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 Order [17,21] with Interpretation:

POL( G₁(x₁) ) = 2x₁ + 3

POL( ok₁(x₁) ) = 3x₁ + 3

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 Order [17,21] with Interpretation:

POL( F₁(x₁) ) = 2x₁ + 3

POL( ok₁(x₁) ) = 3x₁ + 3

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 Order [17,21] with Interpretation:

POL( g₁(x₁) ) = 3x₁ + 1

POL( PROPER₁(x₁) ) = 2x₁

POL( f₁(x₁) ) = 2x₁ + 2

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 Order [17,21] with Interpretation:

POL( mark₁(x₁) ) = x₁ + 1

POL( active₁(x₁) ) = x₁

POL( f₁(x₁) ) = 2x₁

POL( c ) = 3

POL( TOP₁(x₁) ) = x₁

POL( ok₁(x₁) ) = x₁

POL( g₁(x₁) ) = 1

POL( proper₁(x₁) ) = x₁

The following usable rules [14] were oriented:

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

↳ 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 Order [17,21] with Interpretation:

POL( mark₁(x₁) ) = max{0, -3}

POL( active₁(x₁) ) = 1

POL( f₁(x₁) ) = 3

POL( c ) = 1

POL( TOP₁(x₁) ) = max{0, 2x₁ - 3}

POL( ok₁(x₁) ) = 3x₁ + 2

POL( g₁(x₁) ) = 1

The following usable rules [14] were oriented:

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

↳ 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.