The TRS could be proven terminating. The proof took 31 ms.
Problem 1 was processed with processor SubtermCriterion (1ms).
| f#(g(x, y), z) | → | f#(x, z) | f#(f(x, y), z) | → | f#(y, z) | |
| f#(f(x, y), z) | → | f#(x, f(y, z)) | f#(g(x, y), z) | → | f#(y, z) |
| f(0, y) | → | y | f(x, 0) | → | x | |
| f(i(x), y) | → | i(x) | f(f(x, y), z) | → | f(x, f(y, z)) | |
| f(g(x, y), z) | → | g(f(x, z), f(y, z)) | f(1, g(x, y)) | → | x | |
| f(2, g(x, y)) | → | y |
Termination of terms over the following signature is verified: f, g, 2, 1, 0, i
The following projection was used:
Thus, the following dependency pairs are removed:
| f#(g(x, y), z) | → | f#(x, z) | f#(f(x, y), z) | → | f#(y, z) | |
| f#(f(x, y), z) | → | f#(x, f(y, z)) | f#(g(x, y), z) | → | f#(y, z) |