Termination w.r.t. Q proof of TRS/Rubioquick.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

  ↳ 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))))

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:

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))))

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:

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))))

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 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))))

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 oriented strictly and are deleted.

APP₂(cons₂(N, L), Y) -> APP₂(L, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP₂(x₁, x₂)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂

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:

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.
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:

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))))

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 oriented strictly and are deleted.

LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ 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:

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.
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:

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))))

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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(HIGH₂(x₁, x₂)) = 1 + x₁ + x₂
POL(IFHIGH₃(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(cons₂(x₁, x₂)) = 1 + x₁ + x₂
POL(false) = 0
POL(le₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 0
POL(true) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))))

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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(IFLOW₃(x₁, x₂, x₃)) = 1 + x₂ + x₃
POL(LOW₂(x₁, x₂)) = 1 + x₁ + x₂
POL(cons₂(x₁, x₂)) = 1 + x₂
POL(false) = 0
POL(le₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 0
POL(true) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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))))

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 oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(QUICKSORT₁(x₁)) = x₁
POL(cons₂(x₁, x₂)) = 1 + x₂
POL(false) = 0
POL(high₂(x₁, x₂)) = x₂
POL(ifhigh₃(x₁, x₂, x₃)) = x₃
POL(iflow₃(x₁, x₂, x₃)) = x₃
POL(le₂(x₁, x₂)) = 0
POL(low₂(x₁, x₂)) = x₂
POL(nil) = 0
POL(s₁(x₁)) = 0
POL(true) = 0

The following usable rules [14] were oriented:

low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
ifhigh₃(false, N, cons₂(M, L)) -> cons₂(M, high₂(N, L))
ifhigh₃(true, N, cons₂(M, L)) -> high₂(N, L)
high₂(N, cons₂(M, L)) -> ifhigh₃(le₂(M, N), N, cons₂(M, L))
low₂(N, nil) -> nil
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))
high₂(N, nil) -> nil

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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))))

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.