YES
The TRS could be proven terminating. The proof took 17 ms.
Problem 1 was processed with processor SubtermCriterion (1ms).
f#(f(x, y, z), u, f(x, y, v)) | → | f#(z, u, v) | f#(f(x, y, z), u, f(x, y, v)) | → | f#(x, y, f(z, u, v)) |
f(f(x, y, z), u, f(x, y, v)) | → | f(x, y, f(z, u, v)) | f(x, y, y) | → | y | |
f(x, y, g(y)) | → | x | f(x, x, y) | → | x | |
f(g(x), x, y) | → | y |
Termination of terms over the following signature is verified: f, g
The following projection was used:
Thus, the following dependency pairs are removed:
f#(f(x, y, z), u, f(x, y, v)) | → | f#(z, u, v) | f#(f(x, y, z), u, f(x, y, v)) | → | f#(x, y, f(z, u, v)) |