YES
The TRS could be proven terminating. The proof took 650 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (24ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor PolynomialLinearRange4iUR (440ms).
| – Problem 5 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| gcd#(s(x), s(y)) | → | -#(max(x, y), min(x, y)) | | min#(s(x), s(y)) | → | min#(x, y) |
| gcd#(s(x), s(y)) | → | gcd#(-(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) |
| -#(s(x), s(y)) | → | -#(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)) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| gcd(s(x), 0) | → | s(x) | | gcd(0, s(x)) | → | s(x) |
| gcd(s(x), s(y)) | → | gcd(-(max(x, y), min(x, y)), s(min(x, y))) |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, s, -, gcd
Strategy
The following SCCs where found
| min#(s(x), s(y)) → min#(x, y) |
| gcd#(s(x), s(y)) → gcd#(-(max(x, y), min(x, y)), s(min(x, y))) |
| max#(s(x), s(y)) → max#(x, y) |
| -#(s(x), s(y)) → -#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| -#(s(x), s(y)) | → | -#(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)) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| gcd(s(x), 0) | → | s(x) | | gcd(0, s(x)) | → | s(x) |
| gcd(s(x), s(y)) | → | gcd(-(max(x, y), min(x, y)), s(min(x, y))) |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, s, -, gcd
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| -#(s(x), s(y)) | → | -#(x, y) |
Problem 3: 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)) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| gcd(s(x), 0) | → | s(x) | | gcd(0, s(x)) | → | s(x) |
| gcd(s(x), s(y)) | → | gcd(-(max(x, y), min(x, y)), s(min(x, y))) |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, s, -, gcd
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| max#(s(x), s(y)) | → | max#(x, y) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| gcd#(s(x), s(y)) | → | gcd#(-(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)) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| gcd(s(x), 0) | → | s(x) | | gcd(0, s(x)) | → | s(x) |
| gcd(s(x), s(y)) | → | gcd(-(max(x, y), min(x, y)), s(min(x, y))) |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, s, -, gcd
Strategy
Polynomial Interpretation
- -(x,y): x
- 0: 0
- gcd(x,y): 0
- gcd#(x,y): y + 2x
- max(x,y): y + x
- min(x,y): x
- s(x): 3x + 1
Improved Usable rules
| -(s(x), s(y)) | → | -(x, y) | | min(0, y) | → | 0 |
| -(x, 0) | → | x | | max(s(x), s(y)) | → | s(max(x, y)) |
| min(s(x), s(y)) | → | s(min(x, y)) | | max(0, y) | → | y |
| 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#(-(max(x, y), min(x, y)), s(min(x, y))) |
Problem 5: 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)) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| gcd(s(x), 0) | → | s(x) | | gcd(0, s(x)) | → | s(x) |
| gcd(s(x), s(y)) | → | gcd(-(max(x, y), min(x, y)), s(min(x, y))) |
Original Signature
Termination of terms over the following signature is verified: min, 0, max, s, -, gcd
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| min#(s(x), s(y)) | → | min#(x, y) |