Termination w.r.t. Q proof of TRS/TRCSREx1_GL02a_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:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(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:

ACTIVATE(n__length(X)) → LENGTH(activate(X))
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)
LENGTH(cons(X, L)) → S(n__length(activate(L)))
ACTIVATE(n__0) → 0¹
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__inf(X)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
EQ(n__s(X), n__s(Y)) → ACTIVATE(X)
EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y))
ACTIVATE(n__length(X)) → ACTIVATE(X)
ACTIVATE(n__inf(X)) → INF(activate(X))
LENGTH(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
EQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
LENGTH(nil) → 0¹
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(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:

ACTIVATE(n__length(X)) → LENGTH(activate(X))
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)
LENGTH(cons(X, L)) → S(n__length(activate(L)))
ACTIVATE(n__0) → 0¹
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__inf(X)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
EQ(n__s(X), n__s(Y)) → ACTIVATE(X)
EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y))
ACTIVATE(n__length(X)) → ACTIVATE(X)
ACTIVATE(n__inf(X)) → INF(activate(X))
LENGTH(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
EQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
LENGTH(nil) → 0¹
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(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:

ACTIVATE(n__length(X)) → LENGTH(activate(X))
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)
LENGTH(cons(X, L)) → S(n__length(activate(L)))
ACTIVATE(n__0) → 0¹
ACTIVATE(n__inf(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
EQ(n__s(X), n__s(Y)) → ACTIVATE(X)
EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y))
ACTIVATE(n__length(X)) → ACTIVATE(X)
ACTIVATE(n__inf(X)) → INF(activate(X))
LENGTH(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
EQ(n__s(X), n__s(Y)) → ACTIVATE(Y)
LENGTH(nil) → 0¹
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(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 2 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

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

ACTIVATE(n__length(X)) → LENGTH(activate(X))
TAKE(s(X), cons(Y, L)) → ACTIVATE(L)
TAKE(s(X), cons(Y, L)) → ACTIVATE(Y)
ACTIVATE(n__inf(X)) → ACTIVATE(X)
TAKE(s(X), cons(Y, L)) → ACTIVATE(X)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__length(X)) → ACTIVATE(X)
ACTIVATE(n__take(X1, X2)) → TAKE(activate(X1), activate(X2))
LENGTH(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__take(X1, X2)) → ACTIVATE(X1)

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

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

EQ(n__s(X), n__s(Y)) → EQ(activate(X), activate(Y))

The TRS R consists of the following rules:

eq(n__0, n__0) → true
eq(n__s(X), n__s(Y)) → eq(activate(X), activate(Y))
eq(X, Y) → false
inf(X) → cons(X, n__inf(n__s(X)))
take(0, X) → nil
take(s(X), cons(Y, L)) → cons(activate(Y), n__take(activate(X), activate(L)))
length(nil) → 0
length(cons(X, L)) → s(n__length(activate(L)))
0 → n__0
s(X) → n__s(X)
inf(X) → n__inf(X)
take(X1, X2) → n__take(X1, X2)
length(X) → n__length(X)
activate(n__0) → 0
activate(n__s(X)) → s(X)
activate(n__inf(X)) → inf(activate(X))
activate(n__take(X1, X2)) → take(activate(X1), activate(X2))
activate(n__length(X)) → length(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.