YES
The TRS could be proven terminating. The proof took 19 ms.
Problem 1 was processed with processor DependencyGraph (3ms). | Problem 2 was processed with processor SubtermCriterion (0ms).
a__f#(f(a)) | → | a__f#(g(f(a))) | mark#(f(X)) | → | a__f#(mark(X)) | |
mark#(f(X)) | → | mark#(X) |
a__f(f(a)) | → | a__f(g(f(a))) | mark(f(X)) | → | a__f(mark(X)) | |
mark(a) | → | a | mark(g(X)) | → | g(X) | |
a__f(X) | → | f(X) |
Termination of terms over the following signature is verified: f, g, a, mark, a__f
mark#(f(X)) → mark#(X) |
mark#(f(X)) | → | mark#(X) |
a__f(f(a)) | → | a__f(g(f(a))) | mark(f(X)) | → | a__f(mark(X)) | |
mark(a) | → | a | mark(g(X)) | → | g(X) | |
a__f(X) | → | f(X) |
Termination of terms over the following signature is verified: f, g, a, mark, a__f
The following projection was used:
Thus, the following dependency pairs are removed:
mark#(f(X)) | → | mark#(X) |