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(X)) → c(n__f(g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
d(X) → n__d(X)
activate(n__f(X)) → f(X)
activate(n__d(X)) → d(X)
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(f(X)) → c(n__f(g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
d(X) → n__d(X)
activate(n__f(X)) → f(X)
activate(n__d(X)) → d(X)
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:

ACTIVATE(n__f(X)) → F(X)
F(f(X)) → C(n__f(g(n__f(X))))
ACTIVATE(n__d(X)) → D(X)
H(X) → C(n__d(X))
C(X) → ACTIVATE(X)
C(X) → D(activate(X))

The TRS R consists of the following rules:

f(f(X)) → c(n__f(g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
d(X) → n__d(X)
activate(n__f(X)) → f(X)
activate(n__d(X)) → d(X)
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:

ACTIVATE(n__f(X)) → F(X)
F(f(X)) → C(n__f(g(n__f(X))))
ACTIVATE(n__d(X)) → D(X)
H(X) → C(n__d(X))
C(X) → ACTIVATE(X)
C(X) → D(activate(X))

The TRS R consists of the following rules:

f(f(X)) → c(n__f(g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
d(X) → n__d(X)
activate(n__f(X)) → f(X)
activate(n__d(X)) → d(X)
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:

F(f(X)) → C(n__f(g(n__f(X))))
ACTIVATE(n__f(X)) → F(X)
C(X) → ACTIVATE(X)

The TRS R consists of the following rules:

f(f(X)) → c(n__f(g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
d(X) → n__d(X)
activate(n__f(X)) → f(X)
activate(n__d(X)) → d(X)
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.


F(f(X)) → C(n__f(g(n__f(X))))
The remaining pairs can at least be oriented weakly.

ACTIVATE(n__f(X)) → F(X)
C(X) → ACTIVATE(X)
Used ordering: Polynomial interpretation [25,35]:

POL(C(x1)) = (4)x_1   
POL(f(x1)) = 1/4   
POL(g(x1)) = 0   
POL(n__f(x1)) = (4)x_1   
POL(ACTIVATE(x1)) = (1/2)x_1   
POL(F(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



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

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

ACTIVATE(n__f(X)) → F(X)
C(X) → ACTIVATE(X)

The TRS R consists of the following rules:

f(f(X)) → c(n__f(g(n__f(X))))
c(X) → d(activate(X))
h(X) → c(n__d(X))
f(X) → n__f(X)
d(X) → n__d(X)
activate(n__f(X)) → f(X)
activate(n__d(X)) → d(X)
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 0 SCCs with 2 less nodes.