Termination w.r.t. Q proof of TRS/TRCSREx6_15_AEL02

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, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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:

ACTIVATE₁(n__0) -> 0¹
FIRST1₂(s₁(X), cons₂(Y, Z)) -> FIRST1₂(X, activate₁(Z))
ACTIVATE₁(n__s₁(X)) -> S₁(X)
UNQUOTE₁(01) -> 0¹
QUOTE₁(n__sel₂(X, Z)) -> SEL1₂(activate₁(X), activate₁(Z))
QUOTE1₁(n__cons₂(X, Z)) -> QUOTE1₁(activate₁(Z))
ACTIVATE₁(n__nil) -> NIL
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
SEL1₂(0, cons₂(X, Z)) -> QUOTE₁(X)
QUOTE1₁(n__first₂(X, Z)) -> FIRST1₂(activate₁(X), activate₁(Z))
QUOTE₁(n__sel₂(X, Z)) -> ACTIVATE₁(X)
FIRST₂(0, Z) -> NIL
ACTIVATE₁(n__cons₂(X1, X2)) -> CONS₂(X1, X2)
ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)
UNQUOTE1₁(nil1) -> NIL
QUOTE1₁(n__first₂(X, Z)) -> ACTIVATE₁(X)
QUOTE1₁(n__cons₂(X, Z)) -> ACTIVATE₁(Z)
FCONS₂(X, Z) -> CONS₂(X, Z)
QUOTE1₁(n__cons₂(X, Z)) -> QUOTE₁(activate₁(X))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
FROM₁(X) -> CONS₂(X, n__from₁(s₁(X)))
UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE1₁(Z)
QUOTE₁(n__sel₂(X, Z)) -> ACTIVATE₁(Z)
UNQUOTE1₁(cons1₂(X, Z)) -> FCONS₂(unquote₁(X), unquote1₁(Z))
QUOTE1₁(n__cons₂(X, Z)) -> ACTIVATE₁(X)
FROM₁(X) -> S₁(X)
UNQUOTE₁(s1₁(X)) -> S₁(unquote₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
UNQUOTE₁(s1₁(X)) -> UNQUOTE₁(X)
FIRST1₂(s₁(X), cons₂(Y, Z)) -> QUOTE₁(Y)
SEL1₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
FIRST1₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, activate₁(Z))
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
QUOTE₁(n__s₁(X)) -> QUOTE₁(activate₁(X))
QUOTE₁(n__s₁(X)) -> ACTIVATE₁(X)
QUOTE1₁(n__first₂(X, Z)) -> ACTIVATE₁(Z)
UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> CONS₂(Y, n__first₂(X, activate₁(Z)))

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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:

ACTIVATE₁(n__0) -> 0¹
FIRST1₂(s₁(X), cons₂(Y, Z)) -> FIRST1₂(X, activate₁(Z))
ACTIVATE₁(n__s₁(X)) -> S₁(X)
UNQUOTE₁(01) -> 0¹
QUOTE₁(n__sel₂(X, Z)) -> SEL1₂(activate₁(X), activate₁(Z))
QUOTE1₁(n__cons₂(X, Z)) -> QUOTE1₁(activate₁(Z))
ACTIVATE₁(n__nil) -> NIL
ACTIVATE₁(n__from₁(X)) -> FROM₁(X)
SEL1₂(0, cons₂(X, Z)) -> QUOTE₁(X)
QUOTE1₁(n__first₂(X, Z)) -> FIRST1₂(activate₁(X), activate₁(Z))
QUOTE₁(n__sel₂(X, Z)) -> ACTIVATE₁(X)
FIRST₂(0, Z) -> NIL
ACTIVATE₁(n__cons₂(X1, X2)) -> CONS₂(X1, X2)
ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)
UNQUOTE1₁(nil1) -> NIL
QUOTE1₁(n__first₂(X, Z)) -> ACTIVATE₁(X)
QUOTE1₁(n__cons₂(X, Z)) -> ACTIVATE₁(Z)
FCONS₂(X, Z) -> CONS₂(X, Z)
QUOTE1₁(n__cons₂(X, Z)) -> QUOTE₁(activate₁(X))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
FROM₁(X) -> CONS₂(X, n__from₁(s₁(X)))
UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE1₁(Z)
QUOTE₁(n__sel₂(X, Z)) -> ACTIVATE₁(Z)
UNQUOTE1₁(cons1₂(X, Z)) -> FCONS₂(unquote₁(X), unquote1₁(Z))
QUOTE1₁(n__cons₂(X, Z)) -> ACTIVATE₁(X)
FROM₁(X) -> S₁(X)
UNQUOTE₁(s1₁(X)) -> S₁(unquote₁(X))
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
UNQUOTE₁(s1₁(X)) -> UNQUOTE₁(X)
FIRST1₂(s₁(X), cons₂(Y, Z)) -> QUOTE₁(Y)
SEL1₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
FIRST1₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, activate₁(Z))
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
QUOTE₁(n__s₁(X)) -> QUOTE₁(activate₁(X))
QUOTE₁(n__s₁(X)) -> ACTIVATE₁(X)
QUOTE1₁(n__first₂(X, Z)) -> ACTIVATE₁(Z)
UNQUOTE1₁(cons1₂(X, Z)) -> UNQUOTE₁(X)
FIRST₂(s₁(X), cons₂(Y, Z)) -> CONS₂(Y, n__first₂(X, activate₁(Z)))

