YES
The TRS could be proven terminating. The proof took 29 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (5ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| f#(+(x, 0)) | → | f#(x) | | +#(x, +(y, z)) | → | +#(+(x, y), z) |
| +#(x, +(y, z)) | → | +#(x, y) |
Rewrite Rules
| f(+(x, 0)) | → | f(x) | | +(x, +(y, z)) | → | +(+(x, y), z) |
Original Signature
Termination of terms over the following signature is verified: f, 0, +
Strategy
The following SCCs where found
| +#(x, +(y, z)) → +#(+(x, y), z) | +#(x, +(y, z)) → +#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| +#(x, +(y, z)) | → | +#(+(x, y), z) | | +#(x, +(y, z)) | → | +#(x, y) |
Rewrite Rules
| f(+(x, 0)) | → | f(x) | | +(x, +(y, z)) | → | +(+(x, y), z) |
Original Signature
Termination of terms over the following signature is verified: f, 0, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| +#(x, +(y, z)) | → | +#(+(x, y), z) | | +#(x, +(y, z)) | → | +#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| f(+(x, 0)) | → | f(x) | | +(x, +(y, z)) | → | +(+(x, y), z) |
Original Signature
Termination of terms over the following signature is verified: f, 0, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: