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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)
QUICKSORT(cons(N, L)) → APP(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))
IFLOW(true, N, cons(M, L)) → LOW(N, L)
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
LOW(N, cons(M, L)) → LE(M, N)
LE(s(X), s(Y)) → LE(X, Y)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
HIGH(N, cons(M, L)) → LE(M, N)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → HIGH(N, L)
APP(cons(N, L), Y) → APP(L, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)
QUICKSORT(cons(N, L)) → APP(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))
IFLOW(true, N, cons(M, L)) → LOW(N, L)
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
LOW(N, cons(M, L)) → LE(M, N)
LE(s(X), s(Y)) → LE(X, Y)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
HIGH(N, cons(M, L)) → LE(M, N)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → HIGH(N, L)
APP(cons(N, L), Y) → APP(L, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)
QUICKSORT(cons(N, L)) → APP(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
IFLOW(true, N, cons(M, L)) → LOW(N, L)
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
LOW(N, cons(M, L)) → LE(M, N)
LE(s(X), s(Y)) → LE(X, Y)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
HIGH(N, cons(M, L)) → LE(M, N)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → HIGH(N, L)
APP(cons(N, L), Y) → APP(L, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

APP(cons(N, L), Y) → APP(L, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.

APP(cons(N, L), Y) → APP(L, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = x1
cons(x1, x2) = cons(x1, x2)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

LE(s(X), s(Y)) → LE(X, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.

LE(s(X), s(Y)) → LE(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LE(x1, x2) = LE(x2)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > LE₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
HIGH(x1, x2) = HIGH(x2)
cons(x1, x2) = cons(x2)
IFHIGH(x1, x2, x3) = IFHIGH(x1, x3)
le(x1, x2) = le
false = false
true = true
s(x1) = s
0 = 0

Lexicographic Path Order [19].
Precedence:

cons₁ > le > false > HIGH₁ > IFHIGH₂
cons₁ > le > true > IFHIGH₂
s > IFHIGH₂
0 > false > HIGH₁ > IFHIGH₂
0 > true > IFHIGH₂

The following usable rules [14] were oriented:

le(s(X), 0) → false
le(0, Y) → true
le(s(X), s(Y)) → le(X, Y)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

IFLOW(false, N, cons(M, L)) → LOW(N, L)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
IFLOW(true, N, cons(M, L)) → LOW(N, L)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.

IFLOW(false, N, cons(M, L)) → LOW(N, L)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
IFLOW(true, N, cons(M, L)) → LOW(N, L)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
IFLOW(x1, x2, x3) = IFLOW(x1, x3)
false = false
cons(x1, x2) = cons(x1, x2)
LOW(x1, x2) = LOW(x2)
le(x1, x2) = le(x1)
true = true
s(x1) = s(x1)
0 = 0

Lexicographic Path Order [19].
Precedence:

cons₂ > LOW₁ > IFLOW₂ > false
cons₂ > LOW₁ > le₁ > false
s₁ > le₁ > false
0 > true > LOW₁ > IFLOW₂ > false
0 > true > LOW₁ > le₁ > false

The following usable rules [14] were oriented:

le(s(X), 0) → false
le(0, Y) → true
le(s(X), s(Y)) → le(X, Y)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

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

QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.

QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUICKSORT(x1) = QUICKSORT(x1)
cons(x1, x2) = cons(x1, x2)
low(x1, x2) = x2
high(x1, x2) = x2
le(x1, x2) = le(x1, x2)
s(x1) = s
0 = 0
false = false
nil = nil
iflow(x1, x2, x3) = x3
true = true
ifhigh(x1, x2, x3) = x3

Lexicographic Path Order [19].
Precedence:

cons₂ > le₂ > false
cons₂ > le₂ > true
0 > false

The following usable rules [14] were oriented:

low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(false, N, cons(M, L)) → low(N, L)
ifhigh(true, N, cons(M, L)) → high(N, L)
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
high(N, nil) → nil
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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.