Termination w.r.t. Q proof of TRS/TRCSREx24_GM04

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(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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:

ACTIVE₁(g₁(X)) -> ACTIVE₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
ACTIVE₁(f₃(X, g₁(X), Y)) -> F₃(Y, Y, Y)
PROPER₁(g₁(X)) -> PROPER₁(X)
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X3)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
G₁(mark₁(X)) -> G₁(X)
ACTIVE₁(g₁(X)) -> G₁(active₁(X))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X2)
G₁(ok₁(X)) -> G₁(X)
PROPER₁(f₃(X1, X2, X3)) -> F₃(proper₁(X1), proper₁(X2), proper₁(X3))
F₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> F₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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:

ACTIVE₁(g₁(X)) -> ACTIVE₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
ACTIVE₁(f₃(X, g₁(X), Y)) -> F₃(Y, Y, Y)
PROPER₁(g₁(X)) -> PROPER₁(X)
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X3)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
G₁(mark₁(X)) -> G₁(X)
ACTIVE₁(g₁(X)) -> G₁(active₁(X))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X2)
G₁(ok₁(X)) -> G₁(X)
PROPER₁(f₃(X1, X2, X3)) -> F₃(proper₁(X1), proper₁(X2), proper₁(X3))
F₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> F₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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 5 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> F₃(X1, X2, X3)

The TRS R consists of the following rules:

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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.

F₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> F₃(X1, X2, X3)

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

POL(F₃(x₁, x₂, x₃)) = x₁·x₂·x₃ + x₂·x₃
POL(ok₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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

          ↳ QDP

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

G₁(mark₁(X)) -> G₁(X)
G₁(ok₁(X)) -> G₁(X)

The TRS R consists of the following rules:

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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(mark₁(X)) -> G₁(X)
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(mark₁(x₁)) = 3 + x₁
POL(ok₁(x₁)) = 3 + 3·x₁ + 3·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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

          ↳ QDP

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

PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(g₁(X)) -> PROPER₁(X)
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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(f₃(X1, X2, X3)) -> PROPER₁(X1)
PROPER₁(g₁(X)) -> PROPER₁(X)
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X3)
PROPER₁(f₃(X1, X2, X3)) -> PROPER₁(X2)

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

POL(PROPER₁(x₁)) = 3·x₁ + 3·x₁²
POL(f₃(x₁, x₂, x₃)) = 3 + x₁ + 2·x₂ + 2·x₃
POL(g₁(x₁)) = 3 + 3·x₁ + 3·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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

          ↳ QDP

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

ACTIVE₁(g₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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.

ACTIVE₁(g₁(X)) -> ACTIVE₁(X)

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

POL(ACTIVE₁(x₁)) = 3·x₁ + 3·x₁²
POL(g₁(x₁)) = 2 + 3·x₁ + 3·x₁²

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

active₁(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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

          ↳ 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(f₃(X, g₁(X), Y)) -> mark₁(f₃(Y, Y, Y))
active₁(g₁(b)) -> mark₁(c)
active₁(b) -> mark₁(c)
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₃(X1, X2, X3)) -> f₃(proper₁(X1), proper₁(X2), proper₁(X3))
proper₁(g₁(X)) -> g₁(proper₁(X))
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
f₃(ok₁(X1), ok₁(X2), ok₁(X3)) -> ok₁(f₃(X1, X2, X3))
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.