w)))) |
| f(x, f(s(s(y)), nil)) | → | f(s(x), f(y, f(s(0), nil))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, L, nil
Strategy
Polynomial Interpretation
- 0: -1
- L(x): -2
- L#(x): 2x - 2
- f(x,y): y + 2x - 1
- nil: 2
- s(x): x + 1
Improved Usable rules
| f(x, f(s(s(y)), f(z, w))) | → | f(s(x), f(y, f(s(z), w))) | | f(x, f(s(s(y)), nil)) | → | f(s(x), f(y, f(s(0), nil))) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| L#(f(s(s(y)), f(z, w))) | → | L#(f(s(0), f(y, f(s(z), w)))) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| f#(x, f(s(s(y)), nil)) | → | f#(y, f(s(0), nil)) | | f#(x, f(s(s(y)), f(z, w))) | → | f#(s(x), f(y, f(s(z), w))) |
| f#(x, f(s(s(y)), nil)) | → | f#(s(x), f(y, f(s(0), nil))) | | f#(x, f(s(s(y)), f(z, w))) | → | f#(y, f(s(z), w)) |
| f#(x, f(s(s(y)), f(z, w))) | → | f#(s(z), w) |
Rewrite Rules
| f(x, f(s(s(y)), f(z, w))) | → | f(s(x), f(y, f(s(z), w))) | | L(f(s(s(y)), f(z, w))) | → | L(f(s(0), f(y, f(s(z), w)))) |
| f(x, f(s(s(y)), nil)) | → | f(s(x), f(y, f(s(0), nil))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, s, L, nil
Strategy
Polynomial Interpretation
- 0: 0
- L(x): 0
- f(x,y): y + x
- f#(x,y): y
- nil: 0
- s(x): x + 1
Improved Usable rules
| f(x, f(s(s(y)), f(z, w))) | → | f(s(x), f(y, f(s(z), w))) | | f(x, f(s(s(y)), nil)) | → | f(s(x), f(y, f(s(0), nil))) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| f#(x, f(s(s(y)), nil)) | → | f#(y, f(s(0), nil)) | | f#(x, f(s(s(y)), f(z, w))) | → | f#(s(x), f(y, f(s(z), w))) |
| f#(x, f(s(s(y)), nil)) | → | f#(s(x), f(y, f(s(0), nil))) | | f#(x, f(s(s(y)), f(z, w))) | → | f#(y, f(s(z), w)) |
| f#(x, f(s(s(y)), f(z, w))) | → | f#(s(z), w) |