Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/AG01SLASHHASH3.57.trs

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:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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))
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> APP₂(l, sum₁(cons₂(x, cons₂(y, k))))
SUM₁(cons₂(x, cons₂(y, l))) -> PLUS₂(x, y)
MINUS₂(minus₂(x, y), z) -> PLUS₂(y, z)
SUM₁(cons₂(x, cons₂(y, l))) -> SUM₁(cons₂(plus₂(x, y), l))
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(cons₂(x, cons₂(y, k)))
MINUS₂(minus₂(x, y), z) -> MINUS₂(x, plus₂(y, z))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
APP₂(cons₂(x, l), k) -> APP₂(l, k)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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))
PLUS₂(s₁(x), y) -> PLUS₂(x, y)
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> APP₂(l, sum₁(cons₂(x, cons₂(y, k))))
SUM₁(cons₂(x, cons₂(y, l))) -> PLUS₂(x, y)
MINUS₂(minus₂(x, y), z) -> PLUS₂(y, z)
SUM₁(cons₂(x, cons₂(y, l))) -> SUM₁(cons₂(plus₂(x, y), l))
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(cons₂(x, cons₂(y, k)))
MINUS₂(minus₂(x, y), z) -> MINUS₂(x, plus₂(y, z))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
QUOT₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))
APP₂(cons₂(x, l), k) -> APP₂(l, k)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 6 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

APP₂(cons₂(x, l), k) -> APP₂(l, k)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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.

APP₂(cons₂(x, l), k) -> APP₂(l, k)

Used argument filtering: APP₂(x1, x2) = x1
cons₂(x1, x2) = cons₁(x2)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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.

PLUS₂(s₁(x), y) -> PLUS₂(x, y)

Used argument filtering: PLUS₂(x1, x2) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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

          ↳ QDP

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

SUM₁(cons₂(x, cons₂(y, l))) -> SUM₁(cons₂(plus₂(x, y), l))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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.

SUM₁(cons₂(x, cons₂(y, l))) -> SUM₁(cons₂(plus₂(x, y), l))

Used argument filtering: SUM₁(x1) = x1
cons₂(x1, x2) = cons₁(x2)
plus₂(x1, x2) = x2
0 = 0
s₁(x1) = s
Used ordering: Precedence:

cons₁ > s

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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

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

SUM₁(app₂(l, cons₂(x, cons₂(y, k)))) -> SUM₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

The TRS R consists of the following rules:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

MINUS₂(minus₂(x, y), z) -> MINUS₂(x, plus₂(y, z))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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(minus₂(x, y), z) -> MINUS₂(x, plus₂(y, z))

Used argument filtering: MINUS₂(x1, x2) = x1
minus₂(x1, x2) = minus₁(x1)
s₁(x1) = x1
plus₂(x1, x2) = x2
0 = 0
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ 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:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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

            ↳ QDPAfsSolverProof

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:

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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) = s₁(x1)
minus₂(x1, x2) = x1
plus₂(x1, x2) = plus₂(x1, x2)
0 = 0
Used ordering: Precedence:

plus₂ > s₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

minus₂(x, 0) -> x
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)))
plus₂(0, y) -> y
plus₂(s₁(x), y) -> s₁(plus₂(x, y))
minus₂(minus₂(x, y), z) -> minus₂(x, plus₂(y, z))
app₂(nil, k) -> k
app₂(l, nil) -> l
app₂(cons₂(x, l), k) -> cons₂(x, app₂(l, k))
sum₁(cons₂(x, nil)) -> cons₂(x, nil)
sum₁(cons₂(x, cons₂(y, l))) -> sum₁(cons₂(plus₂(x, y), l))
sum₁(app₂(l, cons₂(x, cons₂(y, k)))) -> sum₁(app₂(l, sum₁(cons₂(x, cons₂(y, k)))))

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.