The TRS could be proven terminating. The proof took 35 ms.
Problem 1 was processed with processor SubtermCriterion (4ms). | – Problem 2 was processed with processor DependencyGraph (1ms).
| f#(x, y, s(z)) | → | f#(0, 1, z) | f#(0, 1, x) | → | f#(s(x), x, x) |
| f(0, 1, x) | → | f(s(x), x, x) | f(x, y, s(z)) | → | s(f(0, 1, z)) |
Termination of terms over the following signature is verified: f, 1, 0, s
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(x, y, s(z)) | → | f#(0, 1, z) |
| f#(0, 1, x) | → | f#(s(x), x, x) |
| f(0, 1, x) | → | f(s(x), x, x) | f(x, y, s(z)) | → | s(f(0, 1, z)) |
Termination of terms over the following signature is verified: f, 1, 0, s