YES
The TRS could be proven terminating. The proof took 502 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (4ms).
| Problem 2 was processed with processor SubtermCriterion (1ms).
| Problem 3 was processed with processor PolynomialOrderingProcessor (158ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
fac#(s(x)) | → | p#(s(x)) | | fac#(s(x)) | → | fac#(p(s(x))) |
p#(s(s(x))) | → | p#(s(x)) |
Rewrite Rules
fac(s(x)) | → | *(fac(p(s(x))), s(x)) | | p(s(0)) | → | 0 |
p(s(s(x))) | → | s(p(s(x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, p, fac, *
Strategy
The following SCCs where found
fac#(s(x)) → fac#(p(s(x))) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
fac(s(x)) | → | *(fac(p(s(x))), s(x)) | | p(s(0)) | → | 0 |
p(s(s(x))) | → | s(p(s(x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, p, fac, *
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
fac#(s(x)) | → | fac#(p(s(x))) |
Rewrite Rules
fac(s(x)) | → | *(fac(p(s(x))), s(x)) | | p(s(0)) | → | 0 |
p(s(s(x))) | → | s(p(s(x))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, p, fac, *
Strategy
Polynomial Interpretation
- *(x,y): -2
- 0: -1
- fac(x): -2
- fac#(x): x + 1
- p(x): x - 2
- s(x): 3x + 1
Improved Usable rules
p(s(s(x))) | → | s(p(s(x))) | | p(s(0)) | → | 0 |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
fac#(s(x)) | → | fac#(p(s(x))) |