YES
The TRS could be proven terminating. The proof took 858 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (36ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
| – Problem 5 was processed with processor PolynomialLinearRange4iUR (660ms).
| – Problem 6 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | any#(y) | | min#(s(x), s(y)) | → | min#(x, y) |
| minus#(s(x), s(y)) | → | minus#(x, any(y)) | | any#(s(x)) | → | any#(x) |
| gcd#(s(x), s(y)) | → | gcd#(minus(max(x, y), min(x, y)), s(min(x, y))) | | gcd#(s(x), s(y)) | → | max#(x, y) |
| gcd#(s(x), s(y)) | → | min#(x, y) | | max#(s(x), s(y)) | → | max#(x, y) |
| gcd#(s(x), s(y)) | → | minus#(max(x, y), min(x, y)) |
Rewrite Rules
| min(x, 0) | → | 0 | | min(0, y) | → | 0 |
| min(s(x), s(y)) | → | s(min(x, y)) | | max(x, 0) | → | x |
| max(0, y) | → | y | | max(s(x), s(y)) | → | s(max(x, y)) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | s(minus(x, any(y))) |
| gcd(s(x), s(y)) | → | gcd(minus(max(x, y), min(x, y)), s(min(x, y))) | | any(s(x)) | → | s(s(any(x))) |
| any(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, minus, s, any, gcd
Strategy
The following SCCs where found
| min#(s(x), s(y)) → min#(x, y) |
| minus#(s(x), s(y)) → minus#(x, any(y)) |
| gcd#(s(x), s(y)) → gcd#(minus(max(x, y), min(x, y)), s(min(x, y))) |
| max#(s(x), s(y)) → max#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| max#(s(x), s(y)) | → | max#(x, y) |
Rewrite Rules
| min(x, 0) | → | 0 | | min(0, y) | → | 0 |
| min(s(x), s(y)) | → | s(min(x, y)) | | max(x, 0) | → | x |
| max(0, y) | → | y | | max(s(x), s(y)) | → | s(max(x, y)) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | s(minus(x, any(y))) |
| gcd(s(x), s(y)) | → | gcd(minus(max(x, y), min(x, y)), s(min(x, y))) | | any(s(x)) | → | s(s(any(x))) |
| any(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, minus, s, any, gcd
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| max#(s(x), s(y)) | → | max#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| min#(s(x), s(y)) | → | min#(x, y) |
Rewrite Rules
| min(x, 0) | → | 0 | | min(0, y) | → | 0 |
| min(s(x), s(y)) | → | s(min(x, y)) | | max(x, 0) | → | x |
| max(0, y) | → | y | | max(s(x), s(y)) | → | s(max(x, y)) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | s(minus(x, any(y))) |
| gcd(s(x), s(y)) | → | gcd(minus(max(x, y), min(x, y)), s(min(x, y))) | | any(s(x)) | → | s(s(any(x))) |
| any(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, minus, s, any, gcd
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| min#(s(x), s(y)) | → | min#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, any(y)) |
Rewrite Rules
| min(x, 0) | → | 0 | | min(0, y) | → | 0 |
| min(s(x), s(y)) | → | s(min(x, y)) | | max(x, 0) | → | x |
| max(0, y) | → | y | | max(s(x), s(y)) | → | s(max(x, y)) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | s(minus(x, any(y))) |
| gcd(s(x), s(y)) | → | gcd(minus(max(x, y), min(x, y)), s(min(x, y))) | | any(s(x)) | → | s(s(any(x))) |
| any(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, minus, s, any, gcd
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, any(y)) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| gcd#(s(x), s(y)) | → | gcd#(minus(max(x, y), min(x, y)), s(min(x, y))) |
Rewrite Rules
| min(x, 0) | → | 0 | | min(0, y) | → | 0 |
| min(s(x), s(y)) | → | s(min(x, y)) | | max(x, 0) | → | x |
| max(0, y) | → | y | | max(s(x), s(y)) | → | s(max(x, y)) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | s(minus(x, any(y))) |
| gcd(s(x), s(y)) | → | gcd(minus(max(x, y), min(x, y)), s(min(x, y))) | | any(s(x)) | → | s(s(any(x))) |
| any(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, minus, s, any, gcd
Strategy
Polynomial Interpretation
- 0: 0
- any(x): 2x
- gcd(x,y): 0
- gcd#(x,y): y + 2x
- max(x,y): y + x
- min(x,y): x
- minus(x,y): x
- s(x): 2x + 1
Improved Usable rules
| minus(s(x), s(y)) | → | s(minus(x, any(y))) | | min(0, y) | → | 0 |
| max(s(x), s(y)) | → | s(max(x, y)) | | min(s(x), s(y)) | → | s(min(x, y)) |
| max(0, y) | → | y | | minus(x, 0) | → | x |
| min(x, 0) | → | 0 | | max(x, 0) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| gcd#(s(x), s(y)) | → | gcd#(minus(max(x, y), min(x, y)), s(min(x, y))) |
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| min(x, 0) | → | 0 | | min(0, y) | → | 0 |
| min(s(x), s(y)) | → | s(min(x, y)) | | max(x, 0) | → | x |
| max(0, y) | → | y | | max(s(x), s(y)) | → | s(max(x, y)) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | s(minus(x, any(y))) |
| gcd(s(x), s(y)) | → | gcd(minus(max(x, y), min(x, y)), s(min(x, y))) | | any(s(x)) | → | s(s(any(x))) |
| any(x) | → | x |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, minus, s, any, gcd
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: