Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/nonterminSLASHCSRSLASHEx3_2

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:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

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:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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

INDX₂(cons₂(X, Y), Z) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBL₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)
DBL₁(s₁(X)) -> DBL₁(X)
FROM₁(X) -> FROM₁(s₁(X))
INDX₂(cons₂(X, Y), Z) -> INDX₂(Y, Z)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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:

INDX₂(cons₂(X, Y), Z) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBL₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)
DBL₁(s₁(X)) -> DBL₁(X)
FROM₁(X) -> FROM₁(s₁(X))
INDX₂(cons₂(X, Y), Z) -> INDX₂(Y, Z)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ 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:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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(X), cons₂(Y, Z)) -> SEL₂(X, Z)

Used argument filtering: SEL₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

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

INDX₂(cons₂(X, Y), Z) -> INDX₂(Y, Z)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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.

INDX₂(cons₂(X, Y), Z) -> INDX₂(Y, Z)

Used argument filtering: INDX₂(x1, x2) = x1
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

DBL₁(s₁(X)) -> DBL₁(X)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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.

DBL₁(s₁(X)) -> DBL₁(X)

Used argument filtering: DBL₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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

DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)

The TRS R consists of the following rules:

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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.

DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)

Used argument filtering: DBLS₁(x1) = x1
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

dbl₁(0) -> 0
dbl₁(s₁(X)) -> s₁(s₁(dbl₁(X)))
dbls₁(nil) -> nil
dbls₁(cons₂(X, Y)) -> cons₂(dbl₁(X), dbls₁(Y))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
indx₂(nil, X) -> nil
indx₂(cons₂(X, Y), Z) -> cons₂(sel₂(X, Z), indx₂(Y, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))

The set Q consists of the following terms:

dbl₁(0)
dbl₁(s₁(x0))
dbls₁(nil)
dbls₁(cons₂(x0, x1))
sel₂(0, cons₂(x0, x1))
sel₂(s₁(x0), cons₂(x1, x2))
indx₂(nil, x0)
indx₂(cons₂(x0, x1), x2)
from₁(x0)

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