The TRS could be proven terminating. The proof took 52 ms.
Problem 1 was processed with processor SubtermCriterion (1ms). | – Problem 2 was processed with processor DependencyGraph (1ms).
| g#(h(x), y) | → | f#(x, y) | f#(x, y) | → | g#(x, y) | |
| g#(h(x), y) | → | g#(x, y) |
| f(x, y) | → | g(x, y) | g(h(x), y) | → | h(f(x, y)) | |
| g(h(x), y) | → | h(g(x, y)) |
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) | → | f#(x, y) | g#(h(x), y) | → | g#(x, y) |
| f#(x, y) | → | g#(x, y) |
| f(x, y) | → | g(x, y) | g(h(x), y) | → | h(f(x, y)) | |
| g(h(x), y) | → | h(g(x, y)) |
Termination of terms over the following signature is verified: f, g, h