Termination w.r.t. Q proof of /tmp/tmp1xZSvo/Ex1_Zan97

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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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:

MARK(d) → ACTIVE(d)
ACTIVE(g(X)) → H(X)
H(active(X)) → H(X)
MARK(h(X)) → ACTIVE(h(X))
ACTIVE(g(X)) → MARK(h(X))
ACTIVE(h(d)) → MARK(g(c))
G(active(X)) → G(X)
ACTIVE(h(d)) → G(c)
G(mark(X)) → G(X)
MARK(c) → ACTIVE(c)
ACTIVE(c) → MARK(d)
MARK(g(X)) → ACTIVE(g(X))
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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:

MARK(d) → ACTIVE(d)
ACTIVE(g(X)) → H(X)
H(active(X)) → H(X)
MARK(h(X)) → ACTIVE(h(X))
ACTIVE(g(X)) → MARK(h(X))
ACTIVE(h(d)) → MARK(g(c))
G(active(X)) → G(X)
ACTIVE(h(d)) → G(c)
G(mark(X)) → G(X)
MARK(c) → ACTIVE(c)
ACTIVE(c) → MARK(d)
MARK(g(X)) → ACTIVE(g(X))
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

H(active(X)) → H(X)
H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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.

H(active(X)) → H(X)
H(mark(X)) → H(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(H(x₁)) = (4)x_1
POL(mark(x₁)) = 4 + x_1

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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

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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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)
G(mark(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(mark(x₁)) = 4 + x_1
POL(G(x₁)) = (4)x_1

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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

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

MARK(h(X)) → ACTIVE(h(X))
ACTIVE(g(X)) → MARK(h(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(h(d)) → MARK(g(c))

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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(h(X)) → ACTIVE(h(X))
ACTIVE(g(X)) → MARK(h(X))
MARK(g(X)) → ACTIVE(g(X))
ACTIVE(h(d)) → MARK(g(c))

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

POL(c) = 0
POL(active(x₁)) = x_1
POL(MARK(x₁)) = 4 + (3)x_1
POL(g(x₁)) = 4 + (4)x_1
POL(h(x₁)) = (2)x_1
POL(mark(x₁)) = (4)x_1
POL(d) = 4
POL(ACTIVE(x₁)) = 3 + (2)x_1

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

g(active(X)) → g(X)
g(mark(X)) → g(X)
h(active(X)) → h(X)
h(mark(X)) → h(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
mark(g(X)) → active(g(X))
mark(h(X)) → active(h(X))
mark(c) → active(c)
mark(d) → active(d)
g(mark(X)) → g(X)
g(active(X)) → g(X)
h(mark(X)) → h(X)
h(active(X)) → h(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.