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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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__0) → 01
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))
ACTIVATE(n__s(X)) → S(X)
UNQUOTE(01) → 01
QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))
ACTIVATE(n__nil) → NIL
ACTIVATE(n__from(X)) → FROM(X)
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__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
UNQUOTE1(nil1) → NIL
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)
FROM(X) → CONS(X, n__from(s(X)))
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
QUOTE(n__sel(X, Z)) → ACTIVATE(Z)
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
QUOTE1(n__cons(X, Z)) → ACTIVATE(X)
FROM(X) → S(X)
UNQUOTE(s1(X)) → S(unquote(X))
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
UNQUOTE(s1(X)) → UNQUOTE(X)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
SEL1(s(X), cons(Y, Z)) → ACTIVATE(Z)
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))
QUOTE(n__s(X)) → QUOTE(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)))

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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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__0) → 01
FIRST1(s(X), cons(Y, Z)) → FIRST1(X, activate(Z))
ACTIVATE(n__s(X)) → S(X)
UNQUOTE(01) → 01
QUOTE(n__sel(X, Z)) → SEL1(activate(X), activate(Z))
QUOTE1(n__cons(X, Z)) → QUOTE1(activate(Z))
ACTIVATE(n__nil) → NIL
ACTIVATE(n__from(X)) → FROM(X)
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__cons(X1, X2)) → CONS(X1, X2)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
UNQUOTE1(nil1) → NIL
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)
FROM(X) → CONS(X, n__from(s(X)))
UNQUOTE1(cons1(X, Z)) → UNQUOTE1(Z)
QUOTE(n__sel(X, Z)) → ACTIVATE(Z)
UNQUOTE1(cons1(X, Z)) → FCONS(unquote(X), unquote1(Z))
QUOTE1(n__cons(X, Z)) → ACTIVATE(X)
FROM(X) → S(X)
UNQUOTE(s1(X)) → S(unquote(X))
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
UNQUOTE(s1(X)) → UNQUOTE(X)
FIRST1(s(X), cons(Y, Z)) → QUOTE(Y)
SEL1(s(X), cons(Y, Z)) → ACTIVATE(Z)
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))
QUOTE(n__s(X)) → QUOTE(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)))

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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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:

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

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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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 27 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

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

SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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.


SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The remaining pairs can at least be oriented weakly.

ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1, x2)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)
ACTIVATE(x1)  =  x1
n__sel(x1, x2)  =  n__sel(x1, x2)
activate(x1)  =  activate(x1)
n__first(x1, x2)  =  x2
FIRST(x1, x2)  =  x2
n__s(x1)  =  x1
first(x1, x2)  =  first(x2)
n__0  =  n__0
0  =  0
n__nil  =  n__nil
nil  =  nil
from(x1)  =  from(x1)
n__from(x1)  =  n__from(x1)
sel(x1, x2)  =  sel(x1, x2)
n__cons(x1, x2)  =  n__cons(x1, x2)

Lexicographic path order with status [19].
Quasi-Precedence:
[SEL2, nsel2, sel2] > [activate1, first1, nil] > 0 > n0
[SEL2, nsel2, sel2] > [activate1, first1, nil] > nnil
[SEL2, nsel2, sel2] > [activate1, first1, nil] > from1 > s1 > cons2 > ncons2
[SEL2, nsel2, sel2] > [activate1, first1, nil] > from1 > nfrom1

Status:
sel2: [1,2]
from1: [1]
nnil: multiset
n0: multiset
nfrom1: [1]
0: multiset
nsel2: [1,2]
nil: multiset
cons2: [2,1]
activate1: [1]
SEL2: [1,2]
first1: [1]
s1: [1]
ncons2: [1,2]


The following usable rules [14] were oriented:

s(X) → n__s(X)
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
activate(n__0) → 0
activate(n__nil) → nil
activate(X) → X
first(0, Z) → nil
0n__0
from(X) → n__from(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__cons(X1, X2)) → cons(X1, X2)
first(X1, X2) → n__first(X1, X2)
cons(X1, X2) → n__cons(X1, X2)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
sel(0, cons(X, Z)) → X
from(X) → cons(X, n__from(s(X)))
niln__nil
activate(n__s(X)) → s(X)
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
activate(n__from(X)) → from(X)



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

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

ACTIVATE(n__sel(X1, X2)) → SEL(X1, X2)
ACTIVATE(n__first(X1, X2)) → FIRST(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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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 0 SCCs with 2 less nodes.

↳ 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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)  =  cons
activate(x1)  =  activate
n__s(x1)  =  n__s
first(x1, x2)  =  first
n__first(x1, x2)  =  n__first(x2)
n__0  =  n__0
0  =  0
n__nil  =  n__nil
nil  =  nil
from(x1)  =  x1
n__from(x1)  =  n__from
sel(x1, x2)  =  sel(x1)
n__sel(x1, x2)  =  n__sel(x1, x2)
n__cons(x1, x2)  =  n__cons(x2)

Lexicographic path order with status [19].
Quasi-Precedence:
[cons, sel1] > activate > nil > [nfirst1, nfrom]
[cons, sel1] > nsel2 > [nfirst1, nfrom]
[cons, sel1] > ncons1 > [nfirst1, nfrom]
ns > s1 > activate > nil > [nfirst1, nfrom]
first > activate > nil > [nfirst1, nfrom]
0 > n0 > [nfirst1, nfrom]
0 > nil > [nfirst1, nfrom]
nnil > nil > [nfirst1, nfrom]

Status:
nnil: multiset
n0: multiset
first: []
0: multiset
nsel2: [2,1]
nil: multiset
ncons1: [1]
cons: []
nfrom: []
activate: []
s1: [1]
ns: []
sel1: [1]
nfirst1: [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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, 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(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)
0n__0
cons(X1, X2) → n__cons(X1, X2)
niln__nil
s(X) → n__s(X)
sel(X1, X2) → n__sel(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(n__from(X)) → from(X)
activate(n__0) → 0
activate(n__cons(X1, X2)) → cons(X1, X2)
activate(n__nil) → nil
activate(n__s(X)) → s(X)
activate(n__sel(X1, X2)) → sel(X1, X2)
activate(X) → X

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