YES
The TRS could be proven terminating. The proof took 760 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (12ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (130ms).
| – Problem 3 was processed with processor PolynomialLinearRange4iUR (393ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| log#(s(s(X))) | → | log#(s(quot(X, s(s(0))))) | | quot#(s(X), s(Y)) | → | min#(X, Y) |
| log#(s(s(X))) | → | quot#(X, s(s(0))) | | quot#(s(X), s(Y)) | → | quot#(min(X, Y), s(Y)) |
| min#(s(X), s(Y)) | → | min#(X, Y) |
Rewrite Rules
| min(X, 0) | → | X | | min(s(X), s(Y)) | → | min(X, Y) |
| quot(0, s(Y)) | → | 0 | | quot(s(X), s(Y)) | → | s(quot(min(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: min, 0, s, quot, log
Strategy
The following SCCs where found
| log#(s(s(X))) → log#(s(quot(X, s(s(0))))) |
| quot#(s(X), s(Y)) → quot#(min(X, Y), s(Y)) |
| min#(s(X), s(Y)) → min#(X, Y) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| quot#(s(X), s(Y)) | → | quot#(min(X, Y), s(Y)) |
Rewrite Rules
| min(X, 0) | → | X | | min(s(X), s(Y)) | → | min(X, Y) |
| quot(0, s(Y)) | → | 0 | | quot(s(X), s(Y)) | → | s(quot(min(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: min, 0, s, quot, log
Strategy
Polynomial Interpretation
- 0: 0
- log(x): 0
- min(x,y): 2x
- quot(x,y): 0
- quot#(x,y): 2x
- s(x): 2x + 1
Improved Usable rules
| min(s(X), s(Y)) | → | min(X, Y) | | min(X, 0) | → | X |
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#(min(X, Y), s(Y)) |
Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| log#(s(s(X))) | → | log#(s(quot(X, s(s(0))))) |
Rewrite Rules
| min(X, 0) | → | X | | min(s(X), s(Y)) | → | min(X, Y) |
| quot(0, s(Y)) | → | 0 | | quot(s(X), s(Y)) | → | s(quot(min(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: min, 0, s, quot, log
Strategy
Polynomial Interpretation
- 0: 0
- log(x): 0
- log#(x): x + 1
- min(x,y): x
- quot(x,y): x
- s(x): x + 1
Improved Usable rules
| min(s(X), s(Y)) | → | min(X, Y) | | quot(s(X), s(Y)) | → | s(quot(min(X, Y), s(Y))) |
| quot(0, s(Y)) | → | 0 | | min(X, 0) | → | X |
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))))) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| min#(s(X), s(Y)) | → | min#(X, Y) |
Rewrite Rules
| min(X, 0) | → | X | | min(s(X), s(Y)) | → | min(X, Y) |
| quot(0, s(Y)) | → | 0 | | quot(s(X), s(Y)) | → | s(quot(min(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: min, 0, s, quot, log
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| min#(s(X), s(Y)) | → | min#(X, Y) |