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(n__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)
s1(X) -> n__s1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__nil) -> nil
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(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(n__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)
s1(X) -> n__s1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__nil) -> nil
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(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__first2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__0) -> 01
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, activate1(Z))
UNQUOTE1(01) -> 01
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X1)
QUOTE1(n__sel2(X, Z)) -> SEL12(activate1(X), activate1(Z))
QUOTE11(n__cons2(X, Z)) -> QUOTE11(activate1(Z))
ACTIVATE1(n__nil) -> NIL
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
QUOTE11(n__first2(X, Z)) -> FIRST12(activate1(X), activate1(Z))
QUOTE1(n__sel2(X, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X2)
FIRST2(0, Z) -> NIL
UNQUOTE11(nil1) -> NIL
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(activate1(X1), activate1(X2))
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)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__sel2(X1, X2)) -> SEL2(activate1(X1), activate1(X2))
QUOTE1(n__sel2(X, Z)) -> ACTIVATE1(Z)
UNQUOTE11(cons12(X, Z)) -> FCONS2(unquote1(X), unquote11(Z))
QUOTE11(n__cons2(X, Z)) -> ACTIVATE1(X)
UNQUOTE1(s11(X)) -> S1(unquote1(X))
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
FIRST12(s1(X), cons2(Y, Z)) -> QUOTE1(Y)
SEL12(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
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))
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X2)
FROM1(X) -> CONS2(X, n__from1(n__s1(X)))
QUOTE1(n__s1(X)) -> QUOTE1(activate1(X))
ACTIVATE1(n__from1(X)) -> 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)))
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(activate1(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(n__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)
s1(X) -> n__s1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__nil) -> nil
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(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__first2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__0) -> 01
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, activate1(Z))
UNQUOTE1(01) -> 01
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X1)
QUOTE1(n__sel2(X, Z)) -> SEL12(activate1(X), activate1(Z))
QUOTE11(n__cons2(X, Z)) -> QUOTE11(activate1(Z))
ACTIVATE1(n__nil) -> NIL
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
QUOTE11(n__first2(X, Z)) -> FIRST12(activate1(X), activate1(Z))
QUOTE1(n__sel2(X, Z)) -> ACTIVATE1(X)
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X2)
FIRST2(0, Z) -> NIL
UNQUOTE11(nil1) -> NIL
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(activate1(X1), activate1(X2))
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)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__sel2(X1, X2)) -> SEL2(activate1(X1), activate1(X2))
QUOTE1(n__sel2(X, Z)) -> ACTIVATE1(Z)
UNQUOTE11(cons12(X, Z)) -> FCONS2(unquote1(X), unquote11(Z))
QUOTE11(n__cons2(X, Z)) -> ACTIVATE1(X)
UNQUOTE1(s11(X)) -> S1(unquote1(X))
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
FIRST12(s1(X), cons2(Y, Z)) -> QUOTE1(Y)
SEL12(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
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))
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X2)
FROM1(X) -> CONS2(X, n__from1(n__s1(X)))
QUOTE1(n__s1(X)) -> QUOTE1(activate1(X))
ACTIVATE1(n__from1(X)) -> 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)))
ACTIVATE1(n__cons2(X1, X2)) -> CONS2(activate1(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(n__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)
s1(X) -> n__s1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__nil) -> nil
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(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 26 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(n__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)
s1(X) -> n__s1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__nil) -> nil
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(X2))
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


UNQUOTE1(s11(X)) -> UNQUOTE1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( UNQUOTE1(x1) ) = max{0, x1 - 2}


POL( s11(x1) ) = x1 + 3



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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( UNQUOTE11(x1) ) = max{0, x1 - 2}


POL( cons12(x1, x2) ) = x2 + 3



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

ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__cons2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__first2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__sel2(X1, X2)) -> SEL2(activate1(X1), activate1(X2))
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(activate1(X1), activate1(X2))
ACTIVATE1(n__sel2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(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(n__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)
s1(X) -> n__s1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__nil) -> nil
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(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
          ↳ 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(n__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)
s1(X) -> n__s1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__nil) -> nil
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(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
          ↳ 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(n__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)
s1(X) -> n__s1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__nil) -> nil
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(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
          ↳ 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(n__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)
s1(X) -> n__s1(X)
0 -> n__0
cons2(X1, X2) -> n__cons2(X1, X2)
nil -> n__nil
sel2(X1, X2) -> n__sel2(X1, X2)
activate1(n__first2(X1, X2)) -> first2(activate1(X1), activate1(X2))
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__cons2(X1, X2)) -> cons2(activate1(X1), X2)
activate1(n__nil) -> nil
activate1(n__sel2(X1, X2)) -> sel2(activate1(X1), activate1(X2))
activate1(X) -> X

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