The TRS could be proven terminating. The proof took 31 ms.
Problem 1 was processed with processor SubtermCriterion (1ms). | – Problem 2 was processed with processor SubtermCriterion (1ms).
| *#(+(x, y), z) | → | *#(x, z) | *#(x, oplus(y, z)) | → | *#(x, y) | |
| *#(x, oplus(y, z)) | → | *#(x, z) | *#(+(x, y), z) | → | *#(y, z) | |
| *#(x, *(y, z)) | → | *#(otimes(x, y), z) |
| *(x, *(y, z)) | → | *(otimes(x, y), z) | *(1, y) | → | y | |
| *(+(x, y), z) | → | oplus(*(x, z), *(y, z)) | *(x, oplus(y, z)) | → | oplus(*(x, y), *(x, z)) |
Termination of terms over the following signature is verified: 1, otimes, *, +, oplus
The following projection was used:
Thus, the following dependency pairs are removed:
| *#(x, oplus(y, z)) | → | *#(x, y) | *#(x, oplus(y, z)) | → | *#(x, z) | |
| *#(x, *(y, z)) | → | *#(otimes(x, y), z) |
| *#(+(x, y), z) | → | *#(x, z) | *#(+(x, y), z) | → | *#(y, z) |
| *(x, *(y, z)) | → | *(otimes(x, y), z) | *(1, y) | → | y | |
| *(+(x, y), z) | → | oplus(*(x, z), *(y, z)) | *(x, oplus(y, z)) | → | oplus(*(x, y), *(x, z)) |
Termination of terms over the following signature is verified: 1, otimes, *, +, oplus
The following projection was used:
Thus, the following dependency pairs are removed:
| *#(+(x, y), z) | → | *#(x, z) | *#(+(x, y), z) | → | *#(y, z) |