TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60043 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (33ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (141ms), DependencyGraph (1ms), PolynomialLinearRange8NegiUR (733ms), DependencyGraph (2ms), ReductionPairSAT (341ms), DependencyGraph (1ms), SizeChangePrinciple (17ms), ForwardNarrowing (1ms), BackwardInstantiation (3ms), ForwardInstantiation (2ms), Propagation (0ms)].
| – 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
| mod#(s(x), s(y)) | → | mod#(-(s(x), s(y)), s(y)) |
Rewrite Rules
| leq(0, y) | → | true | | leq(s(x), 0) | → | false |
| leq(s(x), s(y)) | → | leq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, y) | | mod(0, y) | → | 0 |
| mod(s(x), 0) | → | 0 | | mod(s(x), s(y)) | → | if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, leq, mod, false, true, -
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| leq#(s(x), s(y)) | → | leq#(x, y) | | mod#(s(x), s(y)) | → | if#(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x)) |
| mod#(s(x), s(y)) | → | leq#(y, x) | | mod#(s(x), s(y)) | → | -#(s(x), s(y)) |
| mod#(s(x), s(y)) | → | mod#(-(s(x), s(y)), s(y)) | | -#(s(x), s(y)) | → | -#(x, y) |
Rewrite Rules
| leq(0, y) | → | true | | leq(s(x), 0) | → | false |
| leq(s(x), s(y)) | → | leq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, y) | | mod(0, y) | → | 0 |
| mod(s(x), 0) | → | 0 | | mod(s(x), s(y)) | → | if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, leq, if, mod, true, false, -
Strategy
The following SCCs where found
| leq#(s(x), s(y)) → leq#(x, y) |
| mod#(s(x), s(y)) → mod#(-(s(x), s(y)), s(y)) |
| -#(s(x), s(y)) → -#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| leq#(s(x), s(y)) | → | leq#(x, y) |
Rewrite Rules
| leq(0, y) | → | true | | leq(s(x), 0) | → | false |
| leq(s(x), s(y)) | → | leq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, y) | | mod(0, y) | → | 0 |
| mod(s(x), 0) | → | 0 | | mod(s(x), s(y)) | → | if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, leq, if, mod, true, false, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| leq#(s(x), s(y)) | → | leq#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| -#(s(x), s(y)) | → | -#(x, y) |
Rewrite Rules
| leq(0, y) | → | true | | leq(s(x), 0) | → | false |
| leq(s(x), s(y)) | → | leq(x, y) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, y) | | mod(0, y) | → | 0 |
| mod(s(x), 0) | → | 0 | | mod(s(x), s(y)) | → | if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x)) |
Original Signature
Termination of terms over the following signature is verified: 0, s, leq, if, mod, true, false, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| -#(s(x), s(y)) | → | -#(x, y) |