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(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b

Q is empty.

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

a__f(X, X) → a__f(a, b)
a__b → a
mark(f(X1, X2)) → a__f(mark(X1), X2)
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b

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

POL(a) = 0   
POL(a__b) = 0   
POL(a__f(x1, x2)) = 1 + 2·x1 + x2   
POL(b) = 0   
POL(f(x1, x2)) = 1 + 2·x1 + x2   
POL(mark(x1)) = 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

a__f(X, X) → a__f(a, b)
a__b → a
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b

Q is empty.

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

a__f(X, X) → a__f(a, b)
a__b → a
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b

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__b → a
mark(b) → a__b
mark(a) → a
a__f(X1, X2) → f(X1, X2)
a__b → b
Used ordering:
Polynomial interpretation [25]:

POL(a) = 0   
POL(a__b) = 1   
POL(a__f(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(b) = 0   
POL(f(x1, x2)) = 1 + x1 + x2   
POL(mark(x1)) = 2 + x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost

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

a__f(X, X) → a__f(a, b)

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

a__f(X, X) → a__f(a, b)

The signature Sigma is {a__f}

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof

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

a__f(X, X) → a__f(a, b)

The set Q consists of the following terms:

a__f(x0, x0)


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(X, X) → A__F(a, b)

The TRS R consists of the following rules:

a__f(X, X) → a__f(a, b)

The set Q consists of the following terms:

a__f(x0, x0)

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof

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

A__F(X, X) → A__F(a, b)

The TRS R consists of the following rules:

a__f(X, X) → a__f(a, b)

The set Q consists of the following terms:

a__f(x0, x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.