Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/AG01SLASHHASH3.11.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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(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:

IF_HIGH₃(false, n, add₂(m, x)) -> HIGH₂(n, x)
QUICKSORT₁(add₂(n, x)) -> APP₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))
QUICKSORT₁(add₂(n, x)) -> QUICKSORT₁(low₂(n, x))
QUICKSORT₁(add₂(n, x)) -> QUICKSORT₁(high₂(n, x))
QUICKSORT₁(add₂(n, x)) -> HIGH₂(n, x)
LOW₂(n, add₂(m, x)) -> LE₂(m, n)
LOW₂(n, add₂(m, x)) -> IF_LOW₃(le₂(m, n), n, add₂(m, x))
HIGH₂(n, add₂(m, x)) -> IF_HIGH₃(le₂(m, n), n, add₂(m, x))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
HIGH₂(n, add₂(m, x)) -> LE₂(m, n)
QUICKSORT₁(add₂(n, x)) -> LOW₂(n, x)
IF_LOW₃(true, n, add₂(m, x)) -> LOW₂(n, x)
IF_LOW₃(false, n, add₂(m, x)) -> LOW₂(n, x)
IF_HIGH₃(true, n, add₂(m, x)) -> HIGH₂(n, x)
APP₂(add₂(n, x), y) -> APP₂(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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(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:

IF_HIGH₃(false, n, add₂(m, x)) -> HIGH₂(n, x)
QUICKSORT₁(add₂(n, x)) -> APP₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))
QUICKSORT₁(add₂(n, x)) -> QUICKSORT₁(low₂(n, x))
QUICKSORT₁(add₂(n, x)) -> QUICKSORT₁(high₂(n, x))
QUICKSORT₁(add₂(n, x)) -> HIGH₂(n, x)
LOW₂(n, add₂(m, x)) -> LE₂(m, n)
LOW₂(n, add₂(m, x)) -> IF_LOW₃(le₂(m, n), n, add₂(m, x))
HIGH₂(n, add₂(m, x)) -> IF_HIGH₃(le₂(m, n), n, add₂(m, x))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
HIGH₂(n, add₂(m, x)) -> LE₂(m, n)
QUICKSORT₁(add₂(n, x)) -> LOW₂(n, x)
IF_LOW₃(true, n, add₂(m, x)) -> LOW₂(n, x)
IF_LOW₃(false, n, add₂(m, x)) -> LOW₂(n, x)
IF_HIGH₃(true, n, add₂(m, x)) -> HIGH₂(n, x)
APP₂(add₂(n, x), y) -> APP₂(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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(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

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

APP₂(add₂(n, x), y) -> APP₂(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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

APP₂(add₂(n, x), y) -> APP₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP₂(x1, x2) = APP₁(x1)
add₂(x1, x2) = add₁(x2)

Lexicographic Path Order [19].
Precedence:

[APP₁, add_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(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

                ↳ 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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LE₂(x1, x2) = LE₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(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

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

IF_HIGH₃(false, n, add₂(m, x)) -> HIGH₂(n, x)
HIGH₂(n, add₂(m, x)) -> IF_HIGH₃(le₂(m, n), n, add₂(m, x))
IF_HIGH₃(true, n, add₂(m, x)) -> HIGH₂(n, x)

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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

IF_HIGH₃(false, n, add₂(m, x)) -> HIGH₂(n, x)
HIGH₂(n, add₂(m, x)) -> IF_HIGH₃(le₂(m, n), n, add₂(m, x))
IF_HIGH₃(true, n, add₂(m, x)) -> HIGH₂(n, x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
IF_HIGH₃(x1, x2, x3) = x3
false = false
add₂(x1, x2) = add₂(x1, x2)
HIGH₂(x1, x2) = HIGH₁(x2)
le₂(x1, x2) = le₁(x1)
true = true
0 = 0
s₁(x1) = s

Lexicographic Path Order [19].
Precedence:

add₂ > [false, le_1] > HIGH₁
true > HIGH₁
0 > HIGH₁
s > HIGH₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ 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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(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

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

LOW₂(n, add₂(m, x)) -> IF_LOW₃(le₂(m, n), n, add₂(m, x))
IF_LOW₃(true, n, add₂(m, x)) -> LOW₂(n, x)
IF_LOW₃(false, n, add₂(m, x)) -> LOW₂(n, x)

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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

LOW₂(n, add₂(m, x)) -> IF_LOW₃(le₂(m, n), n, add₂(m, x))
IF_LOW₃(true, n, add₂(m, x)) -> LOW₂(n, x)
IF_LOW₃(false, n, add₂(m, x)) -> LOW₂(n, x)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LOW₂(x1, x2) = LOW₁(x2)
add₂(x1, x2) = add₂(x1, x2)
IF_LOW₃(x1, x2, x3) = x3
le₂(x1, x2) = le
true = true
false = false
0 = 0
s₁(x1) = s

Lexicographic Path Order [19].
Precedence:

add₂ > LOW₁
add₂ > le > true
add₂ > le > false
0 > true
0 > false
s > le > true
s > le > false

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ 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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
app₂(nil, y) -> y
app₂(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(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

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

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

QUICKSORT₁(add₂(n, x)) -> QUICKSORT₁(low₂(n, x))
QUICKSORT₁(add₂(n, x)) -> QUICKSORT₁(high₂(n, x))

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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

QUICKSORT₁(add₂(n, x)) -> QUICKSORT₁(low₂(n, x))
QUICKSORT₁(add₂(n, x)) -> QUICKSORT₁(high₂(n, x))

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUICKSORT₁(x1) = QUICKSORT₁(x1)
add₂(x1, x2) = add₁(x2)
low₂(x1, x2) = x2
high₂(x1, x2) = x2
if_high₃(x1, x2, x3) = x3
le₂(x1, x2) = x2
true = true
0 = 0
s₁(x1) = s
false = false
nil = nil
if_low₃(x1, x2, x3) = x3

Lexicographic Path Order [19].
Precedence:

0 > true > add₁
0 > false > add₁

The following usable rules [14] were oriented:

high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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(add₂(n, x), y) -> add₂(n, app₂(x, y))
low₂(n, nil) -> nil
low₂(n, add₂(m, x)) -> if_low₃(le₂(m, n), n, add₂(m, x))
if_low₃(true, n, add₂(m, x)) -> add₂(m, low₂(n, x))
if_low₃(false, n, add₂(m, x)) -> low₂(n, x)
high₂(n, nil) -> nil
high₂(n, add₂(m, x)) -> if_high₃(le₂(m, n), n, add₂(m, x))
if_high₃(true, n, add₂(m, x)) -> high₂(n, x)
if_high₃(false, n, add₂(m, x)) -> add₂(m, high₂(n, x))
quicksort₁(nil) -> nil
quicksort₁(add₂(n, x)) -> app₂(quicksort₁(low₂(n, x)), add₂(n, quicksort₁(high₂(n, x))))

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
app₂(nil, x0)
app₂(add₂(x0, x1), x2)
low₂(x0, nil)
low₂(x0, add₂(x1, x2))
if_low₃(true, x0, add₂(x1, x2))
if_low₃(false, x0, add₂(x1, x2))
high₂(x0, nil)
high₂(x0, add₂(x1, x2))
if_high₃(true, x0, add₂(x1, x2))
if_high₃(false, x0, add₂(x1, x2))
quicksort₁(nil)
quicksort₁(add₂(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.