TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60015 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (66ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (1197ms).
| | – Problem 5 was processed with processor PolynomialLinearRange4iUR (1071ms).
| | | – Problem 6 was processed with processor PolynomialLinearRange4iUR (1337ms).
| | | | – Problem 7 remains open; application of the following processors failed [DependencyGraph (2ms), PolynomialLinearRange4iUR (1148ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (17592ms), DependencyGraph (24ms), ReductionPairSAT (timeout)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor PolynomialLinearRange4iUR (81ms).
The following open problems remain:
Open Dependency Pair Problem 7
Dependency Pairs
| plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(plus(y, s(s(z))), plus(x, s(0))) | | plus#(minus(x, s(0)), minus(y, s(s(z)))) | → | plus#(minus(y, s(s(z))), minus(x, s(0))) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) | | plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| quot#(s(x), s(y)) | → | minus#(x, y) | | plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(plus(y, s(s(z))), plus(x, s(0))) |
| plus#(s(x), y) | → | plus#(x, y) | | quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
| plus#(minus(x, s(0)), minus(y, s(s(z)))) | → | minus#(x, s(0)) | | minus#(s(x), s(y)) | → | minus#(x, y) |
| plus#(minus(x, s(0)), minus(y, s(s(z)))) | → | minus#(y, s(s(z))) | | plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(y, s(s(z))) |
| plus#(minus(x, s(0)), minus(y, s(s(z)))) | → | plus#(minus(y, s(s(z))), minus(x, s(0))) | | plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(x, s(0)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) | | plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Strategy
The following SCCs where found
| quot#(s(x), s(y)) → quot#(minus(x, y), s(y)) |
| minus#(s(x), s(y)) → minus#(x, y) |
| plus#(plus(x, s(0)), plus(y, s(s(z)))) → plus#(plus(y, s(s(z))), plus(x, s(0))) | plus#(s(x), y) → plus#(x, y) |
| plus#(plus(x, s(0)), plus(y, s(s(z)))) → plus#(y, s(s(z))) | plus#(minus(x, s(0)), minus(y, s(s(z)))) → plus#(minus(y, s(s(z))), minus(x, s(0))) |
| plus#(plus(x, s(0)), plus(y, s(s(z)))) → plus#(x, s(0)) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(plus(y, s(s(z))), plus(x, s(0))) | | plus#(s(x), y) | → | plus#(x, y) |
| plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(y, s(s(z))) | | plus#(minus(x, s(0)), minus(y, s(s(z)))) | → | plus#(minus(y, s(s(z))), minus(x, s(0))) |
| plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(x, s(0)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) | | plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Strategy
Polynomial Interpretation
- 0: 1
- minus(x,y): y + x
- plus(x,y): y + x
- plus#(x,y): 2y + 2x
- quot(x,y): 0
- s(x): x
Improved Usable rules
| minus(s(x), s(y)) | → | minus(x, y) | | plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) |
| plus(s(x), y) | → | s(plus(x, y)) | | plus(0, y) | → | y |
| plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) | | minus(x, 0) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(y, s(s(z))) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(plus(y, s(s(z))), plus(x, s(0))) | | plus#(s(x), y) | → | plus#(x, y) |
| plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(x, s(0)) | | plus#(minus(x, s(0)), minus(y, s(s(z)))) | → | plus#(minus(y, s(s(z))), minus(x, s(0))) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) | | plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, minus, s, quot
Strategy
Polynomial Interpretation
- 0: 0
- minus(x,y): 3y + 2x + 1
- plus(x,y): y + x + 1
- plus#(x,y): y + x
- quot(x,y): 0
- s(x): x
Improved Usable rules
| minus(s(x), s(y)) | → | minus(x, y) | | plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) |
| plus(s(x), y) | → | s(plus(x, y)) | | plus(0, y) | → | y |
| plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) | | minus(x, 0) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(x, s(0)) |
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| plus#(plus(x, s(0)), plus(y, s(s(z)))) | → | plus#(plus(y, s(s(z))), plus(x, s(0))) | | plus#(s(x), y) | → | plus#(x, y) |
| plus#(minus(x, s(0)), minus(y, s(s(z)))) | → | plus#(minus(y, s(s(z))), minus(x, s(0))) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) | | plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Strategy
Polynomial Interpretation
- 0: 0
- minus(x,y): x
- plus(x,y): y + x
- plus#(x,y): y + x
- quot(x,y): 0
- s(x): x + 1
Improved Usable rules
| minus(s(x), s(y)) | → | minus(x, y) | | plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) |
| plus(s(x), y) | → | s(plus(x, y)) | | plus(0, y) | → | y |
| plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) | | minus(x, 0) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| plus#(s(x), y) | → | plus#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, y) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) | | plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| quot(0, s(y)) | → | 0 | | quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) |
| plus(0, y) | → | y | | plus(s(x), y) | → | s(plus(x, y)) |
| plus(minus(x, s(0)), minus(y, s(s(z)))) | → | plus(minus(y, s(s(z))), minus(x, s(0))) | | plus(plus(x, s(0)), plus(y, s(s(z)))) | → | plus(plus(y, s(s(z))), plus(x, s(0))) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, quot
Strategy
Polynomial Interpretation
- 0: 3
- minus(x,y): x
- plus(x,y): 0
- quot(x,y): 0
- quot#(x,y): x + 1
- s(x): 2x + 2
Improved Usable rules
| minus(s(x), s(y)) | → | minus(x, y) | | minus(x, 0) | → | x |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |