Termination w.r.t. Q proof of TRS/TRCSREx4_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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [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:

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( mark₁(x₁) ) = x₁ + 1

POL( ok₁(x₁) ) = x₁ + 1

POL( TAIL₁(x₁) ) = x₁ + 1

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(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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( CONS₂(x₁, x₂) ) = x₁ + 1

POL( mark₁(x₁) ) = x₁ + 1

POL( ok₁(x₁) ) = 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:

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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( CONS₂(x₁, x₂) ) = x₁

POL( ok₁(x₁) ) = x₁ + 1

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(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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(tail₁(X)) -> PROPER₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( tail₁(x₁) ) = x₁ + 1

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( PROPER₁(x₁) ) = x₁ + 1

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(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

            ↳ QDPOrderProof

          ↳ 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(tail₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( tail₁(x₁) ) = x₁ + 1

POL( ACTIVE₁(x₁) ) = x₁ + 1

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 1

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(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.