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, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(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:

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__0) → 01
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))
UNQUOTE(01) → 01
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X1)
QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))
ACTIVATE(n__nil) → NIL
SEL1(0, cons(X, Z)) → QUOTE(X)
QUOTE1(n__first(X, Z)) → FIRST1(activate(X), activate(Z))
QUOTE(n__sel(X, Z)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
FIRST(0, Z) → NIL
UNQUOTE1(nil1) → NIL
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
QUOTE1(n__first(X, Z)) → ACTIVATE(X)
QUOTE1(n__cons(X, Z)) → ACTIVATE(Z)
FCONS(X, Z) → CONS(X, Z)
QUOTE1(n__cons(X, Z)) → QUOTE(activate(X))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__sel(X1, X2)) → SEL(activate(X1), activate(X2))
QUOTE(n__sel(X, Z)) → ACTIVATE(Z)
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
QUOTE1(n__cons(X, Z)) → ACTIVATE(X)
UNQUOTE(s1(X)) → S(unquote(X))
UNQUOTE(s1(X)) → UNQUOTE(X)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
SEL1(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__s(X)) → S(activate(X))
FIRST1(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X2)
FROM(X) → CONS(X, n__from(n__s(X)))
QUOTE(n__s(X)) → QUOTE(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
QUOTE(n__s(X)) → ACTIVATE(X)
QUOTE1(n__first(X, Z)) → ACTIVATE(Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)
FIRST(s(X), cons(Y, Z)) → CONS(Y, n__first(X, activate(Z)))
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__0) → 01
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))
UNQUOTE(01) → 01
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X1)
QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))
ACTIVATE(n__nil) → NIL
SEL1(0, cons(X, Z)) → QUOTE(X)
QUOTE1(n__first(X, Z)) → FIRST1(activate(X), activate(Z))
QUOTE(n__sel(X, Z)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
FIRST(0, Z) → NIL
UNQUOTE1(nil1) → NIL
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
QUOTE1(n__first(X, Z)) → ACTIVATE(X)
QUOTE1(n__cons(X, Z)) → ACTIVATE(Z)
FCONS(X, Z) → CONS(X, Z)
QUOTE1(n__cons(X, Z)) → QUOTE(activate(X))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__sel(X1, X2)) → SEL(activate(X1), activate(X2))
QUOTE(n__sel(X, Z)) → ACTIVATE(Z)
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
QUOTE1(n__cons(X, Z)) → ACTIVATE(X)
UNQUOTE(s1(X)) → S(unquote(X))
UNQUOTE(s1(X)) → UNQUOTE(X)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
SEL1(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__s(X)) → S(activate(X))
FIRST1(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X2)
FROM(X) → CONS(X, n__from(n__s(X)))
QUOTE(n__s(X)) → QUOTE(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
QUOTE(n__s(X)) → ACTIVATE(X)
QUOTE1(n__first(X, Z)) → ACTIVATE(Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)
FIRST(s(X), cons(Y, Z)) → CONS(Y, n__first(X, activate(Z)))
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))
ACTIVATE(n__0) → 01
UNQUOTE(01) → 01
QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X1)
QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))
ACTIVATE(n__nil) → NIL
SEL1(0, cons(X, Z)) → QUOTE(X)
QUOTE1(n__first(X, Z)) → FIRST1(activate(X), activate(Z))
QUOTE(n__sel(X, Z)) → ACTIVATE(X)
FIRST(0, Z) → NIL
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
UNQUOTE1(nil1) → NIL
QUOTE1(n__first(X, Z)) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
QUOTE1(n__cons(X, Z)) → ACTIVATE(Z)
QUOTE1(n__cons(X, Z)) → QUOTE(activate(X))
FCONS(X, Z) → CONS(X, Z)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__sel(X1, X2)) → SEL(activate(X1), activate(X2))
QUOTE(n__sel(X, Z)) → ACTIVATE(Z)
QUOTE1(n__cons(X, Z)) → ACTIVATE(X)
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
UNQUOTE(s1(X)) → S(unquote(X))
UNQUOTE(s1(X)) → UNQUOTE(X)
SEL1(s(X), cons(Y, Z)) → ACTIVATE(Z)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
ACTIVATE(n__from(X)) → FROM(activate(X))
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
FIRST1(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__s(X)) → S(activate(X))
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
SEL1(s(X), cons(Y, Z)) → SEL1(X, activate(Z))
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X2)
QUOTE(n__s(X)) → QUOTE(activate(X))
FROM(X) → CONS(X, n__from(n__s(X)))
ACTIVATE(n__from(X)) → ACTIVATE(X)
QUOTE1(n__first(X, Z)) → ACTIVATE(Z)
QUOTE(n__s(X)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → CONS(Y, n__first(X, activate(Z)))
UNQUOTE1(cons1(X, Z)) → UNQUOTE(X)
ACTIVATE(n__cons(X1, X2)) → CONS(activate(X1), X2)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(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 26 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ 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, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(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.