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 6 SCCs with 27 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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)) = 2·x₁
POL(s1₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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

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, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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)) = 2·x₁
POL(cons1₂(x₁, x₂)) = 1 + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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

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

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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.

ACTIVATE₁(n__first₂(X1, X2)) -> FIRST₂(X1, X2)

The remaining pairs can at least be oriented weakly.

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(ACTIVATE₁(x₁)) = 2·x₁
POL(FIRST₂(x₁, x₂)) = 2·x₂
POL(SEL₂(x₁, x₂)) = 2·x₂
POL(activate₁(x₁)) = 2·x₁
POL(cons₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(first₂(x₁, x₂)) = 2 + 2·x₂
POL(from₁(x₁)) = 2·x₁
POL(n__0) = 0
POL(n__cons₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(n__first₂(x₁, x₂)) = 1 + x₂
POL(n__from₁(x₁)) = 2·x₁
POL(n__nil) = 0
POL(n__s₁(x₁)) = 0
POL(n__sel₂(x₁, x₂)) = 2·x₂
POL(nil) = 0
POL(s₁(x₁)) = 0
POL(sel₂(x₁, x₂)) = 2·x₂

The following usable rules [14] were oriented:

activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
sel₂(0, cons₂(X, Z)) -> X
activate₁(n__0) -> 0
from₁(X) -> n__from₁(X)
nil -> n__nil
activate₁(n__from₁(X)) -> from₁(X)
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
activate₁(n__s₁(X)) -> s₁(X)
s₁(X) -> n__s₁(X)
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
0 -> n__0
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__nil) -> nil
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(X) -> X
first₂(0, Z) -> nil
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
FIRST₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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.

ACTIVATE₁(n__sel₂(X1, X2)) -> SEL₂(X1, X2)

The remaining pairs can at least be oriented weakly.

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(ACTIVATE₁(x₁)) = x₁
POL(SEL₂(x₁, x₂)) = 2·x₂
POL(activate₁(x₁)) = 2·x₁
POL(cons₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(first₂(x₁, x₂)) = 2·x₂
POL(from₁(x₁)) = 2·x₁
POL(n__0) = 0
POL(n__cons₂(x₁, x₂)) = 2·x₁ + 2·x₂
POL(n__first₂(x₁, x₂)) = x₂
POL(n__from₁(x₁)) = x₁
POL(n__nil) = 0
POL(n__s₁(x₁)) = 0
POL(n__sel₂(x₁, x₂)) = 2 + 2·x₂
POL(nil) = 0
POL(s₁(x₁)) = 0
POL(sel₂(x₁, x₂)) = 2 + 2·x₂

The following usable rules [14] were oriented:

activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
sel₂(0, cons₂(X, Z)) -> X
activate₁(n__0) -> 0
from₁(X) -> n__from₁(X)
nil -> n__nil
activate₁(n__from₁(X)) -> from₁(X)
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
activate₁(n__s₁(X)) -> s₁(X)
s₁(X) -> n__s₁(X)
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
0 -> n__0
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__nil) -> nil
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(X) -> X
first₂(0, Z) -> nil
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
first₂(X1, X2) -> n__first₂(X1, X2)
cons₂(X1, X2) -> n__cons₂(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ QDP

