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:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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:

ACTIVE₁(g₁(X)) -> H₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
PROPER₁(g₁(X)) -> PROPER₁(X)
ACTIVE₁(h₁(d)) -> G₁(c)
PROPER₁(h₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
G₁(ok₁(X)) -> G₁(X)
PROPER₁(h₁(X)) -> H₁(proper₁(X))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
H₁(ok₁(X)) -> H₁(X)

The TRS R consists of the following rules:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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:

ACTIVE₁(g₁(X)) -> H₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
PROPER₁(g₁(X)) -> PROPER₁(X)
ACTIVE₁(h₁(d)) -> G₁(c)
PROPER₁(h₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
G₁(ok₁(X)) -> G₁(X)
PROPER₁(h₁(X)) -> H₁(proper₁(X))
TOP₁(ok₁(X)) -> ACTIVE₁(X)
H₁(ok₁(X)) -> H₁(X)

The TRS R consists of the following rules:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

H₁(ok₁(X)) -> H₁(X)

The TRS R consists of the following rules:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

H₁(ok₁(X)) -> H₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(H₁(x₁)) = 3·x₁
POL(ok₁(x₁)) = 1 + x₁

The following usable rules [14] 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:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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:

G₁(ok₁(X)) -> G₁(X)

The TRS R consists of the following rules:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

G₁(ok₁(X)) -> G₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(G₁(x₁)) = 3·x₁
POL(ok₁(x₁)) = 1 + x₁

The following usable rules [14] 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:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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:

PROPER₁(g₁(X)) -> PROPER₁(X)
PROPER₁(h₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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.

PROPER₁(g₁(X)) -> PROPER₁(X)
PROPER₁(h₁(X)) -> PROPER₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER₁(x₁)) = 3·x₁
POL(g₁(x₁)) = 3 + x₁
POL(h₁(x₁)) = 3 + 2·x₁

The following usable rules [14] 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:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(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:

TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The TRS R consists of the following rules:

active₁(g₁(X)) -> mark₁(h₁(X))
active₁(c) -> mark₁(d)
active₁(h₁(d)) -> mark₁(g₁(c))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(h₁(X)) -> h₁(proper₁(X))
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
g₁(ok₁(X)) -> ok₁(g₁(X))
h₁(ok₁(X)) -> ok₁(h₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

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