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:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(from(X)) → FROM(s(X))
PROPER(from(X)) → FROM(proper(X))
TOP(mark(X)) → PROPER(X)
ACTIVE(from(X)) → FROM(active(X))
S(mark(X)) → S(X)
PROPER(2nd(X)) → PROPER(X)
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(2nd(X)) → 2ND(active(X))
TOP(mark(X)) → TOP(proper(X))
PROPER(s(X)) → S(proper(X))
2ND(mark(X)) → 2ND(X)
S(ok(X)) → S(X)
ACTIVE(s(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
FROM(mark(X)) → FROM(X)
PROPER(cons(X1, X2)) → PROPER(X1)
2ND(ok(X)) → 2ND(X)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(from(X)) → S(X)
ACTIVE(s(X)) → S(active(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
PROPER(from(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(from(X)) → ACTIVE(X)
PROPER(2nd(X)) → 2ND(proper(X))
FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

ACTIVE(from(X)) → FROM(s(X))
PROPER(from(X)) → FROM(proper(X))
TOP(mark(X)) → PROPER(X)
ACTIVE(from(X)) → FROM(active(X))
S(mark(X)) → S(X)
PROPER(2nd(X)) → PROPER(X)
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(2nd(X)) → 2ND(active(X))
TOP(mark(X)) → TOP(proper(X))
PROPER(s(X)) → S(proper(X))
2ND(mark(X)) → 2ND(X)
S(ok(X)) → S(X)
ACTIVE(s(X)) → ACTIVE(X)
CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
FROM(mark(X)) → FROM(X)
PROPER(cons(X1, X2)) → PROPER(X1)
2ND(ok(X)) → 2ND(X)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(from(X)) → S(X)
ACTIVE(s(X)) → S(active(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
PROPER(from(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(from(X)) → ACTIVE(X)
PROPER(2nd(X)) → 2ND(proper(X))
FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

PROPER(from(X)) → FROM(proper(X))
ACTIVE(from(X)) → FROM(s(X))
TOP(mark(X)) → PROPER(X)
ACTIVE(from(X)) → FROM(active(X))
PROPER(2nd(X)) → PROPER(X)
S(mark(X)) → S(X)
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(2nd(X)) → 2ND(active(X))
PROPER(s(X)) → S(proper(X))
TOP(mark(X)) → TOP(proper(X))
2ND(mark(X)) → 2ND(X)
ACTIVE(s(X)) → ACTIVE(X)
S(ok(X)) → S(X)
ACTIVE(from(X)) → CONS(X, from(s(X)))
CONS(mark(X1), X2) → CONS(X1, X2)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
FROM(mark(X)) → FROM(X)
PROPER(cons(X1, X2)) → PROPER(X1)
2ND(ok(X)) → 2ND(X)
TOP(ok(X)) → ACTIVE(X)
ACTIVE(from(X)) → S(X)
ACTIVE(s(X)) → S(active(X))
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(s(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(from(X)) → ACTIVE(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
PROPER(2nd(X)) → 2ND(proper(X))
FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

S(ok(X)) → S(X)

The remaining pairs can at least be oriented weakly.

S(mark(X)) → S(X)

Used ordering: Combined order from the following AFS and order.
S(x1) = S(x1)
ok(x1) = ok(x1)
mark(x1) = x1

Lexicographic path order with status [19].
Precedence:

ok₁ > S₁

Status:

ok₁: multiset
S₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

S(mark(X)) → S(X)

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

Lexicographic path order with status [19].
Precedence:

mark₁ > S₁

Status:

mark₁: multiset
S₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

FROM(mark(X)) → FROM(X)
FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

FROM(mark(X)) → FROM(X)

The remaining pairs can at least be oriented weakly.

FROM(ok(X)) → FROM(X)

Used ordering: Combined order from the following AFS and order.
FROM(x1) = FROM(x1)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic path order with status [19].
Precedence:

mark₁ > FROM₁

Status:

mark₁: multiset
FROM₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

FROM(ok(X)) → FROM(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

FROM(ok(X)) → FROM(X)

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

Lexicographic path order with status [19].
Precedence:

ok₁ > FROM₁

Status:

ok₁: multiset
FROM₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

CONS(mark(X1), X2) → CONS(X1, X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

The remaining pairs can at least be oriented weakly.

CONS(mark(X1), X2) → CONS(X1, X2)

Used ordering: Combined order from the following AFS and order.
CONS(x1, x2) = CONS(x1)
mark(x1) = x1
ok(x1) = ok(x1)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

CONS₁: [1]
ok₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

CONS(mark(X1), X2) → CONS(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

CONS(mark(X1), X2) → CONS(X1, X2)

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

Lexicographic path order with status [19].
Precedence:

mark₁ > CONS₁

Status:

mark₁: multiset
CONS₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

2ND(mark(X)) → 2ND(X)
2ND(ok(X)) → 2ND(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

2ND(mark(X)) → 2ND(X)

The remaining pairs can at least be oriented weakly.

