Termination w.r.t. Q proof of /tmp/tmpENEYSt/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:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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:

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__from(X)) → FROM(activate(X))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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:

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__from(X)) → FROM(activate(X))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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.

ACTIVATE(n__from(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → ACTIVATE(X)

Used ordering: Polynomial interpretation [25]:

POL(0) = 1
POL(ACTIVATE(x₁)) = x₁
POL(FIRST(x₁, x₂)) = x₂
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = x₂
POL(first(x₁, x₂)) = x₁ + x₂
POL(from(x₁)) = 1 + x₁
POL(n__first(x₁, x₂)) = x₁ + x₂
POL(n__from(x₁)) = 1 + x₁
POL(n__s(x₁)) = x₁
POL(nil) = 0
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
activate(X) → X
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
first(0, X) → nil
from(X) → cons(X, n__from(n__s(X)))
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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.

ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)

The remaining pairs can at least be oriented weakly.

FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__s(X)) → ACTIVATE(X)

Used ordering: Polynomial interpretation [25]:

POL(0) = 1
POL(ACTIVATE(x₁)) = x₁
POL(FIRST(x₁, x₂)) = x₂
POL(activate(x₁)) = x₁
POL(cons(x₁, x₂)) = x₂
POL(first(x₁, x₂)) = 1 + x₁ + x₂
POL(from(x₁)) = 1
POL(n__first(x₁, x₂)) = 1 + x₁ + x₂
POL(n__from(x₁)) = 1
POL(n__s(x₁)) = x₁
POL(nil) = 1
POL(s(x₁)) = x₁

The following usable rules [17] were oriented:

s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
activate(X) → X
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
first(0, X) → nil
from(X) → cons(X, n__from(n__s(X)))
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

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

ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(X) → 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

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

                    ↳ QDP

                      ↳ UsableRulesProof

                        ↳ QDP

                          ↳ QDPSizeChangeProof

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

ACTIVATE(n__s(X)) → ACTIVATE(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:

ACTIVATE(n__s(X)) → ACTIVATE(X)
The graph contains the following edges 1 > 1