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:

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c
activate(n__b) → b
activate(n__c) → c
activate(X) → X

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c
activate(n__b) → b
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:

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c
activate(n__b) → b
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:

activate(n__b) → b
activate(n__c) → c
activate(X) → X
Used ordering:
Polynomial interpretation [25]:

POL(activate(x1)) = 2 + 2·x1   
POL(b) = 1   
POL(c) = 1   
POL(f(x1, x2, x3)) = x1 + 2·x2 + x3   
POL(n__b) = 1   
POL(n__c) = 1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c

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:

C → B
F(n__b, X, n__c) → F(X, c, X)
F(n__b, X, n__c) → C

The TRS R consists of the following rules:

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c

Q is empty.
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:

C → B
F(n__b, X, n__c) → F(X, c, X)
F(n__b, X, n__c) → C

The TRS R consists of the following rules:

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c

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 2 less nodes.

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesReductionPairsProof

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

F(n__b, X, n__c) → F(X, c, X)

The TRS R consists of the following rules:

f(n__b, X, n__c) → f(X, c, X)
c → b
b → n__b
c → n__c

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(F(x1, x2, x3)) = x1 + 2·x2 + x3   
POL(b) = 0   
POL(c) = 0   
POL(n__b) = 0   
POL(n__c) = 0   



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesReductionPairsProof
                ↳ QDP
                  ↳ NonTerminationProof

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

F(n__b, X, n__c) → F(X, c, X)

The TRS R consists of the following rules:

c → b
c → n__c
b → n__b

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__b, X, n__c) → F(X, c, X)

The TRS R consists of the following rules:

c → b
c → n__c
b → n__b


s = F(c, X, c) evaluates to t =F(X, c, X)

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



Rewriting sequence

F(c, c, c) → F(c, c, n__c)
with rule c → n__c at position [2] and matcher [ ]

F(c, c, n__c) → F(b, c, n__c)
with rule c → b at position [0] and matcher [ ]

F(b, c, n__c) → F(n__b, c, n__c)
with rule b → n__b at position [0] and matcher [ ]

F(n__b, c, n__c) → F(c, c, c)
with rule F(n__b, X, n__c) → F(X, c, 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.