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 oriented strictly 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 be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = max{0, x1 - 3}


POL( cons2(x1, x2) ) = 2x1 + x2


POL( ACTIVATE1(x1) ) = 2x1 + 1


POL( n__take2(x1, x2) ) = x1 + 2x2 + 2


POL( TAKE2(x1, x2) ) = 2x2 + 3



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 oriented strictly and are deleted.


SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( from1(x1) ) = 3x1 + 2


POL( SEL2(x1, x2) ) = 3x1 + 3


POL( n__from1(x1) ) = max{0, -3}


POL( s1(x1) ) = 2x1 + 3


POL( 0 ) = 2


POL( nil ) = 2


POL( cons2(x1, x2) ) = max{0, x1 - 3}


POL( activate1(x1) ) = max{0, 2x1 - 3}


POL( take2(x1, x2) ) = max{0, x1 + x2 - 1}


POL( n__take2(x1, x2) ) = 2x1 + 2x2 + 1



The following usable rules [14] were oriented: none



↳ 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.