Termination w.r.t. Q proof of TRS/TRCSRExConc_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(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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:

TOP(mark(X)) → PROPER(X)
ACTIVE(f(X)) → ACTIVE(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X)) → F(active(X))
PROPER(h(X)) → H(proper(X))
TOP(ok(X)) → ACTIVE(X)
H(mark(X)) → H(X)
H(ok(X)) → H(X)
ACTIVE(f(X)) → H(f(X))
F(mark(X)) → F(X)
PROPER(h(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
F(ok(X)) → F(X)
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
ACTIVE(h(X)) → H(active(X))
ACTIVE(f(X)) → G(h(f(X)))
PROPER(f(X)) → F(proper(X))

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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:

TOP(mark(X)) → PROPER(X)
ACTIVE(f(X)) → ACTIVE(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X)) → F(active(X))
PROPER(h(X)) → H(proper(X))
TOP(ok(X)) → ACTIVE(X)
H(mark(X)) → H(X)
H(ok(X)) → H(X)
ACTIVE(f(X)) → H(f(X))
F(mark(X)) → F(X)
PROPER(h(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
F(ok(X)) → F(X)
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
ACTIVE(h(X)) → H(active(X))
ACTIVE(f(X)) → G(h(f(X)))
PROPER(f(X)) → F(proper(X))

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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:

TOP(mark(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
ACTIVE(f(X)) → ACTIVE(X)
PROPER(g(X)) → PROPER(X)
ACTIVE(h(X)) → ACTIVE(X)
ACTIVE(f(X)) → F(active(X))
PROPER(h(X)) → H(proper(X))
H(mark(X)) → H(X)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(f(X)) → H(f(X))
H(ok(X)) → H(X)
F(mark(X)) → F(X)
PROPER(f(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
F(ok(X)) → F(X)
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
ACTIVE(h(X)) → H(active(X))
PROPER(f(X)) → F(proper(X))
ACTIVE(f(X)) → G(h(f(X)))

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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 6 SCCs with 9 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ 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(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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)

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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

              ↳ QDP

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

H(mark(X)) → H(X)
H(ok(X)) → H(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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.

H(ok(X)) → H(X)

The remaining pairs can at least be oriented weakly.

H(mark(X)) → H(X)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

H(mark(X)) → H(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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.

H(mark(X)) → H(X)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

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

              ↳ QDP

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

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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

              ↳ QDP

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

F(mark(X)) → F(X)
F(ok(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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(X)) → F(X)

The remaining pairs can at least be oriented weakly.

F(mark(X)) → F(X)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

F(mark(X)) → F(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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(X)) → F(X)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

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

              ↳ QDP

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

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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

              ↳ QDP

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

PROPER(g(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)
PROPER(f(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.

PROPER(g(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(g(X)) → PROPER(X)
PROPER(h(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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(h(X)) → PROPER(X)

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) = x1
g(x1) = x1
h(x1) = h(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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)

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ 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)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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

              ↳ QDP

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

ACTIVE(f(X)) → ACTIVE(X)
ACTIVE(h(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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(h(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(f(X)) → ACTIVE(X)

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

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(f(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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(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) = x1
f(x1) = f(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ 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(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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

              ↳ 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(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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(mark(X)) → TOP(proper(X))

The remaining pairs can at least be oriented weakly.

TOP(ok(X)) → TOP(active(X))

Used ordering: Combined order from the following AFS and order.
TOP(x1) = x1
ok(x1) = x1
active(x1) = x1
mark(x1) = mark(x1)
proper(x1) = x1
f(x1) = f(x1)
g(x1) = g
h(x1) = x1

Lexicographic path order with status [19].
Quasi-Precedence:

f₁ > mark₁
f₁ > g

Status:

trivial

The following usable rules [14] were oriented:

proper(f(X)) → f(proper(X))
active(f(X)) → f(active(X))
f(mark(X)) → mark(f(X))
proper(g(X)) → g(proper(X))
f(ok(X)) → ok(f(X))
h(mark(X)) → mark(h(X))
active(h(X)) → h(active(X))
active(f(X)) → mark(g(h(f(X))))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(X))
proper(h(X)) → h(proper(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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))

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
active(x1) = active
f(x1) = x1
mark(x1) = mark
h(x1) = x1

Lexicographic path order with status [19].
Quasi-Precedence:

ok > active > mark

Status:

trivial

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

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

active(f(X)) → mark(g(h(f(X))))
active(f(X)) → f(active(X))
active(h(X)) → h(active(X))
f(mark(X)) → mark(f(X))
h(mark(X)) → mark(h(X))
proper(f(X)) → f(proper(X))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
f(ok(X)) → ok(f(X))
g(ok(X)) → ok(g(X))
h(ok(X)) → ok(h(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.