Termination w.r.t. Q proof of TRS/TRCSREx1_Zan97

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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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:

ACTIVE(g(X)) → H(X)
TOP(mark(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
ACTIVE(h(d)) → G(c)
PROPER(h(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
PROPER(h(X)) → H(proper(X))
TOP(ok(X)) → ACTIVE(X)
H(ok(X)) → H(X)

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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:

ACTIVE(g(X)) → H(X)
TOP(mark(X)) → PROPER(X)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
ACTIVE(h(d)) → G(c)
PROPER(h(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)
PROPER(h(X)) → H(proper(X))
TOP(ok(X)) → ACTIVE(X)
H(ok(X)) → H(X)

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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(h(d)) → G(c)
PROPER(g(X)) → PROPER(X)
PROPER(h(X)) → H(proper(X))
TOP(ok(X)) → ACTIVE(X)
H(ok(X)) → H(X)
ACTIVE(g(X)) → H(X)
PROPER(h(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))
G(ok(X)) → G(X)

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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 4 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

H(ok(X)) → H(X)

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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.
none
Used ordering: Combined order from the following AFS and order.
H(x1) = H(x1)
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

ok₁ > H₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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

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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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) = G(x1)
ok(x1) = ok(x1)

Lexicographic Path Order [19].
Precedence:

ok₁ > G₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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

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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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.

PROPER(h(X)) → PROPER(X)

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

Lexicographic Path Order [19].
Precedence:

g₁ > PROPER₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

PROPER(h(X)) → PROPER(X)

The TRS R consists of the following rules:

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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.
none
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
h(x1) = h(x1)

Lexicographic Path Order [19].
Precedence:

h₁ > PROPER₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

              ↳ QDP

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

active(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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

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(g(X)) → mark(h(X))
active(c) → mark(d)
active(h(d)) → mark(g(c))
proper(g(X)) → g(proper(X))
proper(h(X)) → h(proper(X))
proper(c) → ok(c)
proper(d) → ok(d)
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.