Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/nonterminSLASHCSRSLASHEx4_7_37

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: 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: 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: 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: 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.