YES

The TRS could be proven terminating. The proof took 21 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (5ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

activate#(n__a)a#f#(f(a))f#(g(n__f(n__a)))
activate#(n__f(X))f#(activate(X))activate#(n__f(X))activate#(X)

Rewrite Rules

f(f(a))f(g(n__f(n__a)))f(X)n__f(X)
an__aactivate(n__f(X))f(activate(X))
activate(n__a)aactivate(X)X

Original Signature

Termination of terms over the following signature is verified: f, activate, g, n__a, a, n__f

Strategy


The following SCCs where found

activate#(n__f(X)) → activate#(X)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

activate#(n__f(X))activate#(X)

Rewrite Rules

f(f(a))f(g(n__f(n__a)))f(X)n__f(X)
an__aactivate(n__f(X))f(activate(X))
activate(n__a)aactivate(X)X

Original Signature

Termination of terms over the following signature is verified: f, activate, g, n__a, a, n__f

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

activate#(n__f(X))activate#(X)