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(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)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, 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(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)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, 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__take2(X1, X2)) -> TAKE2(X1, X2)
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
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__from1(X)) -> FROM1(X)
2ND1(cons2(X, XS)) -> ACTIVATE1(XS)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(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)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, 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__take2(X1, X2)) -> TAKE2(X1, X2)
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
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__from1(X)) -> FROM1(X)
2ND1(cons2(X, XS)) -> ACTIVATE1(XS)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(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)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, 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 4 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)
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(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)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, 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)) -> TAKE2(X1, X2)
TAKE2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVATE1(x1)  =  x1
n__take2(x1, x2)  =  n__take2(x1, x2)
TAKE2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial

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:

from1(X) -> cons2(X, n__from1(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)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, 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(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)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, 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__take2(x1, x2)  =  x1
take2(x1, x2)  =  x1
0  =  0
nil  =  nil

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

The following usable rules [14] were oriented:

activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, 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)
from1(X) -> cons2(X, n__from1(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(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)
take2(X1, X2) -> n__take2(X1, X2)
activate1(n__from1(X)) -> from1(X)
activate1(n__take2(X1, X2)) -> take2(X1, 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.