TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60287 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (46ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (242ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (135ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (1321ms), DependencyGraph (3ms), ReductionPairSAT (306ms), DependencyGraph (2ms), SizeChangePrinciple (30ms), ForwardNarrowing (3ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| | – Problem 5 was processed with processor PolynomialLinearRange4iUR (30ms).
| – Problem 4 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (805ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (806ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (8265ms), DependencyGraph (3ms), ReductionPairSAT (3032ms), DependencyGraph (0ms), SizeChangePrinciple (106ms), ForwardNarrowing (0ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (0ms)].
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| div#(s(x), s(y)) | → | div#(minus(s(x), s(y)), s(y)) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| minus(x, s(y)) | → | p(minus(x, y)) | | div(0, s(y)) | → | 0 |
| div(s(x), s(y)) | → | s(div(minus(s(x), s(y)), s(y))) | | log(s(0), s(s(y))) | → | 0 |
| log(s(s(x)), s(s(y))) | → | s(log(div(minus(x, y), s(s(y))), s(s(y)))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, p, div, log
Open Dependency Pair Problem 4
Dependency Pairs
| log#(s(s(x)), s(s(y))) | → | log#(div(minus(x, y), s(s(y))), s(s(y))) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| minus(x, s(y)) | → | p(minus(x, y)) | | div(0, s(y)) | → | 0 |
| div(s(x), s(y)) | → | s(div(minus(s(x), s(y)), s(y))) | | log(s(0), s(s(y))) | → | 0 |
| log(s(s(x)), s(s(y))) | → | s(log(div(minus(x, y), s(s(y))), s(s(y)))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, p, div, log
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| div#(s(x), s(y)) | → | minus#(s(x), s(y)) | | log#(s(s(x)), s(s(y))) | → | div#(minus(x, y), s(s(y))) |
| minus#(x, s(y)) | → | minus#(x, y) | | log#(s(s(x)), s(s(y))) | → | minus#(x, y) |
| minus#(s(x), s(y)) | → | minus#(x, y) | | div#(s(x), s(y)) | → | div#(minus(s(x), s(y)), s(y)) |
| minus#(x, s(y)) | → | p#(minus(x, y)) | | log#(s(s(x)), s(s(y))) | → | log#(div(minus(x, y), s(s(y))), s(s(y))) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| minus(x, s(y)) | → | p(minus(x, y)) | | div(0, s(y)) | → | 0 |
| div(s(x), s(y)) | → | s(div(minus(s(x), s(y)), s(y))) | | log(s(0), s(s(y))) | → | 0 |
| log(s(s(x)), s(s(y))) | → | s(log(div(minus(x, y), s(s(y))), s(s(y)))) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, p, div, log
Strategy
The following SCCs where found
| div#(s(x), s(y)) → div#(minus(s(x), s(y)), s(y)) |
| minus#(x, s(y)) → minus#(x, y) | minus#(s(x), s(y)) → minus#(x, y) |
| log#(s(s(x)), s(s(y))) → log#(div(minus(x, y), s(s(y))), s(s(y))) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(x, s(y)) | → | minus#(x, y) | | minus#(s(x), s(y)) | → | minus#(x, y) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| minus(x, s(y)) | → | p(minus(x, y)) | | div(0, s(y)) | → | 0 |
| div(s(x), s(y)) | → | s(div(minus(s(x), s(y)), s(y))) | | log(s(0), s(s(y))) | → | 0 |
| log(s(s(x)), s(s(y))) | → | s(log(div(minus(x, y), s(s(y))), s(s(y)))) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, p, div, log
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| minus#(x, s(y)) | → | minus#(x, y) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| minus(x, s(y)) | → | p(minus(x, y)) | | div(0, s(y)) | → | 0 |
| div(s(x), s(y)) | → | s(div(minus(s(x), s(y)), s(y))) | | log(s(0), s(s(y))) | → | 0 |
| log(s(s(x)), s(s(y))) | → | s(log(div(minus(x, y), s(s(y))), s(s(y)))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, p, div, log
Strategy
Polynomial Interpretation
- 0: 0
- div(x,y): 0
- log(x,y): 0
- minus(x,y): 0
- minus#(x,y): 2y + x
- p(x): 0
- s(x): 2x + 1
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| minus#(x, s(y)) | → | minus#(x, y) |