Termination w.r.t. Q proof of TRS/TRCSREx6_GM04

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:

active₁(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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:

TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
PROPER₁(g₁(X)) -> PROPER₁(X)
PROPER₁(f₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(c) -> F₁(g₁(c))
F₁(ok₁(X)) -> F₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
G₁(ok₁(X)) -> G₁(X)
ACTIVE₁(c) -> G₁(c)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
PROPER₁(f₁(X)) -> F₁(proper₁(X))

The TRS R consists of the following rules:

active₁(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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:

TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
PROPER₁(g₁(X)) -> PROPER₁(X)
PROPER₁(f₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
ACTIVE₁(c) -> F₁(g₁(c))
F₁(ok₁(X)) -> F₁(X)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
G₁(ok₁(X)) -> G₁(X)
ACTIVE₁(c) -> G₁(c)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
PROPER₁(f₁(X)) -> F₁(proper₁(X))

The TRS R consists of the following rules:

active₁(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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:

G₁(ok₁(X)) -> G₁(X)

The TRS R consists of the following rules:

active₁(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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 + 3·x₁²
POL(ok₁(x₁)) = 2 + 3·x₁ + 3·x₁²

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(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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:

F₁(ok₁(X)) -> F₁(X)

The TRS R consists of the following rules:

active₁(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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.

F₁(ok₁(X)) -> F₁(X)

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

POL(F₁(x₁)) = 3·x₁ + 3·x₁²
POL(ok₁(x₁)) = 2 + 3·x₁ + 3·x₁²

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(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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(f₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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(f₁(X)) -> PROPER₁(X)

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

POL(PROPER₁(x₁)) = 3·x₁ + 3·x₁²
POL(f₁(x₁)) = 3 + 3·x₁ + 3·x₁²
POL(g₁(x₁)) = 3 + x₁ + 2·x₁²

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(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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

            ↳ QDPOrderProof

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(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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.

TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The remaining pairs can at least be oriented weakly.

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

Used ordering: Polynomial interpretation [21]:

POL(TOP₁(x₁)) = 3·x₁
POL(active₁(x₁)) = x₁²
POL(c) = 3
POL(f₁(x₁)) = 1
POL(g₁(x₁)) = 0
POL(mark₁(x₁)) = 1 + 3·x₁
POL(ok₁(x₁)) = x₁²
POL(proper₁(x₁)) = 3·x₁

The following usable rules [14] were oriented:

active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(f₁(X)) -> f₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
g₁(ok₁(X)) -> ok₁(g₁(X))
proper₁(c) -> ok₁(c)
active₁(c) -> mark₁(f₁(g₁(c)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

The TRS R consists of the following rules:

active₁(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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.

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

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

POL(TOP₁(x₁)) = 2·x₁ + 2·x₁²
POL(active₁(x₁)) = 0
POL(c) = 0
POL(f₁(x₁)) = 0
POL(g₁(x₁)) = 2·x₁ + 2·x₁²
POL(mark₁(x₁)) = 0
POL(ok₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented:

active₁(f₁(g₁(X))) -> mark₁(g₁(X))
active₁(c) -> mark₁(f₁(g₁(c)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

active₁(c) -> mark₁(f₁(g₁(c)))
active₁(f₁(g₁(X))) -> mark₁(g₁(X))
proper₁(c) -> ok₁(c)
proper₁(f₁(X)) -> f₁(proper₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₁(ok₁(X)) -> ok₁(f₁(X))
g₁(ok₁(X)) -> ok₁(g₁(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.