Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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:

QUOT2(s1(X), s1(Y)) -> QUOT2(minus2(X, Y), s1(Y))
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ACTIVATE1(n__zWquot2(X1, X2)) -> ZWQUOT2(activate1(X1), activate1(X2))
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> QUOT2(X, Y)
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X1)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X2)
QUOT2(s1(X), s1(Y)) -> S1(quot2(minus2(X, Y), s1(Y)))
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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:

QUOT2(s1(X), s1(Y)) -> QUOT2(minus2(X, Y), s1(Y))
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ACTIVATE1(n__zWquot2(X1, X2)) -> ZWQUOT2(activate1(X1), activate1(X2))
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> QUOT2(X, Y)
ACTIVATE1(n__from1(X)) -> FROM1(activate1(X))
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X1)
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X2)
QUOT2(s1(X), s1(Y)) -> S1(quot2(minus2(X, Y), s1(Y)))
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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 4 SCCs with 6 less nodes.

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

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

MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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 strictly oriented and are deleted.


MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS2(x1, x2)  =  MINUS1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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

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

QUOT2(s1(X), s1(Y)) -> QUOT2(minus2(X, Y), s1(Y))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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 strictly oriented and are deleted.


QUOT2(s1(X), s1(Y)) -> QUOT2(minus2(X, Y), s1(Y))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT2(x1, x2)  =  QUOT1(x1)
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  minus
0  =  0
n__s1(x1)  =  n__s

Lexicographic Path Order [19].
Precedence:
ns > [QUOT1, s1, minus, 0]


The following usable rules [14] were oriented:

minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)



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

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

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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
            ↳ QDPOrderProof
          ↳ QDP

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

ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X1)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ACTIVATE1(n__zWquot2(X1, X2)) -> ZWQUOT2(activate1(X1), activate1(X2))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X2)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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 strictly oriented and are deleted.


ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X1)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(XS)
ACTIVATE1(n__zWquot2(X1, X2)) -> ZWQUOT2(activate1(X1), activate1(X2))
ACTIVATE1(n__zWquot2(X1, X2)) -> ACTIVATE1(X2)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ACTIVATE1(YS)
The remaining pairs can at least by weakly be oriented.

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVATE1(x1)  =  x1
n__zWquot2(x1, x2)  =  n__zWquot2(x1, x2)
ZWQUOT2(x1, x2)  =  ZWQUOT2(x1, x2)
cons2(x1, x2)  =  x2
activate1(x1)  =  x1
n__from1(x1)  =  x1
n__s1(x1)  =  x1
quot2(x1, x2)  =  quot1(x2)
0  =  0
s1(x1)  =  x1
minus2(x1, x2)  =  minus2(x1, x2)
zWquot2(x1, x2)  =  zWquot2(x1, x2)
from1(x1)  =  x1
nil  =  nil

Lexicographic Path Order [19].
Precedence:
[nzWquot2, zWquot2] > [ZWQUOT2, nil]
[quot1, minus2] > [ZWQUOT2, nil]
0 > [ZWQUOT2, nil]


The following usable rules [14] were oriented:

activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(activate1(X1), activate1(X2))
activate1(X) -> X
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
s1(X) -> n__s1(X)
from1(X) -> cons2(X, n__from1(n__s1(X)))
from1(X) -> n__from1(X)



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

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

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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 strictly oriented and are deleted.


ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The remaining pairs can at least by weakly be oriented.

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
Used ordering: Combined order from the following AFS and order.
ACTIVATE1(x1)  =  ACTIVATE1(x1)
n__from1(x1)  =  x1
n__s1(x1)  =  n__s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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 strictly oriented and are deleted.


ACTIVATE1(n__from1(X)) -> ACTIVATE1(X)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACTIVATE1(x1)  =  ACTIVATE1(x1)
n__from1(x1)  =  n__from1(x1)

Lexicographic Path Order [19].
Precedence:
nfrom1 > ACTIVATE1


The following usable rules [14] were oriented: none



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

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

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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
            ↳ QDPOrderProof

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

SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))

The TRS R consists of the following rules:

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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 strictly oriented and are deleted.


SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
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)  =  cons2(x1, x2)
activate1(x1)  =  activate1(x1)
quot2(x1, x2)  =  quot1(x1)
0  =  0
n__s1(x1)  =  x1
minus2(x1, x2)  =  minus1(x2)
n__zWquot2(x1, x2)  =  n__zWquot2(x1, x2)
zWquot2(x1, x2)  =  zWquot1(x2)
from1(x1)  =  from
n__from1(x1)  =  n__from1(x1)
nil  =  nil

Lexicographic Path Order [19].
Precedence:
minus1 > [cons2, quot1, 0, zWquot1]
nzWquot2 > [s1, activate1] > [cons2, quot1, 0, zWquot1]
from > [cons2, quot1, 0, zWquot1]
from > nfrom1


The following usable rules [14] were oriented: none



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

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

from1(X) -> cons2(X, n__from1(n__s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
minus2(X, 0) -> 0
minus2(s1(X), s1(Y)) -> minus2(X, Y)
quot2(0, s1(Y)) -> 0
quot2(s1(X), s1(Y)) -> s1(quot2(minus2(X, Y), s1(Y)))
zWquot2(XS, nil) -> nil
zWquot2(nil, XS) -> nil
zWquot2(cons2(X, XS), cons2(Y, YS)) -> cons2(quot2(X, Y), n__zWquot2(activate1(XS), activate1(YS)))
from1(X) -> n__from1(X)
s1(X) -> n__s1(X)
zWquot2(X1, X2) -> n__zWquot2(X1, X2)
activate1(n__from1(X)) -> from1(activate1(X))
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__zWquot2(X1, X2)) -> zWquot2(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.