Termination w.r.t. Q proof of TRS/TRCSREx15_Luc98_FR.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(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(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(X)
activate(n__s(X)) → s(X)
activate(X) → X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(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

      ↳ EdgeDeletionProof

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

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

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(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(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 deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

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(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(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 [13,14,18] contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

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

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

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(activate(X1), X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(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.