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:

eq2(n__0, n__0) -> true
eq2(n__s1(X), n__s1(Y)) -> eq2(activate1(X), activate1(Y))
eq2(X, Y) -> false
inf1(X) -> cons2(X, n__inf1(n__s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(activate1(Y), n__take2(activate1(X), activate1(L)))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(n__length1(activate1(L)))
0 -> n__0
s1(X) -> n__s1(X)
inf1(X) -> n__inf1(X)
take2(X1, X2) -> n__take2(X1, X2)
length1(X) -> n__length1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(n__inf1(X)) -> inf1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
activate1(n__length1(X)) -> length1(activate1(X))
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

eq2(n__0, n__0) -> true
eq2(n__s1(X), n__s1(Y)) -> eq2(activate1(X), activate1(Y))
eq2(X, Y) -> false
inf1(X) -> cons2(X, n__inf1(n__s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(activate1(Y), n__take2(activate1(X), activate1(L)))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(n__length1(activate1(L)))
0 -> n__0
s1(X) -> n__s1(X)
inf1(X) -> n__inf1(X)
take2(X1, X2) -> n__take2(X1, X2)
length1(X) -> n__length1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(n__inf1(X)) -> inf1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
activate1(n__length1(X)) -> length1(activate1(X))
activate1(X) -> X

Q is empty.

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

ACTIVATE1(n__length1(X)) -> LENGTH1(activate1(X))
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(L)
LENGTH1(cons2(X, L)) -> S1(n__length1(activate1(L)))
ACTIVATE1(n__0) -> 01
ACTIVATE1(n__s1(X)) -> S1(X)
ACTIVATE1(n__inf1(X)) -> ACTIVATE1(X)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(X)
EQ2(n__s1(X), n__s1(Y)) -> ACTIVATE1(X)
EQ2(n__s1(X), n__s1(Y)) -> EQ2(activate1(X), activate1(Y))
ACTIVATE1(n__length1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__inf1(X)) -> INF1(activate1(X))
LENGTH1(cons2(X, L)) -> ACTIVATE1(L)
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X1)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(Y)
EQ2(n__s1(X), n__s1(Y)) -> ACTIVATE1(Y)
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X2)
LENGTH1(nil) -> 01
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(activate1(X1), activate1(X2))

The TRS R consists of the following rules:

eq2(n__0, n__0) -> true
eq2(n__s1(X), n__s1(Y)) -> eq2(activate1(X), activate1(Y))
eq2(X, Y) -> false
inf1(X) -> cons2(X, n__inf1(n__s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(activate1(Y), n__take2(activate1(X), activate1(L)))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(n__length1(activate1(L)))
0 -> n__0
s1(X) -> n__s1(X)
inf1(X) -> n__inf1(X)
take2(X1, X2) -> n__take2(X1, X2)
length1(X) -> n__length1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(n__inf1(X)) -> inf1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
activate1(n__length1(X)) -> length1(activate1(X))
activate1(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:

ACTIVATE1(n__length1(X)) -> LENGTH1(activate1(X))
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(L)
LENGTH1(cons2(X, L)) -> S1(n__length1(activate1(L)))
ACTIVATE1(n__0) -> 01
ACTIVATE1(n__s1(X)) -> S1(X)
ACTIVATE1(n__inf1(X)) -> ACTIVATE1(X)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(X)
EQ2(n__s1(X), n__s1(Y)) -> ACTIVATE1(X)
EQ2(n__s1(X), n__s1(Y)) -> EQ2(activate1(X), activate1(Y))
ACTIVATE1(n__length1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__inf1(X)) -> INF1(activate1(X))
LENGTH1(cons2(X, L)) -> ACTIVATE1(L)
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X1)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(Y)
EQ2(n__s1(X), n__s1(Y)) -> ACTIVATE1(Y)
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X2)
LENGTH1(nil) -> 01
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(activate1(X1), activate1(X2))

The TRS R consists of the following rules:

eq2(n__0, n__0) -> true
eq2(n__s1(X), n__s1(Y)) -> eq2(activate1(X), activate1(Y))
eq2(X, Y) -> false
inf1(X) -> cons2(X, n__inf1(n__s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(activate1(Y), n__take2(activate1(X), activate1(L)))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(n__length1(activate1(L)))
0 -> n__0
s1(X) -> n__s1(X)
inf1(X) -> n__inf1(X)
take2(X1, X2) -> n__take2(X1, X2)
length1(X) -> n__length1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(n__inf1(X)) -> inf1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
activate1(n__length1(X)) -> length1(activate1(X))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 7 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
          ↳ QDP

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

ACTIVATE1(n__length1(X)) -> LENGTH1(activate1(X))
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(L)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(Y)
ACTIVATE1(n__inf1(X)) -> ACTIVATE1(X)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(X)
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__length1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(activate1(X1), activate1(X2))
LENGTH1(cons2(X, L)) -> ACTIVATE1(L)
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X1)

The TRS R consists of the following rules:

eq2(n__0, n__0) -> true
eq2(n__s1(X), n__s1(Y)) -> eq2(activate1(X), activate1(Y))
eq2(X, Y) -> false
inf1(X) -> cons2(X, n__inf1(n__s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(activate1(Y), n__take2(activate1(X), activate1(L)))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(n__length1(activate1(L)))
0 -> n__0
s1(X) -> n__s1(X)
inf1(X) -> n__inf1(X)
take2(X1, X2) -> n__take2(X1, X2)
length1(X) -> n__length1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(n__inf1(X)) -> inf1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
activate1(n__length1(X)) -> length1(activate1(X))
activate1(X) -> X

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP

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

EQ2(n__s1(X), n__s1(Y)) -> EQ2(activate1(X), activate1(Y))

The TRS R consists of the following rules:

eq2(n__0, n__0) -> true
eq2(n__s1(X), n__s1(Y)) -> eq2(activate1(X), activate1(Y))
eq2(X, Y) -> false
inf1(X) -> cons2(X, n__inf1(n__s1(X)))
take2(0, X) -> nil
take2(s1(X), cons2(Y, L)) -> cons2(activate1(Y), n__take2(activate1(X), activate1(L)))
length1(nil) -> 0
length1(cons2(X, L)) -> s1(n__length1(activate1(L)))
0 -> n__0
s1(X) -> n__s1(X)
inf1(X) -> n__inf1(X)
take2(X1, X2) -> n__take2(X1, X2)
length1(X) -> n__length1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(n__inf1(X)) -> inf1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
activate1(n__length1(X)) -> length1(activate1(X))
activate1(X) -> X

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