Termination w.r.t. Q proof of /tmp/tmp465lvX/Ex6_9_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(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

CONS(mark(X1), X2) → CONS(X1, X2)
MARK(2nd(X)) → 2ND(mark(X))
FROM(mark(X)) → FROM(X)
ACTIVE(2nd(cons(X, X1))) → 2ND(cons1(X, X1))
CONS1(active(X1), X2) → CONS1(X1, X2)
MARK(s(X)) → MARK(X)
MARK(from(X)) → FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
FROM(active(X)) → FROM(X)
CONS(X1, mark(X2)) → CONS(X1, X2)
2ND(mark(X)) → 2ND(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(from(X)) → MARK(X)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
CONS1(mark(X1), X2) → CONS1(X1, X2)
MARK(cons1(X1, X2)) → CONS1(mark(X1), mark(X2))
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
ACTIVE(from(X)) → S(X)
ACTIVE(2nd(cons(X, X1))) → CONS1(X, X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
2ND(active(X)) → 2ND(X)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(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:

CONS(mark(X1), X2) → CONS(X1, X2)
MARK(2nd(X)) → 2ND(mark(X))
FROM(mark(X)) → FROM(X)
ACTIVE(2nd(cons(X, X1))) → 2ND(cons1(X, X1))
CONS1(active(X1), X2) → CONS1(X1, X2)
MARK(s(X)) → MARK(X)
MARK(from(X)) → FROM(mark(X))
MARK(cons(X1, X2)) → MARK(X1)
FROM(active(X)) → FROM(X)
CONS(X1, mark(X2)) → CONS(X1, X2)
2ND(mark(X)) → 2ND(X)
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(from(X)) → MARK(X)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
CONS1(mark(X1), X2) → CONS1(X1, X2)
MARK(cons1(X1, X2)) → CONS1(mark(X1), mark(X2))
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
ACTIVE(from(X)) → S(X)
ACTIVE(2nd(cons(X, X1))) → CONS1(X, X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
2ND(active(X)) → 2ND(X)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(s(X))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 10 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

S(mark(X)) → S(X)
The graph contains the following edges 1 > 1
S(active(X)) → S(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FROM(mark(X)) → FROM(X)
FROM(active(X)) → FROM(X)

From the DPs we obtained the following set of size-change graphs:

FROM(mark(X)) → FROM(X)
The graph contains the following edges 1 > 1
FROM(active(X)) → FROM(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

From the DPs we obtained the following set of size-change graphs:

CONS(X1, active(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
CONS(mark(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
CONS(active(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
CONS(X1, mark(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

CONS1(active(X1), X2) → CONS1(X1, X2)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

CONS1(active(X1), X2) → CONS1(X1, X2)
CONS1(X1, mark(X2)) → CONS1(X1, X2)
CONS1(mark(X1), X2) → CONS1(X1, X2)
CONS1(X1, active(X2)) → CONS1(X1, X2)

From the DPs we obtained the following set of size-change graphs:

CONS1(active(X1), X2) → CONS1(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
CONS1(X1, mark(X2)) → CONS1(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
CONS1(X1, active(X2)) → CONS1(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
CONS1(mark(X1), X2) → CONS1(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

2ND(active(X)) → 2ND(X)
2ND(mark(X)) → 2ND(X)

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

2ND(active(X)) → 2ND(X)
2ND(mark(X)) → 2ND(X)

From the DPs we obtained the following set of size-change graphs:

2ND(active(X)) → 2ND(X)
The graph contains the following edges 1 > 1
2ND(mark(X)) → 2ND(X)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(cons1(X1, X2)) → ACTIVE(cons1(mark(X1), mark(X2)))
MARK(s(X)) → ACTIVE(s(mark(X)))

The remaining pairs can at least be oriented weakly.

MARK(from(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

Used ordering: Polynomial interpretation [25]:

POL(2nd(x₁)) = 1
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 1
POL(active(x₁)) = 0
POL(cons(x₁, x₂)) = 0
POL(cons1(x₁, x₂)) = 0
POL(from(x₁)) = 1
POL(mark(x₁)) = 0
POL(s(x₁)) = 0

The following usable rules [17] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(2nd(X)) → MARK(X)
MARK(s(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(2nd(X)) → MARK(X)
ACTIVE(2nd(cons1(X, cons(Y, Z)))) → MARK(Y)

The remaining pairs can at least be oriented weakly.

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( from(x₁) ) =

/	0	\
\	0	/

/	1	1	\
\	1	0	/

x₁

M( active(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	1	1	\
\	0	0	/

x₁

/	0	0	\
\	1	0	/

x₂

M( cons1(x₁, x₂) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

/	1	0	\
\	0	0	/

x₂

M( 2nd(x₁) ) =

/	1	\
\	0	/

/	1	1	\
\	1	1	/

x₁

M( mark(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( s(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	0	/

x₁

Tuple symbols:

M( MARK(x₁) ) =

[

]

x₁

M( ACTIVE(x₁) ) =

[

]

x₁

Matrix type:
We used a basic matrix type which is not further parametrizeable.

