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.

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 contains 5 SCCs with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS2(s1(X), s1(Y)) -> MINUS2(X, Y)
Used argument filtering: MINUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

QUOT2(s1(X), s1(Y)) -> QUOT2(minus2(X, Y), s1(Y))
Used argument filtering: QUOT2(x1, x2)  =  x1
s1(x1)  =  s
minus2(x1, x2)  =  minus
0  =  0
Used ordering: Quasi Precedence: s > minus > 0


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ZWQUOT2(cons2(X, XS), cons2(Y, YS)) -> ZWQUOT2(XS, YS)
Used argument filtering: ZWQUOT2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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
            ↳ QDPAfsSolverProof
          ↳ 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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SEL2(s1(N), cons2(X, XS)) -> SEL2(N, XS)
Used argument filtering: SEL2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
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.