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

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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)



↳ 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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)


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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)

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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)


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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)

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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)


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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)

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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)


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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)

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

The set Q consists of the following terms:

a__f(x0, x1, x2)
mark(c)
mark(f(x0, x1, x2))
a__c
mark(b)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.