                            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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, activate₁(Z))

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

POL(0) = 0
POL(SEL₂(x₁, x₂)) = 2·x₁
POL(activate₁(x₁)) = 0
POL(cons₂(x₁, x₂)) = 0
POL(first₂(x₁, x₂)) = 0
POL(from₁(x₁)) = 0
POL(n__0) = 0
POL(n__cons₂(x₁, x₂)) = 0
POL(n__first₂(x₁, x₂)) = 0
POL(n__from₁(x₁)) = 0
POL(n__nil) = 0
POL(n__s₁(x₁)) = 0
POL(n__sel₂(x₁, x₂)) = 0
POL(nil) = 0
POL(s₁(x₁)) = 2 + 2·x₁
POL(sel₂(x₁, x₂)) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ DependencyGraphProof

                          ↳ 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, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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

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

SEL1₂(s₁(X), cons₂(Y, Z)) -> SEL1₂(X, activate₁(Z))
QUOTE₁(n__sel₂(X, Z)) -> SEL1₂(activate₁(X), activate₁(Z))
QUOTE₁(n__s₁(X)) -> QUOTE₁(activate₁(X))
SEL1₂(0, cons₂(X, Z)) -> QUOTE₁(X)

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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

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

FIRST1₂(s₁(X), cons₂(Y, Z)) -> FIRST1₂(X, activate₁(Z))

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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, activate₁(Z))

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

POL(0) = 0
POL(FIRST1₂(x₁, x₂)) = 2·x₁
POL(activate₁(x₁)) = 0
POL(cons₂(x₁, x₂)) = 0
POL(first₂(x₁, x₂)) = 0
POL(from₁(x₁)) = 0
POL(n__0) = 0
POL(n__cons₂(x₁, x₂)) = 0
POL(n__first₂(x₁, x₂)) = 0
POL(n__from₁(x₁)) = 0
POL(n__nil) = 0
POL(n__s₁(x₁)) = 0
POL(n__sel₂(x₁, x₂)) = 0
POL(nil) = 0
POL(s₁(x₁)) = 2 + 2·x₁
POL(sel₂(x₁, x₂)) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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

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

QUOTE1₁(n__cons₂(X, Z)) -> QUOTE1₁(activate₁(Z))

The TRS R consists of the following rules:

sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
sel₂(0, cons₂(X, Z)) -> X
first₂(0, Z) -> nil
first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, n__first₂(X, activate₁(Z)))
from₁(X) -> cons₂(X, n__from₁(s₁(X)))
sel1₂(s₁(X), cons₂(Y, Z)) -> sel1₂(X, activate₁(Z))
sel1₂(0, cons₂(X, Z)) -> quote₁(X)
first1₂(0, Z) -> nil1
first1₂(s₁(X), cons₂(Y, Z)) -> cons1₂(quote₁(Y), first1₂(X, activate₁(Z)))
quote₁(n__0) -> 01
quote1₁(n__cons₂(X, Z)) -> cons1₂(quote₁(activate₁(X)), quote1₁(activate₁(Z)))
quote1₁(n__nil) -> nil1
quote₁(n__s₁(X)) -> s1₁(quote₁(activate₁(X)))
quote₁(n__sel₂(X, Z)) -> sel1₂(activate₁(X), activate₁(Z))
quote1₁(n__first₂(X, Z)) -> first1₂(activate₁(X), activate₁(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)
first₂(X1, X2) -> n__first₂(X1, X2)
from₁(X) -> n__from₁(X)
0 -> n__0
cons₂(X1, X2) -> n__cons₂(X1, X2)
nil -> n__nil
s₁(X) -> n__s₁(X)
sel₂(X1, X2) -> n__sel₂(X1, X2)
activate₁(n__first₂(X1, X2)) -> first₂(X1, X2)
activate₁(n__from₁(X)) -> from₁(X)
activate₁(n__0) -> 0
activate₁(n__cons₂(X1, X2)) -> cons₂(X1, X2)
activate₁(n__nil) -> nil
activate₁(n__s₁(X)) -> s₁(X)
activate₁(n__sel₂(X1, X2)) -> sel₂(X1, X2)
activate₁(X) -> X

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