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
            ↳ UsableRulesProof
          ↳ 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 can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

ACTIVATE(n__g(X)) → ACTIVATE(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

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



↳ 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.