Termination w.r.t. Q proof of TRS/nontermin/CSREx6_15

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:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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:

QUOTE(sel(X, Z)) → SEL1(X, Z)
QUOTE1(first(X, Z)) → FIRST1(X, Z)
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
SEL1(0, cons(X, Z)) → QUOTE(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
UNQUOTE(s1(X)) → UNQUOTE(X)
FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
FROM(X) → FROM(s(X))
QUOTE1(cons(X, Z)) → QUOTE(X)
QUOTE(s(X)) → QUOTE(X)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
QUOTE1(cons(X, Z)) → QUOTE1(Z)
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

QUOTE(sel(X, Z)) → SEL1(X, Z)
QUOTE1(first(X, Z)) → FIRST1(X, Z)
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
SEL1(0, cons(X, Z)) → QUOTE(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
UNQUOTE(s1(X)) → UNQUOTE(X)
FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
FROM(X) → FROM(s(X))
QUOTE1(cons(X, Z)) → QUOTE(X)
QUOTE(s(X)) → QUOTE(X)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
QUOTE1(cons(X, Z)) → QUOTE1(Z)
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

QUOTE(sel(X, Z)) → SEL1(X, Z)
QUOTE1(first(X, Z)) → FIRST1(X, Z)
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
SEL1(0, cons(X, Z)) → QUOTE(X)
UNQUOTE(s1(X)) → UNQUOTE(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
QUOTE1(cons(X, Z)) → QUOTE(X)
FROM(X) → FROM(s(X))
QUOTE(s(X)) → QUOTE(X)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
QUOTE1(cons(X, Z)) → QUOTE1(Z)
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

UNQUOTE(s1(X)) → UNQUOTE(X)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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.

UNQUOTE(s1(X)) → UNQUOTE(X)

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

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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.

UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
UNQUOTE1(x1) = UNQUOTE1(x1)
cons1(x1, x2) = cons1(x1, x2)

Recursive Path Order [2].
Precedence:

cons1₂ > UNQUOTE1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

QUOTE(sel(X, Z)) → SEL1(X, Z)
QUOTE(s(X)) → QUOTE(X)
SEL1(0, cons(X, Z)) → QUOTE(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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.

QUOTE(sel(X, Z)) → SEL1(X, Z)
QUOTE(s(X)) → QUOTE(X)
SEL1(0, cons(X, Z)) → QUOTE(X)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUOTE(x1) = x1
sel(x1, x2) = sel(x2)
SEL1(x1, x2) = x2
s(x1) = s(x1)
0 = 0
cons(x1, x2) = cons(x1, x2)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

FIRST1(s(X), cons(Y, Z)) → FIRST1(X, Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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.

FIRST1(s(X), cons(Y, Z)) → FIRST1(X, Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FIRST1(x1, x2) = FIRST1(x2)
s(x1) = s
cons(x1, x2) = cons(x1, x2)

Recursive Path Order [2].
Precedence:

s > FIRST1₁
cons₂ > FIRST1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

QUOTE1(cons(X, Z)) → QUOTE1(Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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.

QUOTE1(cons(X, Z)) → QUOTE1(Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUOTE1(x1) = QUOTE1(x1)
cons(x1, x2) = cons(x1, x2)

Recursive Path Order [2].
Precedence:

cons₂ > QUOTE1₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

FROM(X) → FROM(s(X))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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.

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

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

Recursive Path Order [2].
Precedence:

s > FIRST₁
cons₂ > FIRST₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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.

SEL(s(X), cons(Y, Z)) → SEL(X, Z)

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

Recursive Path Order [2].
Precedence:

s > SEL₁
cons₂ > SEL₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ 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:

sel(s(X), cons(Y, Z)) → sel(X, Z)
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
from(X) → cons(X, from(s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, Z)
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, Z))
quote(0) → 01
quote1(cons(X, Z)) → cons1(quote(X), quote1(Z))
quote1(nil) → nil1
quote(s(X)) → s1(quote(X))
quote(sel(X, Z)) → sel1(X, Z)
quote1(first(X, Z)) → first1(X, Z)
unquote(01) → 0
unquote(s1(X)) → s(unquote(X))
unquote1(nil1) → nil
unquote1(cons1(X, Z)) → fcons(unquote(X), unquote1(Z))
fcons(X, Z) → cons(X, Z)

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.