Termination w.r.t. Q proof of /tmp/tmp_QCzN9/Ex6_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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)
SEL(s(X), cons(Y, Z)) → SEL(X, Z)
FROM(X) → FROM(s(X))
QUOTE(s(X)) → QUOTE(X)
FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, Z)
UNQUOTE(s1(X)) → UNQUOTE(X)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
QUOTE1(cons(X, Z)) → QUOTE1(Z)
QUOTE1(cons(X, Z)) → QUOTE(X)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
QUOTE(sel(X, Z)) → SEL1(X, Z)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
SEL1(0, cons(X, Z)) → QUOTE(X)
QUOTE1(first(X, 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.
The approximation of the Dependency Graph [15,17,22] contains 8 SCCs with 5 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:

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 [15].

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: Polynomial interpretation [25,35]:

POL(s1(x₁)) = 1/4 + (7/2)x_1
POL(UNQUOTE(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

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

      ↳ 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 [15].

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: Polynomial interpretation [25,35]:

POL(cons1(x₁, x₂)) = 1/4 + (7/2)x_2
POL(UNQUOTE1(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

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

      ↳ 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)
SEL1(s(X), cons(Y, Z)) → SEL1(X, Z)
SEL1(0, cons(X, Z)) → QUOTE(X)
QUOTE(s(X)) → QUOTE(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 [15].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

QUOTE(sel(X, Z)) → SEL1(X, Z)

Used ordering: Polynomial interpretation [25,35]:

POL(sel(x₁, x₂)) = 1/4 + x_1 + x_2
POL(SEL1(x₁, x₂)) = 4 + (1/2)x_2
POL(cons(x₁, x₂)) = 5/4 + (2)x_1 + (4)x_2
POL(s(x₁)) = 4 + (4)x_1
POL(0) = 9/4
POL(QUOTE(x₁)) = 15/4 + x_1

The value of delta used in the strict ordering is 5/8.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

QUOTE(sel(X, 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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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 [15].

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: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = (13/4)x_2
POL(FIRST1(x₁, x₂)) = (3)x_1 + (15/4)x_2
POL(s(x₁)) = 3/4 + (9/4)x_1

The value of delta used in the strict ordering is 9/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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

      ↳ 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 [15].

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: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = 1/4 + (7/2)x_2
POL(QUOTE1(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

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

      ↳ 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

      ↳ 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 [15].

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: Polynomial interpretation [25,35]:

POL(FIRST(x₁, x₂)) = (3)x_1 + (15/4)x_2
POL(cons(x₁, x₂)) = (13/4)x_2
POL(s(x₁)) = 3/4 + (9/4)x_1

The value of delta used in the strict ordering is 9/4.
The following usable rules [17] 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:

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

      ↳ 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 [15].

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: Polynomial interpretation [25,35]:

POL(cons(x₁, x₂)) = (13/4)x_2
POL(s(x₁)) = 1/4 + (9/4)x_1
POL(SEL(x₁, x₂)) = (1/2)x_1 + (15/4)x_2

The value of delta used in the strict ordering is 1/8.
The following usable rules [17] 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:

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.