Termination w.r.t. Q proof of TRS/TRCSREx1_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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(S₁(x₁)) = 3·x₁
POL(mark₁(x₁)) = 3 + 2·x₁
POL(ok₁(x₁)) = 3 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FROM₁(x₁)) = 3·x₁
POL(mark₁(x₁)) = 3 + x₁
POL(ok₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(CONS₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(mark₁(x₁)) = 3 + x₁
POL(ok₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ 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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(2ND₁(x₁)) = 3·x₁
POL(mark₁(x₁)) = 3 + x₁
POL(ok₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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 oriented strictly and are deleted.

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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(2nd₁(x₁)) = 3 + 2·x₁
POL(PROPER₁(x₁)) = 3·x₁
POL(cons₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂
POL(from₁(x₁)) = 3 + 2·x₁
POL(s₁(x₁)) = 3 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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 oriented strictly and are deleted.

ACTIVE₁(2nd₁(X)) -> ACTIVE₁(X)
ACTIVE₁(cons₂(X1, X2)) -> ACTIVE₁(X1)
ACTIVE₁(from₁(X)) -> ACTIVE₁(X)
ACTIVE₁(s₁(X)) -> ACTIVE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(2nd₁(x₁)) = 3 + x₁
POL(ACTIVE₁(x₁)) = 3·x₁
POL(cons₂(x₁, x₂)) = 3 + 2·x₁
POL(from₁(x₁)) = 3 + 2·x₁
POL(s₁(x₁)) = 3 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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 oriented strictly and are deleted.

TOP₁(ok₁(X)) -> TOP₁(active₁(X))

The remaining pairs can at least be oriented weakly.

TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

Used ordering: Polynomial interpretation [21]:

POL(2nd₁(x₁)) = x₁
POL(TOP₁(x₁)) = x₁
POL(active₁(x₁)) = 1 + 3·x₁
POL(cons₂(x₁, x₂)) = x₁
POL(from₁(x₁)) = x₁
POL(mark₁(x₁)) = 0
POL(ok₁(x₁)) = 2 + 3·x₁
POL(proper₁(x₁)) = 0
POL(s₁(x₁)) = x₁

The following usable rules [14] were oriented:

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

TOP₁(mark₁(X)) -> TOP₁(proper₁(X))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(2nd₁(x₁)) = x₁
POL(TOP₁(x₁)) = 2·x₁
POL(cons₂(x₁, x₂)) = 3·x₁ + 3·x₂
POL(from₁(x₁)) = x₁
POL(mark₁(x₁)) = 2 + x₁
POL(ok₁(x₁)) = 3
POL(proper₁(x₁)) = 0
POL(s₁(x₁)) = x₁

The following usable rules [14] were oriented:

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