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

Lexicographic path order with status [19].
Quasi-Precedence:
[UNQUOTE1, s11]

Status:
s11: [1]
UNQUOTE1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ 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, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(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(cons1(X, Z)) → UNQUOTE1(Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
UNQUOTE1(x1)  =  UNQUOTE1(x1)
cons1(x1, x2)  =  cons1(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:
cons12 > UNQUOTE11

Status:
cons12: [2,1]
UNQUOTE11: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

ACTIVATE(n__first(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__cons(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__first(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__sel(X1, X2)) → SEL(activate(X1), activate(X2))
SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__first(X1, X2)) → FIRST(activate(X1), activate(X2))
ACTIVATE(n__sel(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__from(X)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
              ↳ QDP
              ↳ QDP

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

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

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP

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

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

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(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.


FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
FIRST1(x1, x2)  =  x1
s(x1)  =  s(x1)
cons(x1, x2)  =  x1
activate(x1)  =  activate(x1)
first(x1, x2)  =  first
0  =  0
nil  =  nil
n__first(x1, x2)  =  x1
n__s(x1)  =  n__s
n__0  =  n__0
sel(x1, x2)  =  sel(x1)
n__sel(x1, x2)  =  n__sel
from(x1)  =  from(x1)
n__from(x1)  =  n__from(x1)
n__nil  =  n__nil
n__cons(x1, x2)  =  n__cons(x2)

Lexicographic path order with status [19].
Quasi-Precedence:
[nil, nnil]
ns > s1 > activate1 > first
ns > s1 > activate1 > 0
ns > s1 > activate1 > [sel1, nsel]
ns > s1 > activate1 > from1 > nfrom1
n0 > 0

Status:
from1: [1]
nsel: []
nnil: multiset
n0: multiset
nfrom1: [1]
first: []
0: multiset
nil: multiset
ncons1: [1]
activate1: [1]
s1: [1]
ns: []
sel1: [1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ 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:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP

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

QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))

The TRS R consists of the following rules:

sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
sel(0, cons(X, Z)) → X
first(0, Z) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
sel1(s(X), cons(Y, Z)) → sel1(X, activate(Z))
sel1(0, cons(X, Z)) → quote(X)
first1(0, Z) → nil1
first1(s(X), cons(Y, Z)) → cons1(quote(Y), first1(X, activate(Z)))
quote(n__0) → 01
quote1(n__cons(X, Z)) → cons1(quote(activate(X)), quote1(activate(Z)))
quote1(n__nil) → nil1
quote(n__s(X)) → s1(quote(activate(X)))
quote(n__sel(X, Z)) → sel1(activate(X), activate(Z))
quote1(n__first(X, Z)) → first1(activate(X), activate(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)
first(X1, X2) → n__first(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(activate(X1), X2)
activate(n__nil) → nil
activate(n__sel(X1, X2)) → sel(activate(X1), activate(X2))
activate(X) → X

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