YES
The TRS could be proven terminating. The proof took 588 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (20ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor PolynomialOrderingProcessor (174ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
| – Problem 5 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| fac#(s(x)) | → | p#(s(x)) | | times#(s(x), y) | → | times#(x, y) |
| plus#(x, s(y)) | → | plus#(x, y) | | times#(s(x), y) | → | plus#(times(x, y), y) |
| fac#(s(x)) | → | times#(fac(p(s(x))), s(x)) | | p#(s(s(x))) | → | p#(s(x)) |
| fac#(s(x)) | → | fac#(p(s(x))) |
Rewrite Rules
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| times(0, y) | → | 0 | | times(x, 0) | → | 0 |
| times(s(x), y) | → | plus(times(x, y), y) | | p(s(s(x))) | → | s(p(s(x))) |
| p(s(0)) | → | 0 | | fac(s(x)) | → | times(fac(p(s(x))), s(x)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, p, fac
Strategy
The following SCCs where found
| times#(s(x), y) → times#(x, y) |
| plus#(x, s(y)) → plus#(x, y) |
| fac#(s(x)) → fac#(p(s(x))) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| times(0, y) | → | 0 | | times(x, 0) | → | 0 |
| times(s(x), y) | → | plus(times(x, y), y) | | p(s(s(x))) | → | s(p(s(x))) |
| p(s(0)) | → | 0 | | fac(s(x)) | → | times(fac(p(s(x))), s(x)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, 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
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| times(0, y) | → | 0 | | times(x, 0) | → | 0 |
| times(s(x), y) | → | plus(times(x, y), y) | | p(s(s(x))) | → | s(p(s(x))) |
| p(s(0)) | → | 0 | | fac(s(x)) | → | times(fac(p(s(x))), s(x)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, p, fac
Strategy
Polynomial Interpretation
- 0: 1
- fac(x): -2
- fac#(x): 2x - 2
- p(x): x - 2
- plus(x,y): -2
- s(x): x + 2
- times(x,y): -2
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))) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(x, s(y)) | → | plus#(x, y) |
Rewrite Rules
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| times(0, y) | → | 0 | | times(x, 0) | → | 0 |
| times(s(x), y) | → | plus(times(x, y), y) | | p(s(s(x))) | → | s(p(s(x))) |
| p(s(0)) | → | 0 | | fac(s(x)) | → | times(fac(p(s(x))), s(x)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, p, fac
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(x, s(y)) | → | plus#(x, y) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| times#(s(x), y) | → | times#(x, y) |
Rewrite Rules
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| times(0, y) | → | 0 | | times(x, 0) | → | 0 |
| times(s(x), y) | → | plus(times(x, y), y) | | p(s(s(x))) | → | s(p(s(x))) |
| p(s(0)) | → | 0 | | fac(s(x)) | → | times(fac(p(s(x))), s(x)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, p, fac
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| times#(s(x), y) | → | times#(x, y) |