Termination w.r.t. Q proof of TRS/TRCSREx4_4_Luc96b

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(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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(f₂(g₁(X), Y)) -> F₂(X, f₂(g₁(X), Y))
ACTIVE₁(g₁(X)) -> ACTIVE₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
PROPER₁(g₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
F₂(ok₁(X1), ok₁(X2)) -> F₂(X1, X2)
G₁(mark₁(X)) -> G₁(X)
ACTIVE₁(g₁(X)) -> G₁(active₁(X))
PROPER₁(f₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(f₂(X1, X2)) -> F₂(proper₁(X1), proper₁(X2))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
ACTIVE₁(f₂(X1, X2)) -> F₂(active₁(X1), X2)
G₁(ok₁(X)) -> G₁(X)
F₂(mark₁(X1), X2) -> F₂(X1, X2)
ACTIVE₁(f₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(f₂(X1, X2)) -> PROPER₁(X1)

The TRS R consists of the following rules:

active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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(f₂(g₁(X), Y)) -> F₂(X, f₂(g₁(X), Y))
ACTIVE₁(g₁(X)) -> ACTIVE₁(X)
TOP₁(mark₁(X)) -> PROPER₁(X)
PROPER₁(g₁(X)) -> G₁(proper₁(X))
PROPER₁(g₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
F₂(ok₁(X1), ok₁(X2)) -> F₂(X1, X2)
G₁(mark₁(X)) -> G₁(X)
ACTIVE₁(g₁(X)) -> G₁(active₁(X))
PROPER₁(f₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(f₂(X1, X2)) -> F₂(proper₁(X1), proper₁(X2))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
ACTIVE₁(f₂(X1, X2)) -> F₂(active₁(X1), X2)
G₁(ok₁(X)) -> G₁(X)
F₂(mark₁(X1), X2) -> F₂(X1, X2)
ACTIVE₁(f₂(X1, X2)) -> ACTIVE₁(X1)
PROPER₁(f₂(X1, X2)) -> PROPER₁(X1)

The TRS R consists of the following rules:

active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ 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(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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)) = x₁
POL(mark₁(x₁)) = 1 + 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

          ↳ QDP

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

active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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:

F₂(mark₁(X1), X2) -> F₂(X1, X2)
F₂(ok₁(X1), ok₁(X2)) -> F₂(X1, X2)

The TRS R consists of the following rules:

active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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.

F₂(ok₁(X1), ok₁(X2)) -> F₂(X1, X2)

The remaining pairs can at least be oriented weakly.

F₂(mark₁(X1), X2) -> F₂(X1, X2)

Used ordering: Polynomial interpretation [21]:

POL(F₂(x₁, x₂)) = x₂
POL(mark₁(x₁)) = 0
POL(ok₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

F₂(mark₁(X1), X2) -> F₂(X1, X2)

The TRS R consists of the following rules:

active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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.

F₂(mark₁(X1), X2) -> F₂(X1, X2)

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

POL(F₂(x₁, x₂)) = x₁
POL(mark₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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(g₁(X)) -> PROPER₁(X)
PROPER₁(f₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(f₂(X1, X2)) -> PROPER₁(X1)

The TRS R consists of the following rules:

active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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(f₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(f₂(X1, X2)) -> PROPER₁(X1)

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

POL(PROPER₁(x₁)) = x₁
POL(f₂(x₁, x₂)) = 1 + x₁ + x₂
POL(g₁(x₁)) = 1 + x₁

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(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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)
ACTIVE₁(f₂(X1, X2)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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)
ACTIVE₁(f₂(X1, X2)) -> ACTIVE₁(X1)

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

POL(ACTIVE₁(x₁)) = x₁
POL(f₂(x₁, x₂)) = 1 + x₁
POL(g₁(x₁)) = 1 + x₁

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(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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

            ↳ 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(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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)) = x₁
POL(active₁(x₁)) = 1 + x₁
POL(f₂(x₁, x₂)) = x₁ + x₂
POL(g₁(x₁)) = x₁
POL(mark₁(x₁)) = 1
POL(ok₁(x₁)) = 1 + x₁
POL(proper₁(x₁)) = 0

The following usable rules [14] were oriented:

active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
g₁(ok₁(X)) -> ok₁(g₁(X))
proper₁(g₁(X)) -> g₁(proper₁(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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)) = x₁
POL(active₁(x₁)) = 0
POL(f₂(x₁, x₂)) = x₁
POL(g₁(x₁)) = x₁
POL(mark₁(x₁)) = 0
POL(ok₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented:

active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(g₁(X)) -> g₁(active₁(X))
g₁(mark₁(X)) -> mark₁(g₁(X))
g₁(ok₁(X)) -> ok₁(g₁(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

active₁(f₂(g₁(X), Y)) -> mark₁(f₂(X, f₂(g₁(X), Y)))
active₁(f₂(X1, X2)) -> f₂(active₁(X1), X2)
active₁(g₁(X)) -> g₁(active₁(X))
f₂(mark₁(X1), X2) -> mark₁(f₂(X1, X2))
g₁(mark₁(X)) -> mark₁(g₁(X))
proper₁(f₂(X1, X2)) -> f₂(proper₁(X1), proper₁(X2))
proper₁(g₁(X)) -> g₁(proper₁(X))
f₂(ok₁(X1), ok₁(X2)) -> ok₁(f₂(X1, X2))
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.