Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar4/RubioSLASHquick.trs

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:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

Q is empty.

↳ QTRS

  ↳ Non-Overlap Check

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

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

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

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

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

HIGH₂(N, cons₂(M, L)) -> IFHIGH₃(le₂(M, N), N, cons₂(M, L))
QUICKSORT₁(cons₂(N, L)) -> LOW₂(N, L)
IFLOW₃(false, N, cons₂(M, L)) -> LOW₂(N, L)
QUICKSORT₁(cons₂(N, L)) -> APP₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))
IFLOW₃(true, N, cons₂(M, L)) -> LOW₂(N, L)
IFHIGH₃(true, N, cons₂(M, L)) -> HIGH₂(N, L)
QUICKSORT₁(cons₂(N, L)) -> QUICKSORT₁(high₂(N, L))
QUICKSORT₁(cons₂(N, L)) -> QUICKSORT₁(low₂(N, L))
LOW₂(N, cons₂(M, L)) -> LE₂(M, N)
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
IFHIGH₃(false, N, cons₂(M, L)) -> HIGH₂(N, L)
HIGH₂(N, cons₂(M, L)) -> LE₂(M, N)
LOW₂(N, cons₂(M, L)) -> IFLOW₃(le₂(M, N), N, cons₂(M, L))
QUICKSORT₁(cons₂(N, L)) -> HIGH₂(N, L)
APP₂(cons₂(N, L), Y) -> APP₂(L, Y)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

HIGH₂(N, cons₂(M, L)) -> IFHIGH₃(le₂(M, N), N, cons₂(M, L))
QUICKSORT₁(cons₂(N, L)) -> LOW₂(N, L)
IFLOW₃(false, N, cons₂(M, L)) -> LOW₂(N, L)
QUICKSORT₁(cons₂(N, L)) -> APP₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))
IFLOW₃(true, N, cons₂(M, L)) -> LOW₂(N, L)
IFHIGH₃(true, N, cons₂(M, L)) -> HIGH₂(N, L)
QUICKSORT₁(cons₂(N, L)) -> QUICKSORT₁(high₂(N, L))
QUICKSORT₁(cons₂(N, L)) -> QUICKSORT₁(low₂(N, L))
LOW₂(N, cons₂(M, L)) -> LE₂(M, N)
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
IFHIGH₃(false, N, cons₂(M, L)) -> HIGH₂(N, L)
HIGH₂(N, cons₂(M, L)) -> LE₂(M, N)
LOW₂(N, cons₂(M, L)) -> IFLOW₃(le₂(M, N), N, cons₂(M, L))
QUICKSORT₁(cons₂(N, L)) -> HIGH₂(N, L)
APP₂(cons₂(N, L), Y) -> APP₂(L, Y)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

APP₂(cons₂(N, L), Y) -> APP₂(L, Y)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

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(N, L), Y) -> APP₂(L, Y)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

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.

LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)

Used argument filtering: LE₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

              ↳ QDP

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

HIGH₂(N, cons₂(M, L)) -> IFHIGH₃(le₂(M, N), N, cons₂(M, L))
IFHIGH₃(false, N, cons₂(M, L)) -> HIGH₂(N, L)
IFHIGH₃(true, N, cons₂(M, L)) -> HIGH₂(N, L)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

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.

IFHIGH₃(false, N, cons₂(M, L)) -> HIGH₂(N, L)
IFHIGH₃(true, N, cons₂(M, L)) -> HIGH₂(N, L)

Used argument filtering: HIGH₂(x1, x2) = x2
cons₂(x1, x2) = cons₁(x2)
IFHIGH₃(x1, x2, x3) = x3
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

              ↳ QDP

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

HIGH₂(N, cons₂(M, L)) -> IFHIGH₃(le₂(M, N), N, cons₂(M, L))

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ QDP

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

IFLOW₃(false, N, cons₂(M, L)) -> LOW₂(N, L)
LOW₂(N, cons₂(M, L)) -> IFLOW₃(le₂(M, N), N, cons₂(M, L))
IFLOW₃(true, N, cons₂(M, L)) -> LOW₂(N, L)

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

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.

IFLOW₃(false, N, cons₂(M, L)) -> LOW₂(N, L)
IFLOW₃(true, N, cons₂(M, L)) -> LOW₂(N, L)

Used argument filtering: IFLOW₃(x1, x2, x3) = x3
cons₂(x1, x2) = cons₁(x2)
LOW₂(x1, x2) = x2
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ DependencyGraphProof

              ↳ QDP

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

LOW₂(N, cons₂(M, L)) -> IFLOW₃(le₂(M, N), N, cons₂(M, L))

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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

QUICKSORT₁(cons₂(N, L)) -> QUICKSORT₁(low₂(N, L))
QUICKSORT₁(cons₂(N, L)) -> QUICKSORT₁(high₂(N, L))

The TRS R consists of the following rules:

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

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.

QUICKSORT₁(cons₂(N, L)) -> QUICKSORT₁(low₂(N, L))
QUICKSORT₁(cons₂(N, L)) -> QUICKSORT₁(high₂(N, L))

Used argument filtering: QUICKSORT₁(x1) = x1
cons₂(x1, x2) = cons₁(x2)
low₂(x1, x2) = x2
high₂(x1, x2) = x2
nil = nil
ifhigh₃(x1, x2, x3) = x3
iflow₃(x1, x2, x3) = x3
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(cons₂(N, L), Y) -> cons₂(N, app₂(L, Y))
low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
high₂(N, nil) -> nil
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
quicksort₁(nil) -> nil
quicksort₁(cons₂(N, L)) -> app₂(quicksort₁(low₂(N, L)), cons₂(N, quicksort₁(high₂(N, L))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(cons₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, cons₂(x1, x2))
iflow₃(true, x0, cons₂(x1, x2))
iflow₃(false, x0, cons₂(x1, x2))
high₂(x0, nil)
high₂(x0, cons₂(x1, x2))
ifhigh₃(true, x0, cons₂(x1, x2))
ifhigh₃(false, x0, cons₂(x1, x2))
quicksort₁(nil)
quicksort₁(cons₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.