Termination w.r.t. Q proof of TRS/Thiemannquicksort.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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(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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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:

QUICKSORT₁(x) -> LOW₂(head₁(x), tail₁(x))
IF_HIGH₃(false, n, add₂(m, x)) -> HIGH₂(n, x)
QUICKSORT₁(x) -> HEAD₁(x)
IF_QS₄(false, x, n, y) -> APP₂(quicksort₁(x), add₂(n, quicksort₁(y)))
QUICKSORT₁(x) -> TAIL₁(x)
LOW₂(n, add₂(m, x)) -> LE₂(m, n)
QUICKSORT₁(x) -> HIGH₂(head₁(x), tail₁(x))
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))
IF_QS₄(false, x, n, y) -> QUICKSORT₁(y)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
IF_QS₄(false, x, n, y) -> QUICKSORT₁(x)
HIGH₂(n, add₂(m, x)) -> LE₂(m, n)
QUICKSORT₁(x) -> IF_QS₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(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)
QUICKSORT₁(x) -> ISEMPTY₁(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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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:

QUICKSORT₁(x) -> LOW₂(head₁(x), tail₁(x))
IF_HIGH₃(false, n, add₂(m, x)) -> HIGH₂(n, x)
QUICKSORT₁(x) -> HEAD₁(x)
IF_QS₄(false, x, n, y) -> APP₂(quicksort₁(x), add₂(n, quicksort₁(y)))
QUICKSORT₁(x) -> TAIL₁(x)
LOW₂(n, add₂(m, x)) -> LE₂(m, n)
QUICKSORT₁(x) -> HIGH₂(head₁(x), tail₁(x))
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))
IF_QS₄(false, x, n, y) -> QUICKSORT₁(y)
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
IF_QS₄(false, x, n, y) -> QUICKSORT₁(x)
HIGH₂(n, add₂(m, x)) -> LE₂(m, n)
QUICKSORT₁(x) -> IF_QS₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(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)
QUICKSORT₁(x) -> ISEMPTY₁(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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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 8 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(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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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(add₂(n, x), y) -> APP₂(x, y)

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

POL(APP₂(x₁, x₂)) = x₁
POL(add₂(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(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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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(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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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)) = 3·x₁ + 3·x₂
POL(s₁(x₁)) = 2 + 2·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(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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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:

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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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.

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 be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(HIGH₂(x₁, x₂)) = 2·x₂
POL(IF_HIGH₃(x₁, x₂, x₃)) = x₃
POL(add₂(x₁, x₂)) = 1 + 3·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

                ↳ 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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ 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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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.

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 be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(IF_LOW₃(x₁, x₂, x₃)) = x₃
POL(LOW₂(x₁, x₂)) = 1 + 2·x₂
POL(add₂(x₁, x₂)) = 2 + 3·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

                ↳ 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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF_QS₄(false, x, n, y) -> QUICKSORT₁(y)
IF_QS₄(false, x, n, y) -> QUICKSORT₁(x)
QUICKSORT₁(x) -> IF_QS₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(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))
head₁(add₂(n, x)) -> n
tail₁(add₂(n, x)) -> x
isempty₁(nil) -> true
isempty₁(add₂(n, x)) -> false
quicksort₁(x) -> if_qs₄(isempty₁(x), low₂(head₁(x), tail₁(x)), head₁(x), high₂(head₁(x), tail₁(x)))
if_qs₄(true, x, n, y) -> nil
if_qs₄(false, x, n, y) -> app₂(quicksort₁(x), add₂(n, quicksort₁(y)))

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