Termination w.r.t. Q proof of TRS/TRCSREx24_GM04

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, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVE(g(X)) → ACTIVE(X)
TOP(mark(X)) → PROPER(X)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(g(X)) → G(proper(X))
ACTIVE(f(X, g(X), Y)) → F(Y, Y, Y)
PROPER(g(X)) → PROPER(X)
PROPER(f(X1, X2, X3)) → PROPER(X3)
TOP(ok(X)) → ACTIVE(X)
G(mark(X)) → G(X)
ACTIVE(g(X)) → G(active(X))
TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))
PROPER(f(X1, X2, X3)) → PROPER(X2)
G(ok(X)) → G(X)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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

      ↳ EdgeDeletionProof

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

ACTIVE(g(X)) → ACTIVE(X)
TOP(mark(X)) → PROPER(X)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(g(X)) → G(proper(X))
ACTIVE(f(X, g(X), Y)) → F(Y, Y, Y)
PROPER(g(X)) → PROPER(X)
PROPER(f(X1, X2, X3)) → PROPER(X3)
TOP(ok(X)) → ACTIVE(X)
G(mark(X)) → G(X)
ACTIVE(g(X)) → G(active(X))
TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))
PROPER(f(X1, X2, X3)) → PROPER(X2)
G(ok(X)) → G(X)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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 deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

ACTIVE(g(X)) → ACTIVE(X)
PROPER(f(X1, X2, X3)) → PROPER(X1)
TOP(mark(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
ACTIVE(f(X, g(X), Y)) → F(Y, Y, Y)
PROPER(f(X1, X2, X3)) → PROPER(X3)
TOP(ok(X)) → ACTIVE(X)
G(mark(X)) → G(X)
ACTIVE(g(X)) → G(active(X))
TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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 [13,14,18] contains 5 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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 [13].

The following pairs can be oriented strictly and are deleted.

F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3) = F(x2)
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

ok₁ > F₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

G(mark(X)) → G(X)
G(ok(X)) → G(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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 [13].

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(ok(X)) → G(X)

Used ordering: Combined order from the following AFS and order.
G(x1) = G(x1)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic Path Order [19].
Precedence:

[G₁, mark_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

G(ok(X)) → G(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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 [13].

The following pairs can be oriented strictly and are deleted.

G(ok(X)) → G(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1) = G(x1)
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

[G₁, ok_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(g(X)) → PROPER(X)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(f(X1, X2, X3)) → PROPER(X2)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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 [13].

The following pairs can be oriented strictly and are deleted.

PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(f(X1, X2, X3)) → PROPER(X2)

The remaining pairs can at least be oriented weakly.

PROPER(g(X)) → PROPER(X)

Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
f(x1, x2, x3) = f(x1, x2, x3)
g(x1) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(g(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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 [13].

The following pairs can be oriented strictly and are deleted.

PROPER(g(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
g(x1) = g(x1)

Lexicographic Path Order [19].
Precedence:

[PROPER₁, g_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(g(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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 [13].

The following pairs can be oriented strictly and are deleted.

ACTIVE(g(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
g(x1) = g(x1)

Lexicographic Path Order [19].
Precedence:

[ACTIVE₁, g_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

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

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
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.