As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(2nd(X)) → active(2nd(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
MARK(cons1(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons1(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(cons1(X1, X2)) → MARK(X2)
MARK(cons1(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( from(x₁) ) =

/	0	\
\	0	/

/	1	1	\
\	1	0	/

x₁

M( active(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( cons(x₁, x₂) ) =

/	0	\
\	0	/

/	1	1	\
\	0	0	/

x₁

/	0	0	\
\	1	0	/

x₂

M( cons1(x₁, x₂) ) =

/	1	\
\	0	/

/	1	0	\
\	0	0	/

x₁

/	1	0	\
\	1	0	/

x₂

M( 2nd(x₁) ) =

/	0	\
\	0	/

/	0	1	\
\	0	1	/

x₁

M( mark(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( s(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	0	/

x₁

Tuple symbols:

M( MARK(x₁) ) =

[

]

x₁

M( ACTIVE(x₁) ) =

[

]

x₁

from(active(X)) → from(X)
from(mark(X)) → from(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(2nd(X)) → active(2nd(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

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

MARK(from(X)) → MARK(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.

MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( from(x₁) ) =

/	1	\
\	1	/

/	0	1	\
\	1	1	/

x₁

M( active(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( cons(x₁, x₂) ) =

/	0	\
\	1	/

/	0	0	\
\	1	1	/

x₁

/	0	1	\
\	0	0	/

x₂

M( cons1(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	0	1	\
\	0	0	/

x₂

M( 2nd(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	1	0	/

x₁

M( mark(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( s(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	1	/

x₁

Tuple symbols:

M( MARK(x₁) ) =

[

]

x₁

M( ACTIVE(x₁) ) =

[

]

x₁

from(active(X)) → from(X)
from(mark(X)) → from(X)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(2nd(X)) → active(2nd(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

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

MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(from(X)) → ACTIVE(from(mark(X)))

The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

Used ordering: Polynomial interpretation [25]:

POL(2nd(x₁)) = 0
POL(ACTIVE(x₁)) = 0
POL(MARK(x₁)) = x₁
POL(active(x₁)) = 0
POL(cons(x₁, x₂)) = 0
POL(cons1(x₁, x₂)) = 0
POL(from(x₁)) = 1
POL(mark(x₁)) = 0
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

ACTIVE(from(X)) → MARK(cons(X, from(s(X))))

The remaining pairs can at least be oriented weakly.

MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

Used ordering: Polynomial interpretation [25]:

POL(2nd(x₁)) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = 0
POL(active(x₁)) = 0
POL(cons(x₁, x₂)) = 0
POL(cons1(x₁, x₂)) = 0
POL(from(x₁)) = 1 + x₁
POL(mark(x₁)) = 0
POL(s(x₁)) = 0

The following usable rules [17] were oriented:

2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

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

MARK(s(X)) → MARK(X)
ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.

ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

Used ordering: Polynomial interpretation [25]:

POL(2nd(x₁)) = 0
POL(ACTIVE(x₁)) = 0
POL(MARK(x₁)) = x₁
POL(active(x₁)) = 0
POL(cons(x₁, x₂)) = 0
POL(cons1(x₁, x₂)) = 0
POL(from(x₁)) = 0
POL(mark(x₁)) = 0
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

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

ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))
MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

ACTIVE(2nd(cons(X, X1))) → MARK(2nd(cons1(X, X1)))

The remaining pairs can at least be oriented weakly.

MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

Used ordering: Matrix interpretation [3]:
Non-tuple symbols:

M( from(x₁) ) =

/	0	\
\	1	/

/	1	1	\
\	0	0	/

x₁

M( active(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( cons(x₁, x₂) ) =

/	0	\
\	1	/

/	1	1	\
\	0	0	/

x₁

/	1	0	\
\	1	0	/

x₂

M( cons1(x₁, x₂) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

/	1	0	\
\	1	0	/

x₂

M( 2nd(x₁) ) =

/	0	\
\	0	/

/	0	1	\
\	1	0	/

x₁

M( mark(x₁) ) =

/	0	\
\	0	/

/	1	0	\
\	0	1	/

x₁

M( s(x₁) ) =

/	0	\
\	0	/

/	0	0	\
\	0	0	/

x₁

Tuple symbols:

M( MARK(x₁) ) =

[

]

x₁

M( ACTIVE(x₁) ) =

[

]

x₁

from(active(X)) → from(X)
from(mark(X)) → from(X)
active(from(X)) → mark(cons(X, from(s(X))))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(2nd(X)) → active(2nd(mark(X)))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons1(mark(X1), X2) → cons1(X1, X2)
2nd(active(X)) → 2nd(X)
2nd(mark(X)) → 2nd(X)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ QDPOrderProof

                                  ↳ QDP

                                    ↳ QDPOrderProof

                                      ↳ QDP

                                        ↳ QDPOrderProof

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

MARK(2nd(X)) → ACTIVE(2nd(mark(X)))

The TRS R consists of the following rules:

active(2nd(cons1(X, cons(Y, Z)))) → mark(Y)
active(2nd(cons(X, X1))) → mark(2nd(cons1(X, X1)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(2nd(X)) → active(2nd(mark(X)))
mark(cons1(X1, X2)) → active(cons1(mark(X1), mark(X2)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
mark(s(X)) → active(s(mark(X)))
2nd(mark(X)) → 2nd(X)
2nd(active(X)) → 2nd(X)
cons1(mark(X1), X2) → cons1(X1, X2)
cons1(X1, mark(X2)) → cons1(X1, X2)
cons1(active(X1), X2) → cons1(X1, X2)
cons1(X1, active(X2)) → cons1(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
from(mark(X)) → from(X)
from(active(X)) → from(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.