YES
The TRS could be proven terminating. The proof took 21 ms.
Problem 1 was processed with processor DependencyGraph (11ms). | Problem 2 was processed with processor SubtermCriterion (0ms).
| +#(*(x, y), *(a, y)) | → | +#(x, a) | *#(*(x, y), z) | → | *#(x, *(y, z)) | |
| *#(*(x, y), z) | → | *#(y, z) | +#(*(x, y), *(a, y)) | → | *#(+(x, a), y) |
| +(*(x, y), *(a, y)) | → | *(+(x, a), y) | *(*(x, y), z) | → | *(x, *(y, z)) |
Termination of terms over the following signature is verified: a, *, +
| *#(*(x, y), z) → *#(x, *(y, z)) | *#(*(x, y), z) → *#(y, z) |
| *#(*(x, y), z) | → | *#(x, *(y, z)) | *#(*(x, y), z) | → | *#(y, z) |
| +(*(x, y), *(a, y)) | → | *(+(x, a), y) | *(*(x, y), z) | → | *(x, *(y, z)) |
Termination of terms over the following signature is verified: a, *, +
The following projection was used:
Thus, the following dependency pairs are removed:
| *#(*(x, y), z) | → | *#(y, z) | *#(*(x, y), z) | → | *#(x, *(y, z)) |