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(n__f(n__a)) → f(n__g(n__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(activate(X))
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(n__f(n__a)) → F(n__g(n__f(n__a)))
ACTIVATE(n__f(X)) → F(X)
ACTIVATE(n__a) → A
ACTIVATE(n__g(X)) → G(activate(X))
ACTIVATE(n__g(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(n__f(n__a)) → f(n__g(n__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(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__g(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(n__f(n__a)) → f(n__g(n__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(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
n__g(x1)  =  n__g(x1)

From the DPs we obtained the following set of size-change graphs:

• ACTIVATE(n__g(X)) → ACTIVATE(X) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(6) TRUE