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:

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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))
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ZWQUOT2(XS, YS)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> QUOT2(X, Y)
FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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))
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
QUOT2(s1(X), s1(Y)) -> MINUS2(X, Y)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ZWQUOT2(XS, YS)
ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> QUOT2(X, Y)
FROM1(X) -> FROM1(s1(X))

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ 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, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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
          ↳ QDP

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

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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

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, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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)  =  x1
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  minus
0  =  0

Lexicographic Path Order [19].
Precedence:
[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
          ↳ QDP

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

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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

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

ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ZWQUOT2(XS, YS)

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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.


ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ZWQUOT2(XS, YS)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ZWQUOT2(x1, x2)  =  x1
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
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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

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

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

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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, 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)  =  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

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

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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

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

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

The TRS R consists of the following rules:

from1(X) -> cons2(X, from1(s1(X)))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, 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), zWquot2(XS, YS))

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