Termination w.r.t. Q proof of TRS/TRCSREx4_4_Luc96b

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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 7 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:

G(mark(X)) → G(X)
G(ok(X)) → G(X)

The TRS R consists of the following rules:

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

Recursive Path Order [2].
Precedence:

mark₁ > G₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ 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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) = x1
ok(x1) = ok(x1)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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:

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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(mark(X1), X2) → F(X1, X2)

The remaining pairs can at least be oriented weakly.

F(ok(X1), ok(X2)) → F(X1, X2)

Used ordering: Combined order from the following AFS and order.
F(x1, x2) = F(x1, x2)
mark(x1) = mark(x1)
ok(x1) = x1

Recursive Path Order [2].
Precedence:

trivial

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:

F(ok(X1), ok(X2)) → F(X1, X2)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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)) → F(X1, X2)

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

Recursive Path Order [2].
Precedence:

ok₁ > F₁

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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(g(X)) → PROPER(X)
PROPER(f(X1, X2)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)

The TRS R consists of the following rules:

active(f(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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)) → PROPER(X2)
PROPER(f(X1, X2)) → PROPER(X1)

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)
g(x1) = x1
f(x1, x2) = f(x1, x2)

Recursive Path Order [2].
Precedence:

f₂ > PROPER₁

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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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) = x1
g(x1) = g(x1)

Recursive Path Order [2].
Precedence:

trivial

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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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)
ACTIVE(f(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

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

ACTIVE(f(X1, X2)) → ACTIVE(X1)

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

Recursive Path Order [2].
Precedence:

g₁ > ACTIVE₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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(g(X), Y)) → mark(f(X, f(g(X), Y)))
active(f(X1, X2)) → f(active(X1), X2)
active(g(X)) → g(active(X))
f(mark(X1), X2) → mark(f(X1, X2))
g(mark(X)) → mark(g(X))
proper(f(X1, X2)) → f(proper(X1), proper(X2))
proper(g(X)) → g(proper(X))
f(ok(X1), ok(X2)) → ok(f(X1, X2))
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(f(X1, X2)) → ACTIVE(X1)

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

Recursive Path Order [2].
Precedence:

f₂ > ACTIVE₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

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

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

                ↳ QDPOrderProof

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

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TOP(x1) = x1
ok(x1) = ok(x1)
active(x1) = x1
mark(x1) = mark(x1)
proper(x1) = x1
g(x1) = g(x1)
f(x1, x2) = x1

Recursive Path Order [2].
Precedence:

g₁ > ok₁ > mark₁

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

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