YES
The TRS could be proven terminating. The proof took 18 ms.
Problem 1 was processed with processor DependencyGraph (4ms). | Problem 2 was processed with processor SubtermCriterion (1ms).
mark#(f(X)) | → | a__f#(mark(X)) | mark#(f(X)) | → | mark#(X) | |
mark#(h(X)) | → | mark#(X) |
a__f(X) | → | g(h(f(X))) | mark(f(X)) | → | a__f(mark(X)) | |
mark(g(X)) | → | g(X) | mark(h(X)) | → | h(mark(X)) | |
a__f(X) | → | f(X) |
Termination of terms over the following signature is verified: f, g, mark, h, a__f
mark#(f(X)) → mark#(X) | mark#(h(X)) → mark#(X) |
mark#(f(X)) | → | mark#(X) | mark#(h(X)) | → | mark#(X) |
a__f(X) | → | g(h(f(X))) | mark(f(X)) | → | a__f(mark(X)) | |
mark(g(X)) | → | g(X) | mark(h(X)) | → | h(mark(X)) | |
a__f(X) | → | f(X) |
Termination of terms over the following signature is verified: f, g, mark, h, a__f
The following projection was used:
Thus, the following dependency pairs are removed:
mark#(f(X)) | → | mark#(X) | mark#(h(X)) | → | mark#(X) |