YES
The TRS could be proven terminating. The proof took 458 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (4ms).
| – Problem 2 was processed with processor PolynomialOrderingProcessor (143ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| bits#(s(x)) | → | half#(s(x)) | | bits#(s(x)) | → | bits#(half(s(x))) |
| half#(s(s(x))) | → | half#(x) |
Rewrite Rules
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | bits(0) | → | 0 |
| bits(s(x)) | → | s(bits(half(s(x)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, bits, half
Strategy
The following SCCs where found
| bits#(s(x)) → bits#(half(s(x))) |
| half#(s(s(x))) → half#(x) |
Problem 2: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| bits#(s(x)) | → | bits#(half(s(x))) |
Rewrite Rules
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | bits(0) | → | 0 |
| bits(s(x)) | → | s(bits(half(s(x)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, bits, half
Strategy
Polynomial Interpretation
- 0: -1
- bits(x): -2
- bits#(x): 2x - 2
- half(x): x - 2
- s(x): x + 2
Improved Usable rules
| half(s(0)) | → | 0 | | 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:
| bits#(s(x)) | → | bits#(half(s(x))) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| half#(s(s(x))) | → | half#(x) |
Rewrite Rules
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | bits(0) | → | 0 |
| bits(s(x)) | → | s(bits(half(s(x)))) |
Original Signature
Termination of terms over the following signature is verified: 0, s, bits, half
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| half#(s(s(x))) | → | half#(x) |