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

Q is empty.


↳ 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

Q is empty.

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

Q is empty.
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
      ↳ Overlay + Local Confluence

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

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ 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:

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

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

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ 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:

c
f(n__g(x0))
g(x0)
activate(n__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
      ↳ Overlay + Local Confluence
        ↳ 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:

c
f(n__g(x0))
g(x0)
activate(n__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
      ↳ Overlay + Local Confluence
        ↳ 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:

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

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



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ 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:



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
      ↳ Overlay + Local Confluence
        ↳ 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:

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

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



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ Overlay + Local Confluence
        ↳ 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.