YES
The TRS could be proven terminating. The proof took 1647 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (41ms).
| – Problem 2 was processed with processor PolynomialOrderingProcessor (615ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor PolynomialOrderingProcessor (82ms).
| – Problem 5 was processed with processor PolynomialOrderingProcessor (140ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| times#(s(x), y) | → | plus#(y, times(p(s(x)), y)) | | fac#(s(x), y) | → | times#(s(x), y) |
| plus#(s(x), y) | → | plus#(p(s(x)), y) | | times#(s(x), y) | → | p#(s(x)) |
| times#(s(x), y) | → | times#(p(s(x)), y) | | fac#(s(x), y) | → | p#(s(x)) |
| factorial#(x) | → | fac#(x, s(0)) | | plus#(s(x), y) | → | p#(s(x)) |
| p#(s(s(x))) | → | p#(s(x)) | | fac#(s(x), y) | → | fac#(p(s(x)), times(s(x), y)) |
Rewrite Rules
| plus(0, x) | → | x | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| times(0, y) | → | 0 | | times(s(x), y) | → | plus(y, times(p(s(x)), y)) |
| p(s(0)) | → | 0 | | p(s(s(x))) | → | s(p(s(x))) |
| fac(0, x) | → | x | | fac(s(x), y) | → | fac(p(s(x)), times(s(x), y)) |
| factorial(x) | → | fac(x, s(0)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, p, fac, factorial
Strategy
The following SCCs where found
| plus#(s(x), y) → plus#(p(s(x)), y) |
| times#(s(x), y) → times#(p(s(x)), y) |
| fac#(s(x), y) → fac#(p(s(x)), times(s(x), y)) |
Problem 2: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| fac#(s(x), y) | → | fac#(p(s(x)), times(s(x), y)) |
Rewrite Rules
| plus(0, x) | → | x | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| times(0, y) | → | 0 | | times(s(x), y) | → | plus(y, times(p(s(x)), y)) |
| p(s(0)) | → | 0 | | p(s(s(x))) | → | s(p(s(x))) |
| fac(0, x) | → | x | | fac(s(x), y) | → | fac(p(s(x)), times(s(x), y)) |
| factorial(x) | → | fac(x, s(0)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, p, fac, factorial
Strategy
Polynomial Interpretation
- 0: 1
- fac(x,y): -2
- fac#(x,y): 5x - 1
- factorial(x): -2
- p(x): x - 2
- plus(x,y): 3y
- s(x): 3x + 1
- times(x,y): 3y - 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), y) | → | fac#(p(s(x)), times(s(x), y)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| plus(0, x) | → | x | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| times(0, y) | → | 0 | | times(s(x), y) | → | plus(y, times(p(s(x)), y)) |
| p(s(0)) | → | 0 | | p(s(s(x))) | → | s(p(s(x))) |
| fac(0, x) | → | x | | fac(s(x), y) | → | fac(p(s(x)), times(s(x), y)) |
| factorial(x) | → | fac(x, s(0)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, p, fac, factorial
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 4: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| times#(s(x), y) | → | times#(p(s(x)), y) |
Rewrite Rules
| plus(0, x) | → | x | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| times(0, y) | → | 0 | | times(s(x), y) | → | plus(y, times(p(s(x)), y)) |
| p(s(0)) | → | 0 | | p(s(s(x))) | → | s(p(s(x))) |
| fac(0, x) | → | x | | fac(s(x), y) | → | fac(p(s(x)), times(s(x), y)) |
| factorial(x) | → | fac(x, s(0)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, p, fac, factorial
Strategy
Polynomial Interpretation
- 0: -1
- fac(x,y): -2
- factorial(x): -2
- p(x): x - 2
- plus(x,y): -2
- s(x): 2x + 1
- times(x,y): -2
- times#(x,y): 2x + 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:
| times#(s(x), y) | → | times#(p(s(x)), y) |
Problem 5: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(p(s(x)), y) |
Rewrite Rules
| plus(0, x) | → | x | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| times(0, y) | → | 0 | | times(s(x), y) | → | plus(y, times(p(s(x)), y)) |
| p(s(0)) | → | 0 | | p(s(s(x))) | → | s(p(s(x))) |
| fac(0, x) | → | x | | fac(s(x), y) | → | fac(p(s(x)), times(s(x), y)) |
| factorial(x) | → | fac(x, s(0)) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, s, times, p, fac, factorial
Strategy
Polynomial Interpretation
- 0: -2
- fac(x,y): -2
- factorial(x): -2
- p(x): x - 2
- plus(x,y): -2
- plus#(x,y): 2x - 1
- s(x): 3x + 1
- 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:
| plus#(s(x), y) | → | plus#(p(s(x)), y) |