Termination w.r.t. Q proof of /tmp/tmposQsEI/Ex6_Luc98

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(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

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)
FROM(mark(X)) → FROM(X)
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)
CONS(X1, active(X2)) → CONS(X1, X2)
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
MARK(s(X)) → ACTIVE(s(mark(X)))
FIRST(active(X1), X2) → FIRST(X1, X2)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(first(X1, X2)) → FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
MARK(from(X)) → MARK(X)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(first(X1, X2)) → MARK(X2)
ACTIVE(first(0, X)) → MARK(nil)
FIRST(X1, active(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
ACTIVE(from(X)) → S(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(0) → ACTIVE(0)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
MARK(nil) → ACTIVE(nil)
ACTIVE(from(X)) → CONS(X, from(s(X)))
FIRST(X1, mark(X2)) → FIRST(X1, X2)
ACTIVE(from(X)) → FROM(s(X))

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

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)
FROM(mark(X)) → FROM(X)
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)
CONS(X1, active(X2)) → CONS(X1, X2)
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
MARK(s(X)) → ACTIVE(s(mark(X)))
FIRST(active(X1), X2) → FIRST(X1, X2)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))
MARK(first(X1, X2)) → FIRST(mark(X1), mark(X2))
MARK(first(X1, X2)) → MARK(X1)
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
MARK(from(X)) → MARK(X)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(first(X1, X2)) → MARK(X2)
ACTIVE(first(0, X)) → MARK(nil)
FIRST(X1, active(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
ACTIVE(from(X)) → S(X)
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(0) → ACTIVE(0)
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
MARK(nil) → ACTIVE(nil)
ACTIVE(from(X)) → CONS(X, from(s(X)))
FIRST(X1, mark(X2)) → FIRST(X1, X2)
ACTIVE(from(X)) → FROM(s(X))

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 12 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

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

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

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

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

            ↳ 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(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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

            ↳ UsableRulesProof

          ↳ 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(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

S(active(X)) → S(X)
S(mark(X)) → S(X)

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

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

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

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

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

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

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MARK(first(X1, X2)) → MARK(X1)
MARK(from(X)) → MARK(X)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
ACTIVE(from(X)) → MARK(cons(X, from(s(X))))

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

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

The remaining pairs can at least be oriented weakly.

MARK(first(X1, X2)) → MARK(X1)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = x₁
POL(first(x₁, x₂)) = x₁ + x₂
POL(from(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
active(first(0, X)) → mark(nil)
mark(0) → active(0)
active(from(X)) → mark(cons(X, from(s(X))))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

MARK(first(X1, X2)) → MARK(X1)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(from(X)) → ACTIVE(from(mark(X)))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))
MARK(first(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

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(first(X1, X2)) → MARK(X1)
ACTIVE(first(s(X), cons(Y, Z))) → MARK(cons(Y, first(X, Z)))
MARK(first(X1, X2)) → MARK(X2)
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)) → ACTIVE(s(mark(X)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(first(X1, X2)) → ACTIVE(first(mark(X1), mark(X2)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 1
POL(ACTIVE(x₁)) = x₁
POL(MARK(x₁)) = x₁
POL(active(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + x₁
POL(first(x₁, x₂)) = 1 + x₁ + x₂
POL(from(x₁)) = 1 + x₁
POL(mark(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

from(active(X)) → from(X)
from(mark(X)) → from(X)
active(first(0, X)) → mark(nil)
mark(0) → active(0)
active(from(X)) → mark(cons(X, from(s(X))))
mark(s(X)) → active(s(mark(X)))
mark(from(X)) → active(from(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(nil) → active(nil)
first(active(X1), X2) → first(X1, X2)
first(mark(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ UsableRulesProof

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

MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(from(X)) → mark(cons(X, from(s(X))))
mark(first(X1, X2)) → active(first(mark(X1), mark(X2)))
mark(0) → active(0)
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(from(X)) → active(from(mark(X)))
first(mark(X1), X2) → first(X1, X2)
first(X1, mark(X2)) → first(X1, X2)
first(active(X1), X2) → first(X1, X2)
first(X1, active(X2)) → first(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)
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)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ DependencyGraphProof

                      ↳ QDP

                        ↳ UsableRulesProof

                          ↳ QDP

                            ↳ QDPSizeChangeProof

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

MARK(s(X)) → MARK(X)

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

MARK(s(X)) → MARK(X)
The graph contains the following edges 1 > 1