YES Termination w.r.t. Q proof of ../tpdb/TRS/TRCSR/Ex15_Luc06_GM.trs
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(f(a)) → a__f(g(f(a)))
mark(f(X)) → a__f(X)
mark(a) → a
mark(g(X)) → g(mark(X))
a__f(X) → f(X)

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

a__f(f(a)) → a__f(g(f(a)))
mark(f(X)) → a__f(X)
mark(a) → a
mark(g(X)) → g(mark(X))
a__f(X) → f(X)

Q is empty.

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

a__f(f(a)) → a__f(g(f(a)))
mark(f(X)) → a__f(X)
mark(a) → a
mark(g(X)) → g(mark(X))
a__f(X) → f(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:

mark(a) → a
a__f(X) → f(X)
Used ordering:
Polynomial interpretation [25]:

POL(a) = 2   
POL(a__f(x1)) = 2 + 2·x1   
POL(f(x1)) = 1 + 2·x1   
POL(g(x1)) = x1   
POL(mark(x1)) = 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof

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

a__f(f(a)) → a__f(g(f(a)))
mark(f(X)) → a__f(X)
mark(g(X)) → g(mark(X))

Q is empty.

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

a__f(f(a)) → a__f(g(f(a)))
mark(f(X)) → a__f(X)
mark(g(X)) → g(mark(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:

mark(f(X)) → a__f(X)
Used ordering:
Polynomial interpretation [25]:

POL(a) = 0   
POL(a__f(x1)) = 2·x1   
POL(f(x1)) = x1   
POL(g(x1)) = x1   
POL(mark(x1)) = 1 + 2·x1   




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

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

a__f(f(a)) → a__f(g(f(a)))
mark(g(X)) → g(mark(X))

Q is empty.

We have applied [19,8] to switch to innermost. The TRS R 1 is

mark(g(X)) → g(mark(X))

The TRS R 2 is

a__f(f(a)) → a__f(g(f(a)))

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(f(a)) → a__f(g(f(a)))
mark(g(X)) → g(mark(X))

The set Q consists of the following terms:

a__f(f(a))
mark(g(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(f(a)) → A__F(g(f(a)))
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(a)) → a__f(g(f(a)))
mark(g(X)) → g(mark(X))

The set Q consists of the following terms:

a__f(f(a))
mark(g(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(f(a)) → A__F(g(f(a)))
MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(a)) → a__f(g(f(a)))
mark(g(X)) → g(mark(X))

The set Q consists of the following terms:

a__f(f(a))
mark(g(x0))

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
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof

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

MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

a__f(f(a)) → a__f(g(f(a)))
mark(g(X)) → g(mark(X))

The set Q consists of the following terms:

a__f(f(a))
mark(g(x0))

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

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

MARK(g(X)) → MARK(X)

R is empty.
The set Q consists of the following terms:

a__f(f(a))
mark(g(x0))

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.

a__f(f(a))
mark(g(x0))



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ AAECC Innermost
            ↳ QTRS
              ↳ DependencyPairsProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ QDPSizeChangeProof

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

MARK(g(X)) → MARK(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: