Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/nonterminSLASHCSRSLASHEx5_7

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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

Q is empty.

↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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:

QUOTE₁(sel₂(X, Y)) -> SEL1₂(X, Y)
DBL1₁(s₁(X)) -> DBL1₁(X)
QUOTE₁(s₁(X)) -> QUOTE₁(X)
INDX₂(cons₂(X, Y), Z) -> SEL₂(X, Z)
QUOTE₁(dbl₁(X)) -> DBL1₁(X)
DBLS₁(cons₂(X, Y)) -> DBL₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)
SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, Z)
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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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:

QUOTE₁(sel₂(X, Y)) -> SEL1₂(X, Y)
DBL1₁(s₁(X)) -> DBL1₁(X)
QUOTE₁(s₁(X)) -> QUOTE₁(X)
INDX₂(cons₂(X, Y), Z) -> SEL₂(X, Z)
QUOTE₁(dbl₁(X)) -> DBL1₁(X)
DBLS₁(cons₂(X, Y)) -> DBL₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)
SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, Z)
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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 8 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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 strictly oriented and are deleted.

SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, Z)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SEL1₂(x1, x2) = x1
s₁(x1) = s₁(x1)
cons₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DBL1₁(s₁(X)) -> DBL1₁(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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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 strictly oriented and are deleted.

DBL1₁(s₁(X)) -> DBL1₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DBL1₁(x1) = DBL1₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > DBL1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

QUOTE₁(s₁(X)) -> QUOTE₁(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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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 strictly oriented and are deleted.

QUOTE₁(s₁(X)) -> QUOTE₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOTE₁(x1) = QUOTE₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > QUOTE₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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

          ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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 strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SEL₂(x1, x2) = x1
s₁(x1) = s₁(x1)
cons₂(x1, x2) = x1

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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 strictly oriented and are deleted.

INDX₂(cons₂(X, Y), Z) -> INDX₂(Y, Z)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
INDX₂(x1, x2) = INDX₁(x1)
cons₂(x1, x2) = cons₁(x2)

Lexicographic Path Order [19].
Precedence:

[INDX₁, cons_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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 strictly oriented and are deleted.

DBL₁(s₁(X)) -> DBL₁(X)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DBL₁(x1) = DBL₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > DBL₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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 strictly oriented and are deleted.

DBLS₁(cons₂(X, Y)) -> DBLS₁(Y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DBLS₁(x1) = DBLS₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

cons₂ > DBLS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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)))
dbl1₁(0) -> 01
dbl1₁(s₁(X)) -> s1₁(s1₁(dbl1₁(X)))
sel1₂(0, cons₂(X, Y)) -> X
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
quote₁(0) -> 01
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(dbl₁(X)) -> dbl1₁(X)
quote₁(sel₂(X, Y)) -> sel1₂(X, Y)

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.