Termination w.r.t. Q proof of /tmp/tmpDWB5Kt/Ex6_GM04

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

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

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
c → n__c
activate(n__g(X)) → g(X)
activate(n__c) → c
activate(X) → X

The set Q consists of the following terms:

activate(x0)
f(n__g(x0))
g(x0)
c

↳ QTRS

  ↳ RRRPoloQTRSProof

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

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
c → n__c
activate(n__g(X)) → g(X)
activate(n__c) → c
activate(X) → X

The set Q consists of the following terms:

activate(x0)
f(n__g(x0))
g(x0)
c

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

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
c → n__c
activate(n__g(X)) → g(X)
activate(n__c) → c
activate(X) → X

The set Q consists of the following terms:

activate(x0)
f(n__g(x0))
g(x0)
c

The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

c → n__c
activate(n__g(X)) → g(X)
activate(X) → X

Used ordering:
Polynomial interpretation [25]:

POL(activate(x₁)) = 1 + x₁
POL(c) = 1
POL(f(x₁)) = 1 + x₁
POL(g(x₁)) = x₁
POL(n__c) = 0
POL(n__g(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

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

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
activate(n__c) → c

The set Q consists of the following terms:

activate(x0)
f(n__g(x0))
g(x0)
c

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(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))
F(n__g(X)) → G(activate(X))

The TRS R consists of the following rules:

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
activate(n__c) → c

The set Q consists of the following terms:

activate(x0)
f(n__g(x0))
g(x0)
c

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))
F(n__g(X)) → G(activate(X))

The TRS R consists of the following rules:

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
activate(n__c) → c

The set Q consists of the following terms:

activate(x0)
f(n__g(x0))
g(x0)
c

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

              ↳ UsableRulesProof

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

F(n__g(X)) → ACTIVATE(X)
C → F(n__g(n__c))
ACTIVATE(n__c) → C

The TRS R consists of the following rules:

c → f(n__g(n__c))
f(n__g(X)) → g(activate(X))
g(X) → n__g(X)
activate(n__c) → c

The set Q consists of the following terms:

activate(x0)
f(n__g(x0))
g(x0)
c

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

              ↳ UsableRulesProof

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

F(n__g(X)) → ACTIVATE(X)
C → F(n__g(n__c))
ACTIVATE(n__c) → C

R is empty.
The set Q consists of the following terms:

activate(x0)
f(n__g(x0))
g(x0)
c

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

activate(x0)
f(n__g(x0))
g(x0)
c

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

                    ↳ QDP

                      ↳ NonTerminationProof

              ↳ UsableRulesProof

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

F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

The TRS P consists of the following rules:

F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))

The TRS R consists of the following rules:none

s = ACTIVATE(n__c) evaluates to t =ACTIVATE(n__c)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

ACTIVATE(n__c) → C
with rule ACTIVATE(n__c) → C at position [] and matcher [ ]

C → F(n__g(n__c))
with rule C → F(n__g(n__c)) at position [] and matcher [ ]

F(n__g(n__c)) → ACTIVATE(n__c)
with rule F(n__g(X)) → ACTIVATE(X)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

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

F(n__g(X)) → ACTIVATE(X)
C → F(n__g(n__c))
ACTIVATE(n__c) → C

R is empty.
The set Q consists of the following terms:

activate(x0)
f(n__g(x0))
g(x0)
c

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

activate(x0)
f(n__g(x0))
g(x0)
c

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ UsableRulesProof

              ↳ UsableRulesProof

                ↳ QDP

                  ↳ QReductionProof

                    ↳ QDP

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

F(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__c) → C
C → F(n__g(n__c))

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