YES
The TRS could be proven terminating. The proof took 256 ms.
Problem 1 was processed with processor PolynomialOrderingProcessor (96ms).
f#(x, s(s(y))) | → | f#(y, x) | f#(s(x), y) | → | f#(x, s(s(x))) |
f(s(x), y) | → | f(x, s(s(x))) | f(x, s(s(y))) | → | f(y, x) |
Termination of terms over the following signature is verified: f, s
There are no usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
f#(x, s(s(y))) | → | f#(y, x) | f#(s(x), y) | → | f#(x, s(s(x))) |