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:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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__from1(X)) -> FROM1(activate1(X))
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
2ND1(cons2(X, XS)) -> HEAD1(activate1(XS))
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(activate1(X1), activate1(X2))
2ND1(cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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__from1(X)) -> FROM1(activate1(X))
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
2ND1(cons2(X, XS)) -> HEAD1(activate1(XS))
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(activate1(X1), activate1(X2))
2ND1(cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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 5 less nodes.

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

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

TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(activate1(X1), activate1(X2))
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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 strictly oriented and are deleted.


ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__take2(X1, X2)) -> TAKE2(activate1(X1), activate1(X2))
ACTIVATE1(n__take2(X1, X2)) -> ACTIVATE1(X1)
The remaining pairs can at least by weakly be oriented.

TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
Used ordering: Combined order from the following AFS and order.
TAKE2(x1, x2)  =  TAKE1(x2)
s1(x1)  =  x1
cons2(x1, x2)  =  x2
ACTIVATE1(x1)  =  ACTIVATE1(x1)
n__take2(x1, x2)  =  n__take2(x1, x2)
n__from1(x1)  =  x1
activate1(x1)  =  x1
n__s1(x1)  =  x1
from1(x1)  =  x1
take2(x1, x2)  =  take2(x1, x2)
0  =  0
nil  =  nil

Lexicographic Path Order [19].
Precedence:
[ntake2, take2] > [TAKE1, ACTIVATE1]
[ntake2, take2] > [0, nil]


The following usable rules [14] were oriented:

activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
activate1(X) -> X
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
take2(X1, X2) -> n__take2(X1, X2)
s1(X) -> n__s1(X)
from1(X) -> cons2(X, n__from1(n__s1(X)))
from1(X) -> n__from1(X)



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

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

TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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 1 SCC with 1 less node.

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

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

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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 strictly oriented and are deleted.


ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The remaining pairs can at least by weakly be oriented.

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVATE1(x1)  =  ACTIVATE1(x1)
n__from1(x1)  =  x1
n__s1(x1)  =  n__s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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 strictly oriented and are deleted.


ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVATE1(x1)  =  ACTIVATE1(x1)
n__from1(x1)  =  n__from1(x1)

Lexicographic Path Order [19].
Precedence:
nfrom1 > ACTIVATE1


The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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
            ↳ QDPOrderProof

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

SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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 strictly oriented and are deleted.


SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SEL2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
cons2(x1, x2)  =  cons
activate1(x1)  =  x1
n__from1(x1)  =  x1
from1(x1)  =  x1
n__s1(x1)  =  n__s1(x1)
n__take2(x1, x2)  =  x1
take2(x1, x2)  =  x1
0  =  0
nil  =  nil

Lexicographic Path Order [19].
Precedence:
[s1, ns1] > cons
[0, nil] > cons


The following usable rules [14] were oriented:

activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
activate1(X) -> X
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
take2(X1, X2) -> n__take2(X1, X2)
s1(X) -> n__s1(X)
from1(X) -> cons2(X, n__from1(n__s1(X)))
from1(X) -> n__from1(X)



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(activate1(XS))
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, n__take2(N, activate1(XS)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__take2(X1, X2)) -> take2(activate1(X1), activate1(X2))
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.