The TRS could be proven terminating. The proof took 20 ms.
Problem 1 was processed with processor SubtermCriterion (1ms). | – Problem 2 was processed with processor SubtermCriterion (1ms).
| g#(h(x, y), z) | → | g#(x, f(y, z)) | g#(f(x, y), z) | → | g#(y, z) | |
| g#(x, h(y, z)) | → | g#(x, y) |
| g(f(x, y), z) | → | f(x, g(y, z)) | g(h(x, y), z) | → | g(x, f(y, z)) | |
| g(x, h(y, z)) | → | h(g(x, y), z) |
Termination of terms over the following signature is verified: f, g, h
The following projection was used:
Thus, the following dependency pairs are removed:
| g#(h(x, y), z) | → | g#(x, f(y, z)) | g#(f(x, y), z) | → | g#(y, z) |
| g#(x, h(y, z)) | → | g#(x, y) |
| g(f(x, y), z) | → | f(x, g(y, z)) | g(h(x, y), z) | → | g(x, f(y, z)) | |
| g(x, h(y, z)) | → | h(g(x, y), z) |
Termination of terms over the following signature is verified: f, g, h
The following projection was used:
Thus, the following dependency pairs are removed:
| g#(x, h(y, z)) | → | g#(x, y) |