TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60000 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (22ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (146ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (1210ms), DependencyGraph (1ms), ReductionPairSAT (341ms), DependencyGraph (1ms), SizeChangePrinciple (18ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (0ms)].
| – Problem 3 was processed with processor SubtermCriterion (2ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| quot#(s(x), s(y)) | → | quot#(minus(s(x), s(y)), s(y)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(s(x), s(y)), s(y))) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, le, s, false, true, quot
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) | | quot#(s(x), s(y)) | → | minus#(s(x), s(y)) |
| minus#(s(x), s(y)) | → | minus#(x, y) | | quot#(s(x), s(y)) | → | quot#(minus(s(x), s(y)), s(y)) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(s(x), s(y)), s(y))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, le, true, false, quot
Strategy
The following SCCs where found
| le#(s(x), s(y)) → le#(x, y) |
| minus#(s(x), s(y)) → minus#(x, y) |
| quot#(s(x), s(y)) → quot#(minus(s(x), s(y)), s(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) |
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(s(x), s(y)), s(y))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, le, true, false, 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: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) |
Rewrite Rules
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(s(x), s(y)), s(y))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, le, true, false, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |