Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/TRCSRSLASHEx4_7_77_Bor03

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(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(zeros) -> CONS₂(0, zeros)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
TAIL₁(mark₁(X)) -> TAIL₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
TAIL₁(ok₁(X)) -> TAIL₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(tail₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(tail₁(X)) -> TAIL₁(proper₁(X))
ACTIVE₁(tail₁(X)) -> TAIL₁(active₁(X))
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(zeros) -> CONS₂(0, zeros)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
TAIL₁(mark₁(X)) -> TAIL₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
TAIL₁(ok₁(X)) -> TAIL₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(tail₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(tail₁(X)) -> TAIL₁(proper₁(X))
ACTIVE₁(tail₁(X)) -> TAIL₁(active₁(X))
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

TAIL₁(mark₁(X)) -> TAIL₁(X)
TAIL₁(ok₁(X)) -> TAIL₁(X)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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.

TAIL₁(ok₁(X)) -> TAIL₁(X)

Used argument filtering: TAIL₁(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:

TAIL₁(mark₁(X)) -> TAIL₁(X)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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.

TAIL₁(mark₁(X)) -> TAIL₁(X)

Used argument filtering: TAIL₁(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(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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:

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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.

CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)

Used argument filtering: CONS₂(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:

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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.

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

Used argument filtering: CONS₂(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(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(tail₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)

Used argument filtering: PROPER₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
tail₁(x1) = x1
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(tail₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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(tail₁(X)) -> PROPER₁(X)

Used argument filtering: PROPER₁(x1) = x1
tail₁(x1) = tail₁(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(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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(tail₁(X)) -> ACTIVE₁(X)

Used argument filtering: ACTIVE₁(x1) = x1
cons₂(x1, x2) = x1
tail₁(x1) = tail₁(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(cons₂(X1, X2)) -> ACTIVE₁(X1)

The TRS R consists of the following rules:

active₁(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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(cons₂(X1, X2)) -> ACTIVE₁(X1)

Used argument filtering: ACTIVE₁(x1) = x1
cons₂(x1, x2) = cons₁(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(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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(zeros) -> mark₁(cons₂(0, zeros))
active₁(tail₁(cons₂(X, XS))) -> mark₁(XS)
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(tail₁(X)) -> tail₁(active₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
tail₁(mark₁(X)) -> mark₁(tail₁(X))
proper₁(zeros) -> ok₁(zeros)
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
proper₁(tail₁(X)) -> tail₁(proper₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
tail₁(ok₁(X)) -> ok₁(tail₁(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.