TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60001 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (43ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (2ms), PolynomialLinearRange4iUR (286ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (3447ms), DependencyGraph (2ms), ReductionPairSAT (301ms), DependencyGraph (1ms), SizeChangePrinciple (63ms), ForwardNarrowing (0ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (3ms)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| quot#(x, s(y)) | → | if_quot#(le(s(y), x), x, s(y)) | | if_quot#(true, x, y) | → | quot#(minus(x, y), 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(x, s(y)) | → | if_quot(le(s(y), x), x, s(y)) |
| if_quot(true, x, y) | → | s(quot(minus(x, y), y)) | | if_quot(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: 0, minus, le, s, if_quot, false, true, quot
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| quot#(x, s(y)) | → | if_quot#(le(s(y), x), x, s(y)) | | le#(s(x), s(y)) | → | le#(x, y) |
| if_quot#(true, x, y) | → | minus#(x, y) | | if_quot#(true, x, y) | → | quot#(minus(x, y), y) |
| minus#(s(x), s(y)) | → | minus#(x, y) | | quot#(x, s(y)) | → | le#(s(y), x) |
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(x, s(y)) | → | if_quot(le(s(y), x), x, s(y)) |
| if_quot(true, x, y) | → | s(quot(minus(x, y), y)) | | if_quot(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, le, if_quot, true, false, quot
Strategy
The following SCCs where found
| le#(s(x), s(y)) → le#(x, y) |
| quot#(x, s(y)) → if_quot#(le(s(y), x), x, s(y)) | if_quot#(true, x, y) → quot#(minus(x, y), y) |
| minus#(s(x), s(y)) → minus#(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) |
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | quot(x, s(y)) | → | if_quot(le(s(y), x), x, s(y)) |
| if_quot(true, x, y) | → | s(quot(minus(x, y), y)) | | if_quot(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, le, if_quot, 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(x, s(y)) | → | if_quot(le(s(y), x), x, s(y)) |
| if_quot(true, x, y) | → | s(quot(minus(x, y), y)) | | if_quot(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, le, if_quot, 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) |