2ND(ok(X)) → 2ND(X)

Used ordering: Combined order from the following AFS and order.
2ND(x1) = 2ND(x1)
mark(x1) = mark(x1)
ok(x1) = x1

Lexicographic path order with status [19].
Precedence:

mark₁ > 2ND₁

Status:

mark₁: multiset
2ND₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

2ND(ok(X)) → 2ND(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

2ND(ok(X)) → 2ND(X)

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

Lexicographic path order with status [19].
Precedence:

ok₁ > 2ND₁

Status:

ok₁: multiset
2ND₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(s(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(2nd(X)) → PROPER(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)

The remaining pairs can at least be oriented weakly.

PROPER(s(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)
PROPER(2nd(X)) → PROPER(X)

Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
s(x1) = x1
from(x1) = x1
cons(x1, x2) = cons(x1, x2)
2nd(x1) = x1

Lexicographic path order with status [19].
Precedence:

trivial

Status:

PROPER₁: [1]
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

PROPER(s(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)
PROPER(2nd(X)) → PROPER(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

PROPER(2nd(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.

PROPER(s(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)

Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
s(x1) = x1
from(x1) = x1
2nd(x1) = 2nd(x1)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

PROPER₁: [1]
2nd₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

PROPER(s(X)) → PROPER(X)
PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

PROPER(s(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.

PROPER(from(X)) → PROPER(X)

Used ordering: Combined order from the following AFS and order.
PROPER(x1) = PROPER(x1)
s(x1) = s(x1)
from(x1) = x1

Lexicographic path order with status [19].
Precedence:

s₁ > PROPER₁

Status:

PROPER₁: [1]
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

PROPER(from(X)) → PROPER(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

PROPER(from(X)) → PROPER(X)

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

Lexicographic path order with status [19].
Precedence:

from₁ > PROPER₁

Status:

from₁: multiset
PROPER₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

ACTIVE(cons(X1, X2)) → ACTIVE(X1)

The remaining pairs can at least be oriented weakly.

ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
2nd(x1) = x1
cons(x1, x2) = cons(x1, x2)
from(x1) = x1
s(x1) = x1

Lexicographic path order with status [19].
Precedence:

cons₂ > ACTIVE₁

Status:

ACTIVE₁: [1]
cons₂: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

              ↳ QDP

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

ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

ACTIVE(s(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
2nd(x1) = x1
from(x1) = x1
s(x1) = s(x1)

Lexicographic path order with status [19].
Precedence:

trivial

Status:

ACTIVE₁: [1]
s₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

ACTIVE(2nd(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(from(X)) → ACTIVE(X)

Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = ACTIVE(x1)
2nd(x1) = 2nd(x1)
from(x1) = x1

Lexicographic path order with status [19].
Precedence:

2nd₁ > ACTIVE₁

Status:

ACTIVE₁: [1]
2nd₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

ACTIVE(from(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

ACTIVE(from(X)) → ACTIVE(X)

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

Lexicographic path order with status [19].
Precedence:

from₁ > ACTIVE₁

Status:

from₁: multiset
ACTIVE₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
TOP(x1) = x1
ok(x1) = ok
active(x1) = active
mark(x1) = mark
proper(x1) = proper
s(x1) = x1
cons(x1, x2) = x1
2nd(x1) = x1
from(x1) = x1

Lexicographic path order with status [19].
Precedence:

ok > active > mark > proper

Status:

active: []
ok: []
proper: []
mark: []

The following usable rules [14] were oriented:

s(mark(X)) → mark(s(X))
active(cons(X1, X2)) → cons(active(X1), X2)
2nd(ok(X)) → ok(2nd(X))
proper(s(X)) → s(proper(X))
active(2nd(X)) → 2nd(active(X))
s(ok(X)) → ok(s(X))
proper(2nd(X)) → 2nd(proper(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(ok(X)) → ok(from(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
2nd(mark(X)) → mark(2nd(X))
active(from(X)) → mark(cons(X, from(s(X))))
from(mark(X)) → mark(from(X))
active(from(X)) → from(active(X))
active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(s(X)) → s(active(X))
proper(from(X)) → from(proper(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

active(2nd(cons(X, cons(Y, Z)))) → mark(Y)
active(from(X)) → mark(cons(X, from(s(X))))
active(2nd(X)) → 2nd(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(from(X)) → from(active(X))
active(s(X)) → s(active(X))
2nd(mark(X)) → mark(2nd(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
from(mark(X)) → mark(from(X))
s(mark(X)) → mark(s(X))
proper(2nd(X)) → 2nd(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(from(X)) → from(proper(X))
proper(s(X)) → s(proper(X))
2nd(ok(X)) → ok(2nd(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(ok(X)) → ok(from(X))
s(ok(X)) → ok(s(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.