YES
The TRS could be proven terminating. The proof took 20 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (6ms).
| – Problem 2 was processed with processor SubtermCriterion (0ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| half#(s(s(x))) | → | half#(x) | | double#(s(x)) | → | double#(x) |
| -#(s(x), s(y)) | → | -#(x, y) |
Rewrite Rules
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, y) | | if(0, y, z) | → | y |
| if(s(x), y, z) | → | z | | half(double(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, half, double, -
Strategy
The following SCCs where found
| half#(s(s(x))) → half#(x) |
| double#(s(x)) → double#(x) |
| -#(s(x), s(y)) → -#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| -#(s(x), s(y)) | → | -#(x, y) |
Rewrite Rules
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, y) | | if(0, y, z) | → | y |
| if(s(x), y, z) | → | z | | half(double(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, half, double, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| -#(s(x), s(y)) | → | -#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| half#(s(s(x))) | → | half#(x) |
Rewrite Rules
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, y) | | if(0, y, z) | → | y |
| if(s(x), y, z) | → | z | | half(double(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, half, double, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| half#(s(s(x))) | → | half#(x) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| double#(s(x)) | → | double#(x) |
Rewrite Rules
| double(0) | → | 0 | | double(s(x)) | → | s(s(double(x))) |
| half(0) | → | 0 | | half(s(0)) | → | 0 |
| half(s(s(x))) | → | s(half(x)) | | -(x, 0) | → | x |
| -(s(x), s(y)) | → | -(x, y) | | if(0, y, z) | → | y |
| if(s(x), y, z) | → | z | | half(double(x)) | → | x |
Original Signature
Termination of terms over the following signature is verified: 0, s, if, half, double, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| double#(s(x)) | → | double#(x) |