YES
The TRS could be proven terminating. The proof took 32 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (19ms).
| – Problem 2 was processed with processor SubtermCriterion (2ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| +#(x, minus(y)) | → | +#(minus(x), y) | | +#(x, minus(y)) | → | minus#(+(minus(x), y)) |
| +#(x, +(y, z)) | → | +#(+(x, y), z) | | +#(minus(+(x, 1)), 1) | → | minus#(x) |
| +#(x, minus(y)) | → | minus#(x) | | +#(x, +(y, z)) | → | +#(x, y) |
Rewrite Rules
| minus(0) | → | 0 | | +(x, 0) | → | x |
| +(0, y) | → | y | | +(minus(1), 1) | → | 0 |
| minus(minus(x)) | → | x | | +(x, minus(y)) | → | minus(+(minus(x), y)) |
| +(x, +(y, z)) | → | +(+(x, y), z) | | +(minus(+(x, 1)), 1) | → | minus(x) |
Original Signature
Termination of terms over the following signature is verified: 1, minus, 0, +
Strategy
The following SCCs where found
| +#(x, minus(y)) → +#(minus(x), y) | +#(x, +(y, z)) → +#(+(x, y), z) |
| +#(x, +(y, z)) → +#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| +#(x, minus(y)) | → | +#(minus(x), y) | | +#(x, +(y, z)) | → | +#(+(x, y), z) |
| +#(x, +(y, z)) | → | +#(x, y) |
Rewrite Rules
| minus(0) | → | 0 | | +(x, 0) | → | x |
| +(0, y) | → | y | | +(minus(1), 1) | → | 0 |
| minus(minus(x)) | → | x | | +(x, minus(y)) | → | minus(+(minus(x), y)) |
| +(x, +(y, z)) | → | +(+(x, y), z) | | +(minus(+(x, 1)), 1) | → | minus(x) |
Original Signature
Termination of terms over the following signature is verified: 1, minus, 0, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| +#(x, minus(y)) | → | +#(minus(x), y) | | +#(x, +(y, z)) | → | +#(+(x, y), z) |
| +#(x, +(y, z)) | → | +#(x, y) |