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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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:

QUOTE1(sel2(X, Z)) -> SEL12(X, Z)
QUOTE11(first2(X, Z)) -> FIRST12(X, Z)
UNQUOTE11(cons12(X, Z)) -> FCONS2(unquote1(X), unquote11(Z))
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
FIRST12(s1(X), cons2(Y, Z)) -> QUOTE1(Y)
FROM1(X) -> FROM1(s1(X))
QUOTE11(cons2(X, Z)) -> QUOTE1(X)
QUOTE1(s1(X)) -> QUOTE1(X)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, Z)
QUOTE11(cons2(X, Z)) -> QUOTE11(Z)
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE1(X)

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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:

QUOTE1(sel2(X, Z)) -> SEL12(X, Z)
QUOTE11(first2(X, Z)) -> FIRST12(X, Z)
UNQUOTE11(cons12(X, Z)) -> FCONS2(unquote1(X), unquote11(Z))
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
UNQUOTE1(s11(X)) -> UNQUOTE1(X)
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
FIRST12(s1(X), cons2(Y, Z)) -> QUOTE1(Y)
FROM1(X) -> FROM1(s1(X))
QUOTE11(cons2(X, Z)) -> QUOTE1(X)
QUOTE1(s1(X)) -> QUOTE1(X)
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, Z)
QUOTE11(cons2(X, Z)) -> QUOTE11(Z)
FIRST12(s1(X), cons2(Y, Z)) -> FIRST12(X, Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE11(Z)
UNQUOTE11(cons12(X, Z)) -> UNQUOTE1(X)

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 8 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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
          ↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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
          ↳ QDP
          ↳ QDP

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

QUOTE1(sel2(X, Z)) -> SEL12(X, Z)
QUOTE1(s1(X)) -> QUOTE1(X)
SEL12(0, cons2(X, Z)) -> QUOTE1(X)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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.


SEL12(0, cons2(X, Z)) -> QUOTE1(X)
SEL12(s1(X), cons2(Y, Z)) -> SEL12(X, Z)
The remaining pairs can at least by weakly be oriented.

QUOTE1(sel2(X, Z)) -> SEL12(X, Z)
QUOTE1(s1(X)) -> QUOTE1(X)
Used ordering: Combined order from the following AFS and order.
QUOTE1(x1)  =  x1
sel2(x1, x2)  =  x2
SEL12(x1, x2)  =  x2
s1(x1)  =  x1
0  =  0
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

QUOTE1(sel2(X, Z)) -> SEL12(X, Z)
QUOTE1(s1(X)) -> QUOTE1(X)

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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

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

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

QUOTE1(s1(X)) -> QUOTE1(X)

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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.


QUOTE1(s1(X)) -> QUOTE1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOTE1(x1)  =  QUOTE1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > QUOTE1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ DependencyGraphProof
                  ↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

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

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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, 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)  =  x1
s1(x1)  =  s1(x1)
cons2(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

QUOTE11(cons2(X, Z)) -> QUOTE11(Z)

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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.


QUOTE11(cons2(X, Z)) -> QUOTE11(Z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOTE11(x1)  =  QUOTE11(x1)
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
cons2 > QUOTE11


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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
          ↳ QDP
          ↳ QDP

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

FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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

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

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

FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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.


FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
FIRST2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
cons2(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ 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, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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
          ↳ QDP
QDP
            ↳ QDPOrderProof

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

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

The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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)) -> SEL2(X, Z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
SEL2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
cons2(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sel2(s1(X), cons2(Y, Z)) -> sel2(X, Z)
sel2(0, cons2(X, Z)) -> X
first2(0, Z) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, first2(X, Z))
from1(X) -> cons2(X, from1(s1(X)))
sel12(s1(X), cons2(Y, Z)) -> sel12(X, Z)
sel12(0, cons2(X, Z)) -> quote1(X)
first12(0, Z) -> nil1
first12(s1(X), cons2(Y, Z)) -> cons12(quote1(Y), first12(X, Z))
quote1(0) -> 01
quote11(cons2(X, Z)) -> cons12(quote1(X), quote11(Z))
quote11(nil) -> nil1
quote1(s1(X)) -> s11(quote1(X))
quote1(sel2(X, Z)) -> sel12(X, Z)
quote11(first2(X, Z)) -> first12(X, 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)

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.