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:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(X, Z)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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:

QUOTE₁(sel₂(X, Z)) -> SEL1₂(X, Z)
QUOTE1₁(first₂(X, Z)) -> FIRST1₂(X, Z)
UNQUOTE1₁(cons1₂(X, Z)) -> FCONS₂(unquote₁(X), unquote1₁(Z))
SEL1₂(0, cons₂(X, Z)) -> QUOTE₁(X)
SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, Z)
UNQUOTE₁(s1₁(X)) -> UNQUOTE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)
FIRST1₂(s₁(X), cons₂(Y, Z)) -> QUOTE₁(Y)
FROM₁(X) -> FROM₁(s₁(X))
QUOTE1₁(cons₂(X, Z)) -> QUOTE₁(X)
QUOTE₁(s₁(X)) -> QUOTE₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
QUOTE1₁(cons₂(X, Z)) -> QUOTE1₁(Z)
FIRST1₂(s₁(X), cons₂(Y, Z)) -> FIRST1₂(X, Z)
UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE1₁(Z)
UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE₁(X)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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:

QUOTE₁(sel₂(X, Z)) -> SEL1₂(X, Z)
QUOTE1₁(first₂(X, Z)) -> FIRST1₂(X, Z)
UNQUOTE1₁(cons1₂(X, Z)) -> FCONS₂(unquote₁(X), unquote1₁(Z))
SEL1₂(0, cons₂(X, Z)) -> QUOTE₁(X)
SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, Z)
UNQUOTE₁(s1₁(X)) -> UNQUOTE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)
FIRST1₂(s₁(X), cons₂(Y, Z)) -> QUOTE₁(Y)
FROM₁(X) -> FROM₁(s₁(X))
QUOTE1₁(cons₂(X, Z)) -> QUOTE₁(X)
QUOTE₁(s₁(X)) -> QUOTE₁(X)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, Z)
QUOTE1₁(cons₂(X, Z)) -> QUOTE1₁(Z)
FIRST1₂(s₁(X), cons₂(Y, Z)) -> FIRST1₂(X, Z)
UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE1₁(Z)
UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE₁(X)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

UNQUOTE₁(s1₁(X)) -> UNQUOTE₁(X)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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(s1₁(X)) -> UNQUOTE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(UNQUOTE₁(x₁)) = x₁
POL(s1₁(x₁)) = 1 + x₁

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:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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:

UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE1₁(Z)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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.

UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE1₁(Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(UNQUOTE1₁(x₁)) = x₁
POL(cons1₂(x₁, x₂)) = 1 + x₂

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:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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:

QUOTE₁(sel₂(X, Z)) -> SEL1₂(X, Z)
QUOTE₁(s₁(X)) -> QUOTE₁(X)
SEL1₂(0, cons₂(X, Z)) -> QUOTE₁(X)
SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, Z)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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(sel₂(X, Z)) -> SEL1₂(X, Z)

The remaining pairs can at least be oriented weakly.

QUOTE₁(s₁(X)) -> QUOTE₁(X)
SEL1₂(0, cons₂(X, Z)) -> QUOTE₁(X)
SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, Z)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(QUOTE₁(x₁)) = x₁
POL(SEL1₂(x₁, x₂)) = x₂
POL(cons₂(x₁, x₂)) = x₁ + x₂
POL(s₁(x₁)) = x₁
POL(sel₂(x₁, x₂)) = 1 + x₂

The following usable rules [14] 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:

QUOTE₁(s₁(X)) -> QUOTE₁(X)
SEL1₂(0, cons₂(X, Z)) -> QUOTE₁(X)
SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, Z)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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 2 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                      ↳ QDPOrderProof

                    ↳ QDP

          ↳ 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:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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(s₁(X)) -> QUOTE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(QUOTE₁(x₁)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ PisEmptyProof

                    ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

                      ↳ QDPOrderProof

          ↳ 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:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SEL1₂(x₁, x₂)) = x₂
POL(cons₂(x₁, x₂)) = 1 + x₂
POL(s₁(x₁)) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ AND

                    ↳ QDP

                    ↳ QDP

                      ↳ QDPOrderProof

                        ↳ QDP

                          ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

FIRST1₂(s₁(X), cons₂(Y, Z)) -> FIRST1₂(X, Z)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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.

FIRST1₂(s₁(X), cons₂(Y, Z)) -> FIRST1₂(X, Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FIRST1₂(x₁, x₂)) = x₂
POL(cons₂(x₁, x₂)) = 1 + x₂
POL(s₁(x₁)) = 0

The following usable rules [14] 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:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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:

QUOTE1₁(cons₂(X, Z)) -> QUOTE1₁(Z)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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.

QUOTE1₁(cons₂(X, Z)) -> QUOTE1₁(Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(QUOTE1₁(x₁)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

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:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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:

FROM₁(X) -> FROM₁(s₁(X))

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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:

FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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.

FIRST₂(s₁(X), cons₂(Y, Z)) -> FIRST₂(X, Z)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FIRST₂(x₁, x₂)) = x₂
POL(cons₂(x₁, x₂)) = 1 + x₂
POL(s₁(x₁)) = 0

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:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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:

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

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(SEL₂(x₁, x₂)) = x₂
POL(cons₂(x₁, x₂)) = 1 + x₂
POL(s₁(x₁)) = 0

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:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, Z)
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
from₁(X) -> cons₂(X, from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, Z)
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, Z))
quote₁(0) -> 01
quote1₁(cons₂(X, Z)) -> cons1₂(quote₁(X), quote1₁(Z))
quote1₁(nil) -> nil1
quote₁(s₁(X)) -> s1₁(quote₁(X))
quote₁(sel₂(X, Z)) -> sel1₂(X, Z)
quote1₁(first₂(X, Z)) -> first1₂(X, Z)
unquote₁(01) -> 0
unquote₁(s1₁(X)) -> s₁(unquote₁(X))
unquote1₁(nil1) -> nil
unquote1₁(cons1₂(X, Z)) -> fcons₂(unquote₁(X), unquote1₁(Z))
fcons₂(X, Z) -> cons₂(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.