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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

from1(x0)
head1(cons2(x0, x1))
2nd1(cons2(x0, x1))
take2(0, x0)
take2(s1(x0), cons2(x1, x2))
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))


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)

The set Q consists of the following terms:

from1(x0)
head1(cons2(x0, x1))
2nd1(cons2(x0, x1))
take2(0, x0)
take2(s1(x0), cons2(x1, x2))
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

from1(x0)
head1(cons2(x0, x1))
2nd1(cons2(x0, x1))
take2(0, x0)
take2(s1(x0), cons2(x1, x2))
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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)

The set Q consists of the following terms:

from1(x0)
head1(cons2(x0, x1))
2nd1(cons2(x0, x1))
take2(0, x0)
take2(s1(x0), cons2(x1, x2))
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
Used argument filtering: SEL2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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)

The set Q consists of the following terms:

from1(x0)
head1(cons2(x0, x1))
2nd1(cons2(x0, x1))
take2(0, x0)
take2(s1(x0), cons2(x1, x2))
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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)

The set Q consists of the following terms:

from1(x0)
head1(cons2(x0, x1))
2nd1(cons2(x0, x1))
take2(0, x0)
take2(s1(x0), cons2(x1, x2))
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

TAKE2(s1(N), cons2(X, XS)) -> TAKE2(N, XS)
Used argument filtering: TAKE2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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)

The set Q consists of the following terms:

from1(x0)
head1(cons2(x0, x1))
2nd1(cons2(x0, x1))
take2(0, x0)
take2(s1(x0), cons2(x1, x2))
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

from1(x0)
head1(cons2(x0, x1))
2nd1(cons2(x0, x1))
take2(0, x0)
take2(s1(x0), cons2(x1, x2))
sel2(0, cons2(x0, x1))
sel2(s1(x0), cons2(x1, x2))

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