Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(f(a)) → c(n__f(n__g(n__f(n__a))))
f(X) → n__f(X)
g(X) → n__g(X)
a → n__a
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(n__a) → a
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive Path Order [RPO].
Precedence:

activate₁ > f₁ > c₁
activate₁ > f₁ > n_f1
activate₁ > f₁ > n_g1
activate₁ > f₁ > n_a
activate₁ > a > n_a
activate₁ > g₁ > n_g1

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

f(f(a)) → c(n__f(n__g(n__f(n__a))))
f(X) → n__f(X)
g(X) → n__g(X)
a → n__a
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(n__a) → a
activate(X) → X

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE