Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(f(a)) → F(g(n__f(n__a)))
ACTIVATE(n__a) → A
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F(f(a)) → F(g(n__f(n__a)))
ACTIVATE(n__a) → A
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n__f(x1)) = 4 + (4)x_1   
POL(ACTIVATE(x1)) = (2)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.