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:

from1(X) -> cons2(X, from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(XS)
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, take2(N, XS))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

from1(X) -> cons2(X, from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(XS)
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, take2(N, XS))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)

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:

SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
TAKE2(s1(N), cons2(X, XS)) -> TAKE2(N, XS)
2ND1(cons2(X, XS)) -> HEAD1(XS)
FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(XS)
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, take2(N, XS))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)

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:

SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
TAKE2(s1(N), cons2(X, XS)) -> TAKE2(N, XS)
2ND1(cons2(X, XS)) -> HEAD1(XS)
FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(XS)
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, take2(N, XS))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 1 less node.

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

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

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

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(XS)
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, take2(N, XS))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)

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, XS)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( SEL2(x1, x2) ) = max{0, x2 - 1}


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



The following usable rules [14] were oriented: none



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

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

from1(X) -> cons2(X, from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(XS)
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, take2(N, XS))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)

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
          ↳ QDP

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

TAKE2(s1(N), cons2(X, XS)) -> TAKE2(N, XS)

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(XS)
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, take2(N, XS))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)

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.


TAKE2(s1(N), cons2(X, XS)) -> TAKE2(N, XS)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( TAKE2(x1, x2) ) = max{0, x2 - 1}


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



The following usable rules [14] were oriented: none



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

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

from1(X) -> cons2(X, from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(XS)
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, take2(N, XS))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)

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
          ↳ QDP

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

FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
head1(cons2(X, XS)) -> X
2nd1(cons2(X, XS)) -> head1(XS)
take2(0, XS) -> nil
take2(s1(N), cons2(X, XS)) -> cons2(X, take2(N, XS))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, XS)

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