Termination w.r.t. Q proof of /tmp/tmpqNuA_R/Ex15_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:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(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:

ADD(s(X), Y) → S(n__add(activate(X), activate(Y)))
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ACTIVATE(n__s(X)) → S(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
IF(true, X, Y) → ACTIVATE(X)
FROM(X) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
AND(true, X) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(0, X) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(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:

ADD(s(X), Y) → S(n__add(activate(X), activate(Y)))
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ACTIVATE(n__s(X)) → S(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
IF(true, X, Y) → ACTIVATE(X)
FROM(X) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
AND(true, X) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(0, X) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)
FROM(X) → ACTIVATE(X)
ADD(0, X) → ACTIVATE(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(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.

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

The remaining pairs can at least be oriented weakly.

ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(X)
ADD(0, X) → ACTIVATE(X)

Used ordering: Polynomial interpretation [25,35]:

POL(n__first(x₁, x₂)) = 2 + (2)x_1 + (4)x_2
POL(FIRST(x₁, x₂)) = 2 + (4)x_1 + (4)x_2
POL(cons(x₁, x₂)) = (3)x_1 + (3)x_2
POL(n__add(x₁, x₂)) = x_1 + (4)x_2
POL(n__from(x₁)) = 4 + (3)x_1
POL(s(x₁)) = (4)x_1
POL(ADD(x₁, x₂)) = (2)x_1 + (4)x_2
POL(0) = 0
POL(ACTIVATE(x₁)) = (2)x_1
POL(FROM(x₁)) = 4 + (2)x_1

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(0, X) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(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.

ADD(s(X), Y) → ACTIVATE(Y)
ADD(s(X), Y) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(0, X) → ACTIVATE(X)

Used ordering: Polynomial interpretation [25,35]:

POL(n__add(x₁, x₂)) = (4)x_1 + (2)x_2
POL(s(x₁)) = 4 + (2)x_1
POL(ADD(x₁, x₂)) = x_1 + (3)x_2
POL(0) = 0
POL(ACTIVATE(x₁)) = (2)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

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

ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(0, X) → ACTIVATE(X)

The TRS R consists of the following rules:

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(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__add(X1, X2)) → ADD(X1, X2)
ADD(0, X) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(n__add(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(ADD(x₁, x₂)) = 1 + (4)x_1 + x_2
POL(0) = 4
POL(ACTIVATE(x₁)) = 4 + x_1

The value of delta used in the strict ordering is 7.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ PisEmptyProof

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

and(true, X) → activate(X)
and(false, Y) → false
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
add(0, X) → activate(X)
add(s(X), Y) → s(n__add(activate(X), activate(Y)))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(activate(Y), n__first(activate(X), activate(Z)))
from(X) → cons(activate(X), n__from(n__s(activate(X))))
add(X1, X2) → n__add(X1, X2)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → 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.