YES
The TRS could be proven terminating. The proof took 33 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (16ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| +#(*(x, y), +(*(x, z), u)) | → | +#(y, z) | | +#(*(x, y), *(x, z)) | → | +#(y, z) |
| +#(+(x, y), z) | → | +#(y, z) | | +#(*(x, y), +(*(x, z), u)) | → | +#(*(x, +(y, z)), u) |
| +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| +(*(x, y), *(x, z)) | → | *(x, +(y, z)) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| +(*(x, y), +(*(x, z), u)) | → | +(*(x, +(y, z)), u) |
Original Signature
Termination of terms over the following signature is verified: u, *, +
Strategy
The following SCCs where found
| +#(*(x, y), +(*(x, z), u)) → +#(y, z) | +#(+(x, y), z) → +#(y, z) |
| +#(*(x, y), *(x, z)) → +#(y, z) | +#(+(x, y), z) → +#(x, +(y, z)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| +#(*(x, y), +(*(x, z), u)) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(y, z) |
| +#(*(x, y), *(x, z)) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite Rules
| +(*(x, y), *(x, z)) | → | *(x, +(y, z)) | | +(+(x, y), z) | → | +(x, +(y, z)) |
| +(*(x, y), +(*(x, z), u)) | → | +(*(x, +(y, z)), u) |
Original Signature
Termination of terms over the following signature is verified: u, *, +
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| +#(*(x, y), +(*(x, z), u)) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(y, z) |
| +#(*(x, y), *(x, z)) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |