MAYBE
The TRS could not be proven terminating. The proof attempt took 15009 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (0ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (9ms), PolynomialLinearRange4iUR (602ms), DependencyGraph (5ms), PolynomialLinearRange8NegiUR (2683ms), DependencyGraph (5ms), ReductionPairSAT (11368ms), DependencyGraph (5ms), SizeChangePrinciple (81ms)].
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| times#(x, plus(y, s(z))) | → | times#(x, plus(y, times(s(z), 0))) | | times#(x, plus(y, s(z))) | → | times#(x, s(z)) |
| times#(x, plus(y, s(z))) | → | times#(s(z), 0) | | times#(x, s(y)) | → | times#(x, y) |
Rewrite Rules
| times(x, plus(y, s(z))) | → | plus(times(x, plus(y, times(s(z), 0))), times(x, s(z))) | | times(x, 0) | → | 0 |
| times(x, s(y)) | → | plus(times(x, y), x) | | plus(x, 0) | → | x |
| plus(x, s(y)) | → | s(plus(x, y)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| times#(x, plus(y, s(z))) | → | times#(x, plus(y, times(s(z), 0))) | | times#(x, plus(y, s(z))) | → | times#(x, s(z)) |
| times#(x, plus(y, s(z))) | → | plus#(y, times(s(z), 0)) | | times#(x, s(y)) | → | plus#(times(x, y), x) |
| plus#(x, s(y)) | → | plus#(x, y) | | times#(x, plus(y, s(z))) | → | times#(s(z), 0) |
| times#(x, s(y)) | → | times#(x, y) | | times#(x, plus(y, s(z))) | → | plus#(times(x, plus(y, times(s(z), 0))), times(x, s(z))) |
Rewrite Rules
| times(x, plus(y, s(z))) | → | plus(times(x, plus(y, times(s(z), 0))), times(x, s(z))) | | times(x, 0) | → | 0 |
| times(x, s(y)) | → | plus(times(x, y), x) | | plus(x, 0) | → | x |
| plus(x, s(y)) | → | s(plus(x, y)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times
Strategy
The following SCCs where found
| plus#(x, s(y)) → plus#(x, y) |
| times#(x, plus(y, s(z))) → times#(x, plus(y, times(s(z), 0))) | times#(x, plus(y, s(z))) → times#(x, s(z)) |
| times#(x, plus(y, s(z))) → times#(s(z), 0) | times#(x, s(y)) → times#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(x, s(y)) | → | plus#(x, y) |
Rewrite Rules
| times(x, plus(y, s(z))) | → | plus(times(x, plus(y, times(s(z), 0))), times(x, s(z))) | | times(x, 0) | → | 0 |
| times(x, s(y)) | → | plus(times(x, y), x) | | plus(x, 0) | → | x |
| plus(x, s(y)) | → | s(plus(x, y)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(x, s(y)) | → | plus#(x, y) |