Termination w.r.t. Q proof of /tmp/tmpCIkep2/Ex14_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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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:

F(active(X1), X2) → F(X1, X2)
ACTIVE(h(X)) → G(X, X)
H(active(X)) → H(X)
G(mark(X1), X2) → G(X1, X2)
MARK(g(X1, X2)) → G(mark(X1), X2)
MARK(b) → ACTIVE(b)
ACTIVE(f(X, X)) → H(a)
ACTIVE(g(a, X)) → F(b, X)
ACTIVE(h(X)) → MARK(g(X, X))
MARK(f(X1, X2)) → F(mark(X1), X2)
ACTIVE(a) → MARK(b)
F(X1, active(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
MARK(h(X)) → ACTIVE(h(mark(X)))
F(X1, mark(X2)) → F(X1, X2)
ACTIVE(f(X, X)) → MARK(h(a))
MARK(g(X1, X2)) → MARK(X1)
G(active(X1), X2) → G(X1, X2)
MARK(g(X1, X2)) → ACTIVE(g(mark(X1), X2))
MARK(a) → ACTIVE(a)
MARK(f(X1, X2)) → MARK(X1)
G(X1, mark(X2)) → G(X1, X2)
G(X1, active(X2)) → G(X1, X2)
MARK(h(X)) → MARK(X)
ACTIVE(g(a, X)) → MARK(f(b, X))
H(mark(X)) → H(X)
MARK(h(X)) → H(mark(X))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

The TRS R consists of the following rules:

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

F(active(X1), X2) → F(X1, X2)
ACTIVE(h(X)) → G(X, X)
H(active(X)) → H(X)
G(mark(X1), X2) → G(X1, X2)
MARK(g(X1, X2)) → G(mark(X1), X2)
MARK(b) → ACTIVE(b)
ACTIVE(f(X, X)) → H(a)
ACTIVE(g(a, X)) → F(b, X)
ACTIVE(h(X)) → MARK(g(X, X))
MARK(f(X1, X2)) → F(mark(X1), X2)
ACTIVE(a) → MARK(b)
F(X1, active(X2)) → F(X1, X2)
F(mark(X1), X2) → F(X1, X2)
MARK(h(X)) → ACTIVE(h(mark(X)))
F(X1, mark(X2)) → F(X1, X2)
ACTIVE(f(X, X)) → MARK(h(a))
MARK(g(X1, X2)) → MARK(X1)
G(active(X1), X2) → G(X1, X2)
MARK(g(X1, X2)) → ACTIVE(g(mark(X1), X2))
MARK(a) → ACTIVE(a)
MARK(f(X1, X2)) → MARK(X1)
G(X1, mark(X2)) → G(X1, X2)
G(X1, active(X2)) → G(X1, X2)
MARK(h(X)) → MARK(X)
ACTIVE(g(a, X)) → MARK(f(b, X))
H(mark(X)) → H(X)
MARK(h(X)) → H(mark(X))
MARK(f(X1, X2)) → ACTIVE(f(mark(X1), X2))

The TRS R consists of the following rules:

active(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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 4 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ 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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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)) = 1/4 + (2)x_1
POL(mark(x₁)) = 13/4 + (2)x_1
POL(F(x₁, x₂)) = (1/2)x_1 + (1/4)x_2

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

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

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 4 + (4)x_1
POL(mark(x₁)) = 1/4 + (3/2)x_1
POL(G(x₁, x₂)) = (4)x_1

The value of delta used in the strict ordering is 1.
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:

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

The TRS R consists of the following rules:

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

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

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

POL(active(x₁)) = 1/2 + (3/2)x_1
POL(mark(x₁)) = 9/4 + x_1
POL(G(x₁, x₂)) = (1/2)x_2

The value of delta used in the strict ordering is 1/4.
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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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(h(X)) → mark(g(X, X))
active(g(a, X)) → mark(f(b, X))
active(f(X, X)) → mark(h(a))
active(a) → mark(b)
mark(h(X)) → active(h(mark(X)))
mark(g(X1, X2)) → active(g(mark(X1), X2))
mark(a) → active(a)
mark(f(X1, X2)) → active(f(mark(X1), X2))
mark(b) → active(b)
h(mark(X)) → h(X)
h(active(X)) → h(X)
g(mark(X1), X2) → g(X1, X2)
g(X1, mark(X2)) → g(X1, X2)
g(active(X1), X2) → g(X1, X2)
g(X1, active(X2)) → g(X1, X2)
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.

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)) = 1/2 + (3/2)x_1
POL(H(x₁)) = (1/2)x_1
POL(mark(x₁)) = 9/4 + x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

Used ordering: Polynomial interpretation [25,35]:

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

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [25,35]:

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

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

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

The TRS R consists of the following rules:

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