Termination w.r.t. Q proof of /tmp/tmp2zRMUi/Ex16_Luc06

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:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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(a) → ACTIVE(a)
MARK(f(X1, X2)) → MARK(X1)
F(active(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
MARK(f(X1, X2)) → F(mark(X1), X2)
ACTIVE(b) → MARK(a)
ACTIVE(f(X, X)) → F(a, b)
F(X1, active(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
MARK(b) → ACTIVE(b)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(X, X)) → MARK(f(a, b))

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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(a) → ACTIVE(a)
MARK(f(X1, X2)) → MARK(X1)
F(active(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
MARK(f(X1, X2)) → F(mark(X1), X2)
ACTIVE(b) → MARK(a)
ACTIVE(f(X, X)) → F(a, b)
F(X1, active(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
MARK(b) → ACTIVE(b)
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))
ACTIVE(f(X, X)) → MARK(f(a, b))

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

F(active(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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.

F(active(X1), X2) → F(X1, X2)
F(X1, mark(X2)) → F(X1, X2)
F(X1, active(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 7/4 + x_1
POL(mark(x₁)) = 13/4 + (2)x_1
POL(F(x₁, x₂)) = (1/4)x_1 + (1/2)x_2

The value of delta used in the strict ordering is 7/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 2 + (3/2)x_1
POL(MARK(x₁)) = 2 + (1/2)x_1
POL(a) = 0
POL(f(x₁, x₂)) = 4 + x_1
POL(mark(x₁)) = (1/4)x_1
POL(b) = 0
POL(ACTIVE(x₁)) = 4

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

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

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(a) → active(a)
mark(b) → active(b)
f(mark(X1), X2) → f(X1, X2)
f(X1, mark(X2)) → f(X1, X2)
f(active(X1), X2) → f(X1, X2)
f(X1, active(X2)) → f(X1, X2)

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