Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/TRCSRSLASHEx1_2_Luc02c

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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACTIVE₁(from₁(X)) -> FROM₁(s₁(X))
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
S₁(mark₁(X)) -> S₁(X)
PROPER₁(2nd₁(X)) -> PROPER₁(X)
ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(2nd₁(X)) -> 2ND₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
2ND₁(mark₁(X)) -> 2ND₁(X)
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
2ND₁(ok₁(X)) -> 2ND₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
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)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
PROPER₁(2nd₁(X)) -> 2ND₁(proper₁(X))
FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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(from₁(X)) -> FROM₁(s₁(X))
PROPER₁(from₁(X)) -> FROM₁(proper₁(X))
TOP₁(mark₁(X)) -> PROPER₁(X)
ACTIVE₁(from₁(X)) -> FROM₁(active₁(X))
S₁(mark₁(X)) -> S₁(X)
PROPER₁(2nd₁(X)) -> PROPER₁(X)
ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(2nd₁(X)) -> 2ND₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))
PROPER₁(s₁(X)) -> S₁(proper₁(X))
2ND₁(mark₁(X)) -> 2ND₁(X)
S₁(ok₁(X)) -> S₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)
CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> CONS₂(X, from₁(s₁(X)))
ACTIVE₁(cons₂(X1, X2)) -> CONS₂(active₁(X1), X2)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
FROM₁(mark₁(X)) -> FROM₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
2ND₁(ok₁(X)) -> 2ND₁(X)
TOP₁(ok₁(X)) -> ACTIVE₁(X)
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)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
CONS₂(ok₁(X1), ok₁(X2)) -> CONS₂(X1, X2)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
PROPER₁(2nd₁(X)) -> 2ND₁(proper₁(X))
FROM₁(ok₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 7 SCCs with 13 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:

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

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

S₁(mark₁(X)) -> S₁(X)

The remaining pairs can at least by weakly be oriented.

S₁(ok₁(X)) -> S₁(X)

Used ordering: Combined order from the following AFS and order.
S₁(x1) = S₁(x1)
ok₁(x1) = x1
mark₁(x1) = mark₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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:

S₁(ok₁(X)) -> S₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

S₁(ok₁(X)) -> S₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
S₁(x1) = S₁(x1)
ok₁(x1) = ok₁(x1)

Lexicographic Path Order [19].
Precedence:

ok₁ > S₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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

          ↳ 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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

FROM₁(ok₁(X)) -> FROM₁(X)

The remaining pairs can at least by weakly be oriented.

FROM₁(mark₁(X)) -> FROM₁(X)

Used ordering: Combined order from the following AFS and order.
FROM₁(x1) = FROM₁(x1)
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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:

FROM₁(mark₁(X)) -> FROM₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

FROM₁(mark₁(X)) -> FROM₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
FROM₁(x1) = FROM₁(x1)
mark₁(x1) = mark₁(x1)

Lexicographic Path Order [19].
Precedence:

mark₁ > FROM₁

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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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

          ↳ 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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

Used ordering: Combined order from the following AFS and order.
CONS₂(x1, x2) = x2
mark₁(x1) = mark
ok₁(x1) = ok₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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(mark₁(X1), X2) -> CONS₂(X1, X2)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

CONS₂(mark₁(X1), X2) -> CONS₂(X1, X2)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
CONS₂(x1, x2) = CONS₁(x1)
mark₁(x1) = mark₁(x1)

Lexicographic Path Order [19].
Precedence:

[CONS₁, mark_1]

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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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

          ↳ QDP

          ↳ QDP

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

2ND₁(mark₁(X)) -> 2ND₁(X)
2ND₁(ok₁(X)) -> 2ND₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

2ND₁(ok₁(X)) -> 2ND₁(X)

The remaining pairs can at least by weakly be oriented.

2ND₁(mark₁(X)) -> 2ND₁(X)

Used ordering: Combined order from the following AFS and order.
2ND₁(x1) = 2ND₁(x1)
mark₁(x1) = x1
ok₁(x1) = ok₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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:

2ND₁(mark₁(X)) -> 2ND₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

2ND₁(mark₁(X)) -> 2ND₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
2ND₁(x1) = 2ND₁(x1)
mark₁(x1) = mark₁(x1)

Lexicographic Path Order [19].
Precedence:

mark₁ > 2ND₁

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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
top₁(mark₁(X)) -> top₁(proper₁(X))
top₁(ok₁(X)) -> top₁(active₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(from₁(X)) -> PROPER₁(X)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X1)
PROPER₁(cons₂(X1, X2)) -> PROPER₁(X2)
PROPER₁(2nd₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

PROPER₁(s₁(X)) -> PROPER₁(X)
PROPER₁(2nd₁(X)) -> PROPER₁(X)

Used ordering: Combined order from the following AFS and order.
PROPER₁(x1) = PROPER₁(x1)
s₁(x1) = x1
from₁(x1) = from₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)
2nd₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

cons₂ > PROPER₁

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(s₁(X)) -> PROPER₁(X)
PROPER₁(2nd₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

PROPER₁(2nd₁(X)) -> PROPER₁(X)

The remaining pairs can at least by weakly be oriented.

