Termination w.r.t. Q proof of /tmp/tmpDGKCus/Ex9_Luc04

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

MARK(f(X1, X2, X3)) → MARK(X3)
A__F(a, b, X) → MARK(X)
MARK(c) → A__C
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3))
A__F(a, b, X) → A__F(mark(X), X, mark(X))

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

MARK(f(X1, X2, X3)) → MARK(X3)
A__F(a, b, X) → MARK(X)
MARK(c) → A__C
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3))
A__F(a, b, X) → A__F(mark(X), X, mark(X))

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

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

MARK(f(X1, X2, X3)) → MARK(X3)
A__F(a, b, X) → MARK(X)
MARK(f(X1, X2, X3)) → MARK(X1)
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MARK(f(X1, X2, X3)) → A__F(mark(X1), X2, mark(X3)) at position [0] we obtained the following new rules:

MARK(f(a, y1, y2)) → A__F(a, y1, mark(y2))
MARK(f(f(x0, x1, x2), y1, y2)) → A__F(a__f(mark(x0), x1, mark(x2)), y1, mark(y2))
MARK(f(c, y1, y2)) → A__F(a__c, y1, mark(y2))
MARK(f(b, y1, y2)) → A__F(b, y1, mark(y2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

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

MARK(f(X1, X2, X3)) → MARK(X3)
MARK(f(f(x0, x1, x2), y1, y2)) → A__F(a__f(mark(x0), x1, mark(x2)), y1, mark(y2))
MARK(f(a, y1, y2)) → A__F(a, y1, mark(y2))
A__F(a, b, X) → MARK(X)
MARK(f(b, y1, y2)) → A__F(b, y1, mark(y2))
MARK(f(c, y1, y2)) → A__F(a__c, y1, mark(y2))
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

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

MARK(f(X1, X2, X3)) → MARK(X3)
MARK(f(f(x0, x1, x2), y1, y2)) → A__F(a__f(mark(x0), x1, mark(x2)), y1, mark(y2))
MARK(f(a, y1, y2)) → A__F(a, y1, mark(y2))
A__F(a, b, X) → MARK(X)
MARK(f(c, y1, y2)) → A__F(a__c, y1, mark(y2))
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(X1, X2, X3)) → MARK(X1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MARK(f(c, y1, y2)) → A__F(a__c, y1, mark(y2)) at position [0] we obtained the following new rules:

MARK(f(c, y0, y1)) → A__F(b, y0, mark(y1))
MARK(f(c, y0, y1)) → A__F(c, y0, mark(y1))
MARK(f(c, y0, y1)) → A__F(a, y0, mark(y1))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ DependencyGraphProof

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

MARK(f(c, y0, y1)) → A__F(b, y0, mark(y1))
MARK(f(X1, X2, X3)) → MARK(X3)
MARK(f(a, y1, y2)) → A__F(a, y1, mark(y2))
MARK(f(f(x0, x1, x2), y1, y2)) → A__F(a__f(mark(x0), x1, mark(x2)), y1, mark(y2))
A__F(a, b, X) → MARK(X)
MARK(f(c, y0, y1)) → A__F(c, y0, mark(y1))
MARK(f(X1, X2, X3)) → MARK(X1)
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(c, y0, y1)) → A__F(a, y0, mark(y1))

The TRS R consists of the following rules:

a__f(a, b, X) → a__f(mark(X), X, mark(X))
a__c → a
a__c → b
mark(f(X1, X2, X3)) → a__f(mark(X1), X2, mark(X3))
mark(c) → a__c
mark(a) → a
mark(b) → b
a__f(X1, X2, X3) → f(X1, X2, X3)
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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

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

MARK(f(X1, X2, X3)) → MARK(X3)
MARK(f(f(x0, x1, x2), y1, y2)) → A__F(a__f(mark(x0), x1, mark(x2)), y1, mark(y2))
MARK(f(a, y1, y2)) → A__F(a, y1, mark(y2))
A__F(a, b, X) → MARK(X)
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(c, y0, y1)) → A__F(a, y0, mark(y1))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(f(f(x0, x1, x2), y1, y2)) → A__F(a__f(mark(x0), x1, mark(x2)), y1, mark(y2))

The remaining pairs can at least be oriented weakly.

MARK(f(X1, X2, X3)) → MARK(X3)
MARK(f(a, y1, y2)) → A__F(a, y1, mark(y2))
A__F(a, b, X) → MARK(X)
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(X1, X2, X3)) → MARK(X1)
MARK(f(c, y0, y1)) → A__F(a, y0, mark(y1))

Used ordering: Polynomial interpretation with max and min functions [25]:

POL(A__F(x₁, x₂, x₃)) = x₁
POL(MARK(x₁)) = 1
POL(a) = 1
POL(a__c) = 1
POL(a__f(x₁, x₂, x₃)) = 0
POL(b) = 0
POL(c) = 0
POL(f(x₁, x₂, x₃)) = 0
POL(mark(x₁)) = 1

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ Narrowing

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ Narrowing

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ QDP

                          ↳ QDPOrderProof

                            ↳ QDP

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

MARK(f(X1, X2, X3)) → MARK(X3)
MARK(f(a, y1, y2)) → A__F(a, y1, mark(y2))
A__F(a, b, X) → MARK(X)
MARK(f(X1, X2, X3)) → MARK(X1)
A__F(a, b, X) → A__F(mark(X), X, mark(X))
MARK(f(c, y0, y1)) → A__F(a, y0, mark(y1))

The TRS R consists of the following rules:

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

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