YES
The TRS could be proven terminating. The proof took 21 ms.
Problem 1 was processed with processor DependencyGraph (11ms).
f#(n__f(n__a)) | → | f#(n__a) | activate#(n__a) | → | a# | |
activate#(n__f(X)) | → | f#(X) | activate#(n__g(X)) | → | g#(X) | |
f#(n__f(n__a)) | → | f#(n__g(f(n__a))) |
f(n__f(n__a)) | → | f(n__g(f(n__a))) | f(X) | → | n__f(X) | |
a | → | n__a | g(X) | → | n__g(X) | |
activate(n__f(X)) | → | f(X) | activate(n__a) | → | a | |
activate(n__g(X)) | → | g(X) | activate(X) | → | X |
Termination of terms over the following signature is verified: f, activate, g, n__a, a, n__f, n__g