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:

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
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__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
2ND(cons(X, XS)) → HEAD(activate(XS))
ACTIVATE(n__from(X)) → FROM(X)
2ND(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
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__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
2ND(cons(X, XS)) → HEAD(activate(XS))
ACTIVATE(n__from(X)) → FROM(X)
2ND(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
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__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
SEL(s(N), cons(X, XS)) → ACTIVATE(XS)
SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
2ND(cons(X, XS)) → HEAD(activate(XS))
ACTIVATE(n__from(X)) → FROM(X)
2ND(cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
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 4 less nodes.

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

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

ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(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.


TAKE(s(N), cons(X, XS)) → ACTIVATE(XS)
The remaining pairs can at least be oriented weakly.

ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__take(x1, x2)  =  x2
TAKE(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
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 0 SCCs with 1 less node.

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

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

SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))

The TRS R consists of the following rules:

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(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.


SEL(s(N), cons(X, XS)) → SEL(N, activate(XS))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



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

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

from(X) → cons(X, n__from(s(X)))
head(cons(X, XS)) → X
2nd(cons(X, XS)) → head(activate(XS))
take(0, XS) → nil
take(s(N), cons(X, XS)) → cons(X, n__take(N, activate(XS)))
sel(0, cons(X, XS)) → X
sel(s(N), cons(X, XS)) → sel(N, activate(XS))
from(X) → n__from(X)
take(X1, X2) → n__take(X1, X2)
activate(n__from(X)) → from(X)
activate(n__take(X1, X2)) → take(X1, X2)
activate(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.