Termination w.r.t. Q proof of TRS/TRCSREx3_12_Luc96a

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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(from₁(X)) -> FROM₁(s₁(X))
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
PROPER₁(sel₂(X1, X2)) -> SEL₂(proper₁(X1), proper₁(X2))
S₁(mark₁(X)) -> S₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(X1, active₁(X2))
ACTIVE₁(sel₂(s₁(X), cons₂(Y, Z))) -> SEL₂(X, Z)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)
ACTIVE₁(from₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(active₁(X1), X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(from₁(X)) -> FROM₁(s₁(X))
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
PROPER₁(sel₂(X1, X2)) -> SEL₂(proper₁(X1), proper₁(X2))
S₁(mark₁(X)) -> S₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(X1, active₁(X2))
ACTIVE₁(sel₂(s₁(X), cons₂(Y, Z))) -> SEL₂(X, Z)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)
ACTIVE₁(from₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> S₁(active₁(X))
PROPER₁(cons₂(X1, X2)) -> CONS₂(proper₁(X1), proper₁(X2))
PROPER₁(s₁(X)) -> PROPER₁(X)
TOP₁(ok₁(X)) -> TOP₁(active₁(X))
PROPER₁(from₁(X)) -> PROPER₁(X)
ACTIVE₁(sel₂(X1, X2)) -> SEL₂(active₁(X1), X2)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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 7 SCCs with 15 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)
SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

SEL₂(X1, mark₁(X2)) -> SEL₂(X1, X2)

The remaining pairs can at least be oriented weakly.

SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

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

POL( SEL₂(x₁, x₂) ) = max{0, x₂ - 2}

POL( mark₁(x₁) ) = x₁ + 3

POL( ok₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)
SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

SEL₂(ok₁(X1), ok₁(X2)) -> SEL₂(X1, X2)

The remaining pairs can at least be oriented weakly.

SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

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

POL( SEL₂(x₁, x₂) ) = max{0, x₂ - 2}

POL( ok₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

SEL₂(mark₁(X1), X2) -> SEL₂(X1, X2)

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

POL( SEL₂(x₁, x₂) ) = max{0, x₁ - 2}

POL( mark₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

          ↳ QDP

          ↳ QDP

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

S₁(ok₁(X)) -> S₁(X)
S₁(mark₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

S₁(ok₁(X)) -> S₁(X)

The remaining pairs can at least be oriented weakly.

S₁(mark₁(X)) -> S₁(X)

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

POL( S₁(x₁) ) = max{0, x₁ - 2}

POL( ok₁(x₁) ) = x₁ + 3

POL( mark₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S₁(mark₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

S₁(mark₁(X)) -> S₁(X)

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

POL( S₁(x₁) ) = max{0, x₁ - 2}

POL( mark₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

          ↳ 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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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) ) = max{0, x₁ - 2}

POL( mark₁(x₁) ) = x₁ + 3

POL( ok₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ 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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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) ) = max{0, x₂ - 2}

POL( ok₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

          ↳ QDP

          ↳ QDP

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

FROM₁(mark₁(X)) -> FROM₁(X)
FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

FROM₁(mark₁(X)) -> FROM₁(X)

The remaining pairs can at least be oriented weakly.

FROM₁(ok₁(X)) -> FROM₁(X)

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

POL( FROM₁(x₁) ) = max{0, x₁ - 2}

POL( mark₁(x₁) ) = x₁ + 3

POL( ok₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.

FROM₁(ok₁(X)) -> FROM₁(X)

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

POL( FROM₁(x₁) ) = max{0, x₁ - 2}

POL( ok₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ 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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

          ↳ QDP

          ↳ QDP

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

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(s₁(X)) -> PROPER₁(X)

The remaining pairs can at least be oriented weakly.

PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

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

POL( PROPER₁(x₁) ) = max{0, x₁ - 2}

POL( s₁(x₁) ) = x₁ + 3

POL( sel₂(x₁, x₂) ) = x₁ + x₂ + 2

POL( from₁(x₁) ) = x₁

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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)

The remaining pairs can at least be oriented weakly.

PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

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

POL( PROPER₁(x₁) ) = max{0, x₁ - 2}

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 3

POL( from₁(x₁) ) = x₁

POL( sel₂(x₁, x₂) ) = x₁ + x₂ + 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(sel₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(sel₂(X1, X2)) -> PROPER₁(X2)

The remaining pairs can at least be oriented weakly.

PROPER₁(from₁(X)) -> PROPER₁(X)

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

POL( PROPER₁(x₁) ) = max{0, x₁ - 2}

POL( sel₂(x₁, x₂) ) = x₁ + x₂ + 3

POL( from₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(from₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(from₁(X)) -> PROPER₁(X)

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

POL( PROPER₁(x₁) ) = max{0, x₁ - 2}

POL( from₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(sel₂(X1, X2)) -> ACTIVE₁(X2)
ACTIVE₁(sel₂(X1, X2)) -> ACTIVE₁(X1)

The remaining pairs can at least be oriented weakly.

ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

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

POL( ACTIVE₁(x₁) ) = max{0, x₁ - 2}

POL( sel₂(x₁, x₂) ) = x₁ + x₂ + 3

POL( cons₂(x₁, x₂) ) = x₁

POL( from₁(x₁) ) = x₁

POL( s₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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)

The remaining pairs can at least be oriented weakly.

ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

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

POL( ACTIVE₁(x₁) ) = max{0, x₁ - 2}

POL( cons₂(x₁, x₂) ) = x₁ + 3

POL( from₁(x₁) ) = x₁

POL( s₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(from₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least be oriented weakly.

ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

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

POL( ACTIVE₁(x₁) ) = max{0, x₁ - 2}

POL( from₁(x₁) ) = x₁ + 3

POL( s₁(x₁) ) = x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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(s₁(X)) -> ACTIVE₁(X)

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

POL( ACTIVE₁(x₁) ) = max{0, x₁ - 2}

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

          ↳ QDP

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

active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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

          ↳ 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(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(sel₂(0, cons₂(X, Y))) -> mark₁(X)
active₁(sel₂(s₁(X), cons₂(Y, Z))) -> mark₁(sel₂(X, Z))
active₁(from₁(X)) -> from₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(s₁(X)) -> s₁(active₁(X))
active₁(sel₂(X1, X2)) -> sel₂(active₁(X1), X2)
active₁(sel₂(X1, X2)) -> sel₂(X1, active₁(X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
s₁(mark₁(X)) -> mark₁(s₁(X))
sel₂(mark₁(X1), X2) -> mark₁(sel₂(X1, X2))
sel₂(X1, mark₁(X2)) -> mark₁(sel₂(X1, X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(s₁(X)) -> s₁(proper₁(X))
proper₁(sel₂(X1, X2)) -> sel₂(proper₁(X1), proper₁(X2))
proper₁(0) -> ok₁(0)
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
s₁(ok₁(X)) -> ok₁(s₁(X))
sel₂(ok₁(X1), ok₁(X2)) -> ok₁(sel₂(X1, X2))
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.