Termination w.r.t. Q proof of /tmp/tmpyfNBr7/Ex15_Luc06

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:

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

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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:

F(mark(X)) → F(X)
MARK(a) → ACTIVE(a)
MARK(f(X)) → ACTIVE(f(X))
ACTIVE(f(f(a))) → G(f(a))
ACTIVE(f(f(a))) → MARK(f(g(f(a))))
MARK(g(X)) → ACTIVE(g(mark(X)))
G(active(X)) → G(X)
G(mark(X)) → G(X)
ACTIVE(f(f(a))) → F(g(f(a)))
F(active(X)) → F(X)
MARK(g(X)) → MARK(X)
MARK(g(X)) → G(mark(X))

The TRS R consists of the following rules:

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

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:

F(mark(X)) → F(X)
MARK(a) → ACTIVE(a)
MARK(f(X)) → ACTIVE(f(X))
ACTIVE(f(f(a))) → G(f(a))
ACTIVE(f(f(a))) → MARK(f(g(f(a))))
MARK(g(X)) → ACTIVE(g(mark(X)))
G(active(X)) → G(X)
G(mark(X)) → G(X)
ACTIVE(f(f(a))) → F(g(f(a)))
F(active(X)) → F(X)
MARK(g(X)) → MARK(X)
MARK(g(X)) → G(mark(X))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

G(active(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

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

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.

G(mark(X)) → G(X)

The remaining pairs can at least be oriented weakly.

G(active(X)) → G(X)

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = (2)x_1
POL(mark(x₁)) = 4 + (2)x_1
POL(G(x₁)) = (2)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

G(active(X)) → G(X)

The TRS R consists of the following rules:

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

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.

G(active(X)) → G(X)

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

POL(active(x₁)) = 4 + (4)x_1
POL(G(x₁)) = (2)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

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

          ↳ QDP

          ↳ QDP

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

F(mark(X)) → F(X)
F(active(X)) → F(X)

The TRS R consists of the following rules:

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

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(X)) → F(X)

The remaining pairs can at least be oriented weakly.

F(mark(X)) → F(X)

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 4 + (2)x_1
POL(mark(x₁)) = (2)x_1
POL(F(x₁)) = (2)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

F(mark(X)) → F(X)

The TRS R consists of the following rules:

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

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(mark(X)) → F(X)

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

POL(mark(x₁)) = 4 + (4)x_1
POL(F(x₁)) = (2)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

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

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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

MARK(f(X)) → ACTIVE(f(X))
ACTIVE(f(f(a))) → MARK(f(g(f(a))))

The TRS R consists of the following rules:

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

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.

ACTIVE(f(f(a))) → MARK(f(g(f(a))))

The remaining pairs can at least be oriented weakly.

MARK(f(X)) → ACTIVE(f(X))

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = (4)x_1
POL(MARK(x₁)) = (1/4)x_1
POL(f(x₁)) = (4)x_1
POL(a) = 1/2
POL(g(x₁)) = 0
POL(mark(x₁)) = 1 + (2)x_1
POL(ACTIVE(x₁)) = (1/4)x_1

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

f(mark(X)) → f(X)
f(active(X)) → f(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

MARK(f(X)) → ACTIVE(f(X))

The TRS R consists of the following rules:

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

Q is empty.
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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MARK(g(X)) → MARK(X)

The TRS R consists of the following rules:

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

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(g(X)) → MARK(X)

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

POL(MARK(x₁)) = (2)x_1
POL(g(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

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

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.