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 (49ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (1ms), PolynomialLinearRange4iUR (933ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (12524ms), DependencyGraph (1ms), ReductionPairSAT (693ms), DependencyGraph (1ms), SizeChangePrinciple (187ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (0ms), Propagation (0ms)].
| – Problem 4 was processed with processor SubtermCriterion (0ms).
| – Problem 5 was processed with processor SubtermCriterion (1ms).
| – Problem 6 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| f#(s(x)) | → | f#(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))) |
Rewrite Rules
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| *(x, 0) | → | 0 | | *(x, s(y)) | → | +(x, *(x, y)) |
| twice(0) | → | 0 | | twice(s(x)) | → | s(s(twice(x))) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| f(s(x)) | → | f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))) |
Original Signature
Termination of terms over the following signature is verified: f, twice, 0, s, *, +, -
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(s(x)) | → | *#(s(x), s(s(x))) | | f#(s(x)) | → | +#(*(s(x), s(s(x))), s(s(0))) |
| *#(x, s(y)) | → | +#(x, *(x, y)) | | *#(x, s(y)) | → | *#(x, y) |
| f#(s(x)) | → | f#(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))) | | f#(s(x)) | → | -#(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0)))) |
| twice#(s(x)) | → | twice#(x) | | f#(s(x)) | → | *#(s(s(x)), s(s(x))) |
| +#(s(x), y) | → | +#(x, y) | | -#(s(x), s(y)) | → | -#(x, y) |
Rewrite Rules
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| *(x, 0) | → | 0 | | *(x, s(y)) | → | +(x, *(x, y)) |
| twice(0) | → | 0 | | twice(s(x)) | → | s(s(twice(x))) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| f(s(x)) | → | f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, twice, s, *, +, -
Strategy
The following SCCs where found
| f#(s(x)) → f#(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))) |
| -#(s(x), s(y)) → -#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| *(x, 0) | → | 0 | | *(x, s(y)) | → | +(x, *(x, y)) |
| twice(0) | → | 0 | | twice(s(x)) | → | s(s(twice(x))) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| f(s(x)) | → | f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, twice, s, *, +, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| -#(s(x), s(y)) | → | -#(x, y) |
Rewrite Rules
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| *(x, 0) | → | 0 | | *(x, s(y)) | → | +(x, *(x, y)) |
| twice(0) | → | 0 | | twice(s(x)) | → | s(s(twice(x))) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| f(s(x)) | → | f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, twice, s, *, +, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| -#(s(x), s(y)) | → | -#(x, y) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| *(x, 0) | → | 0 | | *(x, s(y)) | → | +(x, *(x, y)) |
| twice(0) | → | 0 | | twice(s(x)) | → | s(s(twice(x))) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| f(s(x)) | → | f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, twice, s, *, +, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 6: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| +(0, y) | → | y | | +(s(x), y) | → | s(+(x, y)) |
| *(x, 0) | → | 0 | | *(x, s(y)) | → | +(x, *(x, y)) |
| twice(0) | → | 0 | | twice(s(x)) | → | s(s(twice(x))) |
| -(x, 0) | → | x | | -(s(x), s(y)) | → | -(x, y) |
| f(s(x)) | → | f(-(*(s(s(x)), s(s(x))), +(*(s(x), s(s(x))), s(s(0))))) |
Original Signature
Termination of terms over the following signature is verified: f, 0, twice, s, *, +, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: