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 strictly oriented and are deleted.


UNQUOTE1(s11(X)) -> UNQUOTE1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
UNQUOTE1(x1)  =  UNQUOTE1(x1)
s11(x1)  =  s11(x1)

Lexicographic Path Order [19].
Precedence:
s11 > UNQUOTE1


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 strictly oriented and are deleted.


UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
UNQUOTE11(x1)  =  UNQUOTE11(x1)
cons12(x1, x2)  =  cons12(x1, x2)

Lexicographic Path Order [19].
Precedence:
cons12 > UNQUOTE11


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 strictly oriented and are deleted.


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 remaining pairs can at least by weakly be oriented.

ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
Used ordering: Combined order from the following AFS and order.
SEL2(x1, x2)  =  SEL2(x1, x2)
s1(x1)  =  s1(x1)
cons2(x1, x2)  =  cons2(x1, x2)
ACTIVATE1(x1)  =  x1
n__sel2(x1, x2)  =  n__sel2(x1, x2)
activate1(x1)  =  activate1(x1)
n__first2(x1, x2)  =  x2
FIRST2(x1, x2)  =  x2
first2(x1, x2)  =  first1(x2)
n__from1(x1)  =  n__from1(x1)
from1(x1)  =  from1(x1)
n__0  =  n__0
0  =  0
n__cons2(x1, x2)  =  n__cons2(x1, x2)
n__nil  =  n__nil
nil  =  nil
n__s1(x1)  =  x1
sel2(x1, x2)  =  sel2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[nsel2, sel2] > SEL2 > [activate1, first1, from1]


The following usable rules [14] were oriented:

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

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

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 0 SCCs with 1 less node.

↳ 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 strictly oriented and are deleted.


FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, activate1(Z))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
FIRST12(x1, x2)  =  FIRST12(x1, x2)
s1(x1)  =  s1(x1)
cons2(x1, x2)  =  cons2(x1, x2)
activate1(x1)  =  activate1(x1)
n__first2(x1, x2)  =  n__first2(x1, x2)
first2(x1, x2)  =  first2(x1, x2)
n__from1(x1)  =  x1
from1(x1)  =  from1(x1)
n__0  =  n__0
0  =  0
n__cons2(x1, x2)  =  n__cons2(x1, x2)
n__nil  =  n__nil
nil  =  nil
n__s1(x1)  =  x1
n__sel2(x1, x2)  =  n__sel2(x1, x2)
sel2(x1, x2)  =  sel2(x1, x2)

Lexicographic Path Order [19].
Precedence:
FIRST12 > [activate1, from1] > [s1, cons2, ncons2]
[nfirst2, first2] > [activate1, from1] > [s1, cons2, ncons2]
[n0, 0, nnil, nil] > [s1, cons2, ncons2]
[nsel2, sel2] > [activate1, from1] > [s1, cons2, ncons2]


The following usable rules [14] were oriented:

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



↳ 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.