Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/TRCSRSLASHEx4_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.

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 contains 5 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

G₁(ok₁(X)) -> G₁(X)

Used argument filtering: G₁(x1) = x1
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

G₁(mark₁(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

G₁(mark₁(X)) -> G₁(X)

Used argument filtering: G₁(x1) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₂(ok₁(X1), ok₁(X2)) -> F₂(X1, X2)

Used argument filtering: F₂(x1, x2) = x2
ok₁(x1) = ok₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

F₂(mark₁(X1), X2) -> F₂(X1, X2)

Used argument filtering: F₂(x1, x2) = x1
mark₁(x1) = mark₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(f₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(f₂(X1, X2)) -> PROPER₁(X1)

Used argument filtering: PROPER₁(x1) = x1
g₁(x1) = x1
f₂(x1, x2) = f₂(x1, x2)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(g₁(X)) -> 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PROPER₁(g₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
g₁(x1) = g₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(f₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
g₁(x1) = x1
f₂(x1, x2) = f₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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(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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACTIVE₁(g₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
g₁(x1) = g₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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

            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

Used argument filtering: TOP₁(x1) = x1
ok₁(x1) = ok₁(x1)
active₁(x1) = x1
mark₁(x1) = mark
proper₁(x1) = proper
f₂(x1, x2) = x2
g₁(x1) = x1
Used ordering: Quasi Precedence: ok_1 > [mark, proper]

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

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

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

Used argument filtering: TOP₁(x1) = x1
mark₁(x1) = mark
proper₁(x1) = proper
f₂(x1, x2) = x1
g₁(x1) = x1
ok₁(x1) = ok
Used ordering: Quasi Precedence: mark > proper

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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.