YES
The TRS could be proven terminating. The proof took 403 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (4ms).
| – Problem 2 was processed with processor PolynomialLinearRange4iUR (230ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| log#(s(s(x))) | → | half#(x) | | log#(s(s(x))) | → | log#(s(half(x))) |
| half#(s(s(x))) | → | half#(x) |
Rewrite Rules
| half(0) | → | 0 | | half(s(s(x))) | → | s(half(x)) |
| log(s(0)) | → | 0 | | log(s(s(x))) | → | s(log(s(half(x)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, half, log
Strategy
The following SCCs where found
| log#(s(s(x))) → log#(s(half(x))) |
| half#(s(s(x))) → half#(x) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| log#(s(s(x))) | → | log#(s(half(x))) |
Rewrite Rules
| half(0) | → | 0 | | half(s(s(x))) | → | s(half(x)) |
| log(s(0)) | → | 0 | | log(s(s(x))) | → | s(log(s(half(x)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, half, log
Strategy
Polynomial Interpretation
- 0: 0
- half(x): x
- log(x): 0
- log#(x): x
- s(x): 2x + 1
Improved Usable rules
| half(0) | → | 0 | | half(s(s(x))) | → | s(half(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(half(x))) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| half#(s(s(x))) | → | half#(x) |
Rewrite Rules
| half(0) | → | 0 | | half(s(s(x))) | → | s(half(x)) |
| log(s(0)) | → | 0 | | log(s(s(x))) | → | s(log(s(half(x)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, half, log
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| half#(s(s(x))) | → | half#(x) |