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:

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(c) → a__c
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(c) → a__c
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

Q is empty.

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
mark(c) → a__c
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

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

mark(f(X1, X2, X3)) → a__f(X1, mark(X2), X3)
Used ordering:
Polynomial interpretation [25]:

POL(a__c) = 0   
POL(a__f(x1, x2, x3)) = 1 + x1 + 2·x2 + x3   
POL(b) = 0   
POL(c) = 0   
POL(f(x1, x2, x3)) = 1 + x1 + 2·x2 + x3   
POL(mark(x1)) = 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(c) → a__c
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

Q is empty.

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
mark(c) → a__c
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

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

mark(c) → a__c
mark(b) → b
Used ordering:
Polynomial interpretation [25]:

POL(a__c) = 0   
POL(a__f(x1, x2, x3)) = 1 + x1 + 2·x2 + x3   
POL(b) = 0   
POL(c) = 0   
POL(f(x1, x2, x3)) = 1 + x1 + 2·x2 + x3   
POL(mark(x1)) = 1 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

Q is empty.

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__f(X1, X2, X3) → f(X1, X2, X3)
a__c → c

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

a__f(X1, X2, X3) → f(X1, X2, X3)
Used ordering:
Polynomial interpretation [25]:

POL(a__c) = 0   
POL(a__f(x1, x2, x3)) = 2 + x1 + 2·x2 + x3   
POL(b) = 0   
POL(c) = 0   
POL(f(x1, x2, x3)) = x1 + x2 + x3   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ RRRPoloQTRSProof
            ↳ QTRS
              ↳ DependencyPairsProof

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

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__c → 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:

A__F(b, X, c) → A__F(X, a__c, X)
A__F(b, X, c) → A__C

The TRS R consists of the following rules:

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__c → c

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

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

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

A__F(b, X, c) → A__F(X, a__c, X)
A__F(b, X, c) → A__C

The TRS R consists of the following rules:

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__c → 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 1 less node.

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

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

A__F(b, X, c) → A__F(X, a__c, X)

The TRS R consists of the following rules:

a__f(b, X, c) → a__f(X, a__c, X)
a__c → b
a__c → 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(A__F(x1, x2, x3)) = x1 + 2·x2 + x3   
POL(a__c) = 0   
POL(b) = 0   
POL(c) = 0   



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

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

A__F(b, X, c) → A__F(X, a__c, X)

The TRS R consists of the following rules:

a__c → b
a__c → c

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:

A__F(b, X, c) → A__F(X, a__c, X)

The TRS R consists of the following rules:

a__c → b
a__c → c


s = A__F(a__c, X, a__c) evaluates to t =A__F(X, a__c, X)

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



Rewriting sequence

A__F(a__c, a__c, a__c) → A__F(a__c, a__c, c)
with rule a__c → c at position [2] and matcher [ ]

A__F(a__c, a__c, c) → A__F(b, a__c, c)
with rule a__c → b at position [0] and matcher [ ]

A__F(b, a__c, c) → A__F(a__c, a__c, a__c)
with rule A__F(b, X, c) → A__F(X, a__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.