YES
The TRS could be proven terminating. The proof took 236 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (14ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor PolynomialOrderingProcessor (78ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| *#(s(x), y) | → | *#(x, y) | | fact#(s(x)) | → | fact#(p(s(x))) |
| *#(s(x), y) | → | +#(*(x, y), y) | | fact#(s(x)) | → | *#(s(x), fact(p(s(x)))) |
| +#(x, s(y)) | → | +#(x, y) | | fact#(s(x)) | → | p#(s(x)) |
Rewrite Rules
| p(s(x)) | → | x | | fact(0) | → | s(0) |
| fact(s(x)) | → | *(s(x), fact(p(s(x)))) | | *(0, y) | → | 0 |
| *(s(x), y) | → | +(*(x, y), y) | | +(x, 0) | → | x |
| +(x, s(y)) | → | s(+(x, y)) |
Original Signature
Termination of terms over the following signature is verified: fact, 0, s, p, *, +
Strategy
The following SCCs where found
| fact#(s(x)) → fact#(p(s(x))) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| p(s(x)) | → | x | | fact(0) | → | s(0) |
| fact(s(x)) | → | *(s(x), fact(p(s(x)))) | | *(0, y) | → | 0 |
| *(s(x), y) | → | +(*(x, y), y) | | +(x, 0) | → | x |
| +(x, s(y)) | → | s(+(x, y)) |
Original Signature
Termination of terms over the following signature is verified: fact, 0, s, p, *, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| p(s(x)) | → | x | | fact(0) | → | s(0) |
| fact(s(x)) | → | *(s(x), fact(p(s(x)))) | | *(0, y) | → | 0 |
| *(s(x), y) | → | +(*(x, y), y) | | +(x, 0) | → | x |
| +(x, s(y)) | → | s(+(x, y)) |
Original Signature
Termination of terms over the following signature is verified: fact, 0, s, p, *, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 4: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| fact#(s(x)) | → | fact#(p(s(x))) |
Rewrite Rules
| p(s(x)) | → | x | | fact(0) | → | s(0) |
| fact(s(x)) | → | *(s(x), fact(p(s(x)))) | | *(0, y) | → | 0 |
| *(s(x), y) | → | +(*(x, y), y) | | +(x, 0) | → | x |
| +(x, s(y)) | → | s(+(x, y)) |
Original Signature
Termination of terms over the following signature is verified: fact, 0, s, p, *, +
Strategy
Polynomial Interpretation
- *(x,y): -2
- +(x,y): -2
- 0: -2
- fact(x): -2
- fact#(x): 6x - 1
- p(x): x - 1
- s(x): 4x + 1
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| fact#(s(x)) | → | fact#(p(s(x))) |