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(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(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
activate1(n__length1(X)) -> length1(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(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(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
activate1(n__length1(X)) -> length1(X)
activate1(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:

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

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(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(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
activate1(n__length1(X)) -> length1(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__take2(X1, X2)) -> TAKE2(X1, X2)
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)
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))
LENGTH1(cons2(X, L)) -> ACTIVATE1(L)
INF1(X) -> S1(X)
ACTIVATE1(n__length1(X)) -> LENGTH1(X)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(Y)
ACTIVATE1(n__inf1(X)) -> INF1(X)
EQ2(n__s1(X), n__s1(Y)) -> ACTIVATE1(Y)
LENGTH1(nil) -> 01

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(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(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
activate1(n__length1(X)) -> length1(X)
activate1(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 8 less nodes.

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

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

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

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(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(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
activate1(n__length1(X)) -> length1(X)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__take2(X1, X2)) -> TAKE2(X1, X2)
ACTIVATE1(n__length1(X)) -> LENGTH1(X)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(L)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(Y)
TAKE2(s1(X), cons2(Y, L)) -> ACTIVATE1(X)
LENGTH1(cons2(X, L)) -> ACTIVATE1(L)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE1(x1)) = 1 + 2·x1   
POL(LENGTH1(x1)) = 2 + 2·x1   
POL(TAKE2(x1, x2)) = 2 + 2·x1 + x2   
POL(cons2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(n__length1(x1)) = 1 + 2·x1   
POL(n__take2(x1, x2)) = 2 + x1 + 2·x2   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
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(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(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
activate1(n__length1(X)) -> length1(X)
activate1(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.

↳ 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(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(X)
activate1(n__take2(X1, X2)) -> take2(X1, X2)
activate1(n__length1(X)) -> length1(X)
activate1(X) -> X

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