Termination w.r.t. Q proof of /tmp/tmpn_PeNi/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

          ↳ UsableRulesProof

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 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

          ↳ UsableRulesProof

            ↳ QDP

              ↳ QDPSizeChangeProof

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

FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
ADD(s(X), Y) → ACTIVATE(Y)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ACTIVATE(n__from(X)) → FROM(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(0, X) → ACTIVATE(X)
FROM(X) → ACTIVATE(X)
ADD(s(X), Y) → 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__first(X1, X2)) → FIRST(X1, X2)
The graph contains the following edges 1 > 1, 1 > 2
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
The graph contains the following edges 1 > 1, 1 > 2
ACTIVATE(n__from(X)) → FROM(X)
The graph contains the following edges 1 > 1
FROM(X) → ACTIVATE(X)
The graph contains the following edges 1 >= 1
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Y)
The graph contains the following edges 2 > 1
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The graph contains the following edges 2 > 1
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
The graph contains the following edges 1 > 1
ADD(s(X), Y) → ACTIVATE(Y)
The graph contains the following edges 2 >= 1
ADD(s(X), Y) → ACTIVATE(X)
The graph contains the following edges 1 > 1
ADD(0, X) → ACTIVATE(X)
The graph contains the following edges 2 >= 1