YES
The TRS could be proven terminating. The proof took 21 ms.
Problem 1 was processed with processor DependencyGraph (3ms). | Problem 2 was processed with processor SubtermCriterion (0ms).
a#(a(x)) | → | a#(b(a(x))) | a#(a(x)) | → | a#(x) |
a(a(x)) | → | a(b(a(x))) |
Termination of terms over the following signature is verified: b, a
a#(a(x)) → a#(x) |
a#(a(x)) | → | a#(x) |
a(a(x)) | → | a(b(a(x))) |
Termination of terms over the following signature is verified: b, a
The following projection was used:
Thus, the following dependency pairs are removed:
a#(a(x)) | → | a#(x) |