YES
The TRS could be proven terminating. The proof took 414 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (10ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor PolynomialOrderingProcessor (133ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| half#(s(s(x))) | → | half#(x) | | lastbit#(s(s(x))) | → | lastbit#(x) |
| conv#(s(x)) | → | lastbit#(s(x)) | | conv#(s(x)) | → | conv#(half(s(x))) |
| conv#(s(x)) | → | half#(s(x)) |
Rewrite Rules
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | lastbit(0) | → | 0 |
| lastbit(s(0)) | → | s(0) | | lastbit(s(s(x))) | → | lastbit(x) |
| conv(0) | → | cons(nil, 0) | | conv(s(x)) | → | cons(conv(half(s(x))), lastbit(s(x))) |
Original Signature
Termination of terms over the following signature is verified: lastbit, 0, s, half, conv, cons, nil
Strategy
The following SCCs where found
| half#(s(s(x))) → half#(x) |
| lastbit#(s(s(x))) → lastbit#(x) |
| conv#(s(x)) → conv#(half(s(x))) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| lastbit#(s(s(x))) | → | lastbit#(x) |
Rewrite Rules
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | lastbit(0) | → | 0 |
| lastbit(s(0)) | → | s(0) | | lastbit(s(s(x))) | → | lastbit(x) |
| conv(0) | → | cons(nil, 0) | | conv(s(x)) | → | cons(conv(half(s(x))), lastbit(s(x))) |
Original Signature
Termination of terms over the following signature is verified: lastbit, 0, s, half, conv, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| lastbit#(s(s(x))) | → | lastbit#(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)) | | lastbit(0) | → | 0 |
| lastbit(s(0)) | → | s(0) | | lastbit(s(s(x))) | → | lastbit(x) |
| conv(0) | → | cons(nil, 0) | | conv(s(x)) | → | cons(conv(half(s(x))), lastbit(s(x))) |
Original Signature
Termination of terms over the following signature is verified: lastbit, 0, s, half, conv, cons, nil
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| half#(s(s(x))) | → | half#(x) |
Problem 4: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| conv#(s(x)) | → | conv#(half(s(x))) |
Rewrite Rules
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | lastbit(0) | → | 0 |
| lastbit(s(0)) | → | s(0) | | lastbit(s(s(x))) | → | lastbit(x) |
| conv(0) | → | cons(nil, 0) | | conv(s(x)) | → | cons(conv(half(s(x))), lastbit(s(x))) |
Original Signature
Termination of terms over the following signature is verified: lastbit, 0, s, half, conv, cons, nil
Strategy
Polynomial Interpretation
- 0: -2
- cons(x,y): -2
- conv(x): -2
- conv#(x): 2x - 1
- half(x): x - 2
- lastbit(x): -2
- nil: -2
- s(x): 3x + 1
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:
| conv#(s(x)) | → | conv#(half(s(x))) |