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

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(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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(f₃(x, y, z)) -> ACTIVE₁(z)
PROPER₁(f₃(x, y, z)) -> PROPER₁(x)
TOP₁(mark₁(x)) -> PROPER₁(x)
PROPER₁(f₃(x, y, z)) -> PROPER₁(z)
TOP₁(ok₁(x)) -> ACTIVE₁(x)
ACTIVE₁(f₃(x, y, z)) -> F₃(x, y, active₁(z))
ACTIVE₁(f₃(b, c, x)) -> F₃(x, x, x)
TOP₁(ok₁(x)) -> TOP₁(active₁(x))
TOP₁(mark₁(x)) -> TOP₁(proper₁(x))
PROPER₁(f₃(x, y, z)) -> PROPER₁(y)
PROPER₁(f₃(x, y, z)) -> F₃(proper₁(x), proper₁(y), proper₁(z))
F₃(x, y, mark₁(z)) -> F₃(x, y, z)
F₃(ok₁(x), ok₁(y), ok₁(z)) -> F₃(x, y, z)

The TRS R consists of the following rules:

active₁(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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(f₃(x, y, z)) -> ACTIVE₁(z)
PROPER₁(f₃(x, y, z)) -> PROPER₁(x)
TOP₁(mark₁(x)) -> PROPER₁(x)
PROPER₁(f₃(x, y, z)) -> PROPER₁(z)
TOP₁(ok₁(x)) -> ACTIVE₁(x)
ACTIVE₁(f₃(x, y, z)) -> F₃(x, y, active₁(z))
ACTIVE₁(f₃(b, c, x)) -> F₃(x, x, x)
TOP₁(ok₁(x)) -> TOP₁(active₁(x))
TOP₁(mark₁(x)) -> TOP₁(proper₁(x))
PROPER₁(f₃(x, y, z)) -> PROPER₁(y)
PROPER₁(f₃(x, y, z)) -> F₃(proper₁(x), proper₁(y), proper₁(z))
F₃(x, y, mark₁(z)) -> F₃(x, y, z)
F₃(ok₁(x), ok₁(y), ok₁(z)) -> F₃(x, y, z)

The TRS R consists of the following rules:

active₁(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₃(x, y, mark₁(z)) -> F₃(x, y, z)
F₃(ok₁(x), ok₁(y), ok₁(z)) -> F₃(x, y, z)

The TRS R consists of the following rules:

active₁(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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(x), ok₁(y), ok₁(z)) -> F₃(x, y, z)

The remaining pairs can at least be oriented weakly.

F₃(x, y, mark₁(z)) -> F₃(x, y, z)

Used ordering: Polynomial interpretation [21]:

POL(F₃(x₁, x₂, x₃)) = 2·x₂
POL(mark₁(x₁)) = 0
POL(ok₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₃(x, y, mark₁(z)) -> F₃(x, y, z)

The TRS R consists of the following rules:

active₁(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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(x, y, mark₁(z)) -> F₃(x, y, z)

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

POL(F₃(x₁, x₂, x₃)) = 2·x₃
POL(mark₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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:

PROPER₁(f₃(x, y, z)) -> PROPER₁(x)
PROPER₁(f₃(x, y, z)) -> PROPER₁(z)
PROPER₁(f₃(x, y, z)) -> PROPER₁(y)

The TRS R consists of the following rules:

active₁(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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(x, y, z)) -> PROPER₁(x)
PROPER₁(f₃(x, y, z)) -> PROPER₁(z)
PROPER₁(f₃(x, y, z)) -> PROPER₁(y)

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

POL(PROPER₁(x₁)) = 2·x₁
POL(f₃(x₁, x₂, x₃)) = 1 + 2·x₁ + 2·x₂ + 2·x₃

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(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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:

ACTIVE₁(f₃(x, y, z)) -> ACTIVE₁(z)

The TRS R consists of the following rules:

active₁(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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(f₃(x, y, z)) -> ACTIVE₁(z)

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

POL(ACTIVE₁(x₁)) = 2·x₁
POL(f₃(x₁, x₂, x₃)) = 1 + 2·x₃

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(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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(f₃(b, c, x)) -> mark₁(f₃(x, x, x))
active₁(f₃(x, y, z)) -> f₃(x, y, active₁(z))
active₁(d) -> m₁(b)
f₃(x, y, mark₁(z)) -> mark₁(f₃(x, y, z))
active₁(d) -> mark₁(c)
proper₁(b) -> ok₁(b)
proper₁(c) -> ok₁(c)
proper₁(d) -> ok₁(d)
proper₁(f₃(x, y, z)) -> f₃(proper₁(x), proper₁(y), proper₁(z))
f₃(ok₁(x), ok₁(y), ok₁(z)) -> ok₁(f₃(x, y, z))
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.