PROPER₁(s₁(X)) -> PROPER₁(X)

Used ordering: Combined order from the following AFS and order.
PROPER₁(x1) = PROPER₁(x1)
s₁(x1) = x1
2nd₁(x1) = 2nd₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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(s₁(X)) -> PROPER₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

PROPER₁(s₁(X)) -> PROPER₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
PROPER₁(x1) = PROPER₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > PROPER₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)

The remaining pairs can at least by weakly be oriented.

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE₁(x1) = ACTIVE₁(x1)
2nd₁(x1) = x1
cons₂(x1, x2) = cons₂(x1, x2)
from₁(x1) = x1
s₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

trivial

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(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

ACTIVE₁(from₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least by weakly be oriented.

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE₁(x1) = ACTIVE₁(x1)
2nd₁(x1) = x1
from₁(x1) = from₁(x1)
s₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

from₁ > ACTIVE₁

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(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least by weakly be oriented.

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE₁(x1) = ACTIVE₁(x1)
2nd₁(x1) = x1
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

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(2nd₁(X)) -> ACTIVE₁(X)

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVE₁(x1) = ACTIVE₁(x1)
2nd₁(x1) = 2nd₁(x1)

Lexicographic Path Order [19].
Precedence:

2nd₁ > ACTIVE₁

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(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

TOP₁(ok₁(X)) -> TOP₁(active₁(X))
TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The remaining pairs can at least by weakly be oriented.

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

Used ordering: Combined order from the following AFS and order.
TOP₁(x1) = x1
ok₁(x1) = ok
active₁(x1) = active
mark₁(x1) = mark
proper₁(x1) = proper
2nd₁(x1) = x1
cons₂(x1, x2) = x1
from₁(x1) = x1
s₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

[ok, active] > mark > proper

The following usable rules [14] were oriented:

proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

The TRS R consists of the following rules:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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 strictly oriented and are deleted.

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
TOP₁(x1) = TOP₁(x1)
ok₁(x1) = ok₁(x1)
active₁(x1) = x1
2nd₁(x1) = 2nd₁(x1)
cons₂(x1, x2) = x1
mark₁(x1) = mark
from₁(x1) = from₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

from₁ > [ok₁, 2nd₁, mark, s_1]

The following usable rules [14] were oriented:

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
s₁(ok₁(X)) -> ok₁(s₁(X))
from₁(mark₁(X)) -> mark₁(from₁(X))
from₁(ok₁(X)) -> ok₁(from₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

active₁(2nd₁(cons₂(X, cons₂(Y, Z)))) -> mark₁(Y)
active₁(from₁(X)) -> mark₁(cons₂(X, from₁(s₁(X))))
active₁(2nd₁(X)) -> 2nd₁(active₁(X))
active₁(cons₂(X1, X2)) -> cons₂(active₁(X1), X2)
active₁(from₁(X)) -> from₁(active₁(X))
active₁(s₁(X)) -> s₁(active₁(X))
2nd₁(mark₁(X)) -> mark₁(2nd₁(X))
cons₂(mark₁(X1), X2) -> mark₁(cons₂(X1, X2))
from₁(mark₁(X)) -> mark₁(from₁(X))
s₁(mark₁(X)) -> mark₁(s₁(X))
proper₁(2nd₁(X)) -> 2nd₁(proper₁(X))
proper₁(cons₂(X1, X2)) -> cons₂(proper₁(X1), proper₁(X2))
proper₁(from₁(X)) -> from₁(proper₁(X))
proper₁(s₁(X)) -> s₁(proper₁(X))
2nd₁(ok₁(X)) -> ok₁(2nd₁(X))
cons₂(ok₁(X1), ok₁(X2)) -> ok₁(cons₂(X1, X2))
from₁(ok₁(X)) -> ok₁(from₁(X))
s₁(ok₁(X)) -> ok₁(s₁(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.