TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60019 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (34ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (4ms), PolynomialLinearRange4iUR (270ms), DependencyGraph (3ms), PolynomialLinearRange8NegiUR (7242ms), DependencyGraph (2ms), ReductionPairSAT (491ms), DependencyGraph (1ms), SizeChangePrinciple (106ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
| – Problem 3 was processed with processor SubtermCriterion (29ms).
| – 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)) | → | if_mod#(le(y, x), s(x), s(y)) | | if_mod#(true, x, y) | → | mod#(minus(x, y), y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | mod(0, y) | → | 0 |
| mod(s(x), 0) | → | 0 | | mod(s(x), s(y)) | → | if_mod(le(y, x), s(x), s(y)) |
| if_mod(true, x, y) | → | mod(minus(x, y), y) | | if_mod(false, s(x), s(y)) | → | s(x) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, le, mod, false, true, if_mod
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| mod#(s(x), s(y)) | → | le#(y, x) | | le#(s(x), s(y)) | → | le#(x, y) |
| mod#(s(x), s(y)) | → | if_mod#(le(y, x), s(x), s(y)) | | if_mod#(true, x, y) | → | mod#(minus(x, y), y) |
| minus#(s(x), s(y)) | → | minus#(x, y) | | if_mod#(true, x, y) | → | minus#(x, y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | mod(0, y) | → | 0 |
| mod(s(x), 0) | → | 0 | | mod(s(x), s(y)) | → | if_mod(le(y, x), s(x), s(y)) |
| if_mod(true, x, y) | → | mod(minus(x, y), y) | | if_mod(false, s(x), s(y)) | → | s(x) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, le, s, mod, true, false, if_mod
Strategy
The following SCCs where found
| le#(s(x), s(y)) → le#(x, y) |
| minus#(s(x), s(y)) → minus#(x, y) |
| mod#(s(x), s(y)) → if_mod#(le(y, x), s(x), s(y)) | if_mod#(true, x, y) → mod#(minus(x, y), y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | mod(0, y) | → | 0 |
| mod(s(x), 0) | → | 0 | | mod(s(x), s(y)) | → | if_mod(le(y, x), s(x), s(y)) |
| if_mod(true, x, y) | → | mod(minus(x, y), y) | | if_mod(false, s(x), s(y)) | → | s(x) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, le, s, mod, true, false, if_mod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | mod(0, y) | → | 0 |
| mod(s(x), 0) | → | 0 | | mod(s(x), s(y)) | → | if_mod(le(y, x), s(x), s(y)) |
| if_mod(true, x, y) | → | mod(minus(x, y), y) | | if_mod(false, s(x), s(y)) | → | s(x) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, le, s, mod, true, false, if_mod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) |