Termination w.r.t. Q proof of /tmp/tmp7aXK5t/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)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

PROPER(f(X1, X2)) → PROPER(X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X1, X2)) → F(active(X1), X2)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(f(X, X)) → F(a, b)
F(mark(X1), X2) → F(X1, X2)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
F(ok(X1), ok(X2)) → F(X1, X2)
PROPER(f(X1, X2)) → F(proper(X1), proper(X2))
TOP(ok(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
PROPER(f(X1, X2)) → PROPER(X1)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

PROPER(f(X1, X2)) → PROPER(X2)
TOP(mark(X)) → PROPER(X)
ACTIVE(f(X1, X2)) → F(active(X1), X2)
TOP(mark(X)) → TOP(proper(X))
ACTIVE(f(X, X)) → F(a, b)
F(mark(X1), X2) → F(X1, X2)
ACTIVE(f(X1, X2)) → ACTIVE(X1)
F(ok(X1), ok(X2)) → F(X1, X2)
PROPER(f(X1, X2)) → F(proper(X1), proper(X2))
TOP(ok(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
PROPER(f(X1, X2)) → PROPER(X1)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

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

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(X1), X2) → F(X1, X2)
F(ok(X1), ok(X2)) → F(X1, X2)

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

POL(mark(x₁)) = 9/4 + (2)x_1
POL(ok(x₁)) = 4 + x_1
POL(F(x₁, x₂)) = (7/2)x_1

The value of delta used in the strict ordering is 63/8.
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(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

PROPER(f(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

PROPER(f(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)

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

POL(PROPER(x₁)) = (2)x_1
POL(f(x₁, x₂)) = 1/4 + (5/2)x_1 + (5/2)x_2

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ 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)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(f(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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(X1, X2)) → ACTIVE(X1)

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

POL(f(x₁, x₂)) = 1/4 + (7/2)x_1
POL(ACTIVE(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
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(f(X, X)) → mark(f(a, b))
active(b) → mark(a)
active(f(X1, X2)) → f(active(X1), X2)
f(mark(X1), X2) → mark(f(X1, X2))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(a) → ok(a)
proper(b) → ok(b)
f(ok(X1), ok(X2)) → ok(f(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

          ↳ QDP

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

TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

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

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