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(X) → g(n__h(n__f(X)))
h(X) → n__h(X)
f(X) → n__f(X)
activate(n__h(X)) → h(activate(X))
activate(n__f(X)) → f(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:

ACTIVATE(n__h(X)) → H(activate(X))
ACTIVATE(n__h(X)) → ACTIVATE(X)
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → g(n__h(n__f(X)))
h(X) → n__h(X)
f(X) → n__f(X)
activate(n__h(X)) → h(activate(X))
activate(n__f(X)) → f(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 2 less nodes.

(4) Obligation:

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

ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__h(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → g(n__h(n__f(X)))
h(X) → n__h(X)
f(X) → n__f(X)
activate(n__h(X)) → h(activate(X))
activate(n__f(X)) → f(activate(X))
activate(X) → X

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__h(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__f(x1)  =  x1
n__h(x1)  =  n__h(x1)
f(x1)  =  x1
g(x1)  =  g
h(x1)  =  h(x1)
activate(x1)  =  activate(x1)

Recursive Path Order [RPO].
Precedence:

activate₁ > h₁ > n_h1 > g

The following usable rules [FROCOS05] were oriented:

f(X) → g(n__h(n__f(X)))
h(X) → n__h(X)
f(X) → n__f(X)
activate(n__h(X)) → h(activate(X))
activate(n__f(X)) → f(activate(X))
activate(X) → X

(6) Obligation:

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

ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → g(n__h(n__f(X)))
h(X) → n__h(X)
f(X) → n__f(X)
activate(n__h(X)) → h(activate(X))
activate(n__f(X)) → f(activate(X))
activate(X) → X

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__f(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__f(x1)  =  n__f(x1)
f(x1)  =  f(x1)
g(x1)  =  g
n__h(x1)  =  n__h
h(x1)  =  h
activate(x1)  =  activate(x1)

Recursive Path Order [RPO].
Precedence:

activate₁ > f₁ > n_f1
activate₁ > f₁ > g
activate₁ > h > n_h

The following usable rules [FROCOS05] were oriented:

f(X) → g(n__h(n__f(X)))
h(X) → n__h(X)
f(X) → n__f(X)
activate(n__h(X)) → h(activate(X))
activate(n__f(X)) → f(activate(X))
activate(X) → X

(8) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(X) → g(n__h(n__f(X)))
h(X) → n__h(X)
f(X) → n__f(X)
activate(n__h(X)) → h(activate(X))
activate(n__f(X)) → f(activate(X))
activate(X) → X

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

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE