YES
The TRS could be proven terminating. The proof took 57 ms.
Problem 1 was processed with processor DependencyGraph (5ms). | Problem 2 was processed with processor SubtermCriterion (0ms).
f#(f(X)) | → | b#(f(X)) | f#(a(g(X))) | → | b#(X) | |
f#(f(X)) | → | f#(X) | f#(f(X)) | → | f#(a(b(f(X)))) |
f(f(X)) | → | f(a(b(f(X)))) | f(a(g(X))) | → | b(X) | |
b(X) | → | a(X) |
Termination of terms over the following signature is verified: f, g, b, a
f#(f(X)) → f#(X) |
f#(f(X)) | → | f#(X) |
f(f(X)) | → | f(a(b(f(X)))) | f(a(g(X))) | → | b(X) | |
b(X) | → | a(X) |
Termination of terms over the following signature is verified: f, g, b, a
The following projection was used:
Thus, the following dependency pairs are removed:
f#(f(X)) | → | f#(X) |