Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(X))
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(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__g(X)) → G(activate(X))
ACTIVATE(n__g(X)) → ACTIVATE(X)
F(X, n__g(X), Y) → ACTIVATE(Y)
F(X, n__g(X), Y) → F(activate(Y), activate(Y), activate(Y))

The TRS R consists of the following rules:

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(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__g(X)) → G(activate(X))
ACTIVATE(n__g(X)) → ACTIVATE(X)
F(X, n__g(X), Y) → ACTIVATE(Y)
F(X, n__g(X), Y) → F(activate(Y), activate(Y), activate(Y))

The TRS R consists of the following rules:

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(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 2 SCCs with 2 less nodes.

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

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(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(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.


ACTIVATE(n__g(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n__g(x1)) = 1/4 + (7/2)x_1   
POL(ACTIVATE(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
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(X))
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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

F(X, n__g(X), Y) → F(activate(Y), activate(Y), activate(Y))

The TRS R consists of the following rules:

f(X, n__g(X), Y) → f(activate(Y), activate(Y), activate(Y))
g(b) → c
b → c
g(X) → n__g(X)
activate(n__g(X)) → g(activate(X))
activate(X) → X

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