YES
The TRS could be proven terminating. The proof took 868 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (16ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (234ms).
| – Problem 4 was processed with processor PolynomialLinearRange4iUR (565ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| quot#(s(x), s(y)) | → | minus#(x, y) | | log#(s(s(x))) | → | quot#(x, s(s(0))) |
| minus#(x, s(y)) | → | minus#(x, y) | | quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
| minus#(x, s(y)) | → | pred#(minus(x, y)) | | log#(s(s(x))) | → | log#(s(quot(x, s(s(0))))) |
Rewrite Rules
| pred(s(x)) | → | x | | minus(x, 0) | → | x |
| minus(x, s(y)) | → | pred(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, pred, quot, log
Strategy
The following SCCs where found
| quot#(s(x), s(y)) → quot#(minus(x, y), s(y)) |
| minus#(x, s(y)) → minus#(x, y) |
| log#(s(s(x))) → log#(s(quot(x, s(s(0))))) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(x, s(y)) | → | minus#(x, y) |
Rewrite Rules
| pred(s(x)) | → | x | | minus(x, 0) | → | x |
| minus(x, s(y)) | → | pred(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, pred, quot, log
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(x, s(y)) | → | minus#(x, y) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
Rewrite Rules
| pred(s(x)) | → | x | | minus(x, 0) | → | x |
| minus(x, s(y)) | → | pred(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, pred, quot, log
Strategy
Polynomial Interpretation
- 0: 3
- log(x): 0
- minus(x,y): x
- pred(x): x
- quot(x,y): 0
- quot#(x,y): 2x
- s(x): 2x + 1
Improved Usable rules
| pred(s(x)) | → | x | | minus(x, 0) | → | x |
| minus(x, s(y)) | → | pred(minus(x, y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| quot#(s(x), s(y)) | → | quot#(minus(x, y), s(y)) |
Problem 4: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| log#(s(s(x))) | → | log#(s(quot(x, s(s(0))))) |
Rewrite Rules
| pred(s(x)) | → | x | | minus(x, 0) | → | x |
| minus(x, s(y)) | → | pred(minus(x, y)) | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | log(s(0)) | → | 0 |
| log(s(s(x))) | → | s(log(s(quot(x, s(s(0)))))) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, pred, quot, log
Strategy
Polynomial Interpretation
- 0: 0
- log(x): 0
- log#(x): x
- minus(x,y): x
- pred(x): x
- quot(x,y): x
- s(x): 2x + 1
Improved Usable rules
| pred(s(x)) | → | x | | quot(0, s(y)) | → | 0 |
| quot(s(x), s(y)) | → | s(quot(minus(x, y), s(y))) | | minus(x, 0) | → | x |
| minus(x, s(y)) | → | pred(minus(x, y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| log#(s(s(x))) | → | log#(s(quot(x, s(s(0))))) |