Termination w.r.t. Q proof of /tmp/tmp8tIlnI/quicksort.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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

Q is empty.

↳ 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(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

QUICKSORT(x) → TAIL(x)
QUICKSORT(x) → HEAD(x)
LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))
IF_LOW(false, n, add(m, x)) → LOW(n, x)
IF_QS(false, x, n, y) → APP(quicksort(x), add(n, quicksort(y)))
QUICKSORT(x) → IF_QS(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
HIGH(n, add(m, x)) → LE(m, n)
LE(s(x), s(y)) → LE(x, y)
LOW(n, add(m, x)) → LE(m, n)
QUICKSORT(x) → HIGH(head(x), tail(x))
APP(add(n, x), y) → APP(x, y)
IF_HIGH(false, n, add(m, x)) → HIGH(n, x)
QUICKSORT(x) → LOW(head(x), tail(x))
HIGH(n, add(m, x)) → IF_HIGH(le(m, n), n, add(m, x))
IF_LOW(true, n, add(m, x)) → LOW(n, x)
IF_HIGH(true, n, add(m, x)) → HIGH(n, x)
IF_QS(false, x, n, y) → QUICKSORT(x)
IF_QS(false, x, n, y) → QUICKSORT(y)
QUICKSORT(x) → ISEMPTY(x)

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(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(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:

QUICKSORT(x) → TAIL(x)
QUICKSORT(x) → HEAD(x)
LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))
IF_LOW(false, n, add(m, x)) → LOW(n, x)
IF_QS(false, x, n, y) → APP(quicksort(x), add(n, quicksort(y)))
QUICKSORT(x) → IF_QS(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
HIGH(n, add(m, x)) → LE(m, n)
LE(s(x), s(y)) → LE(x, y)
LOW(n, add(m, x)) → LE(m, n)
QUICKSORT(x) → HIGH(head(x), tail(x))
APP(add(n, x), y) → APP(x, y)
IF_HIGH(false, n, add(m, x)) → HIGH(n, x)
QUICKSORT(x) → LOW(head(x), tail(x))
HIGH(n, add(m, x)) → IF_HIGH(le(m, n), n, add(m, x))
IF_LOW(true, n, add(m, x)) → LOW(n, x)
IF_HIGH(true, n, add(m, x)) → HIGH(n, x)
IF_QS(false, x, n, y) → QUICKSORT(x)
IF_QS(false, x, n, y) → QUICKSORT(y)
QUICKSORT(x) → ISEMPTY(x)

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(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

APP(add(n, x), y) → APP(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(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

APP(add(n, x), y) → APP(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(APP(x₁, x₂)) = (4)x_1
POL(add(x₁, x₂)) = 1 + (4)x_2

The value of delta used in the strict ordering is 4.
The following usable rules [17] 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(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:

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(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

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: Polynomial interpretation [25,35]:

POL(s(x₁)) = 4 + (2)x_1
POL(LE(x₁, x₂)) = (3)x_2

The value of delta used in the strict ordering is 12.
The following usable rules [17] 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(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_HIGH(false, n, add(m, x)) → HIGH(n, x)
HIGH(n, add(m, x)) → IF_HIGH(le(m, n), n, add(m, x))
IF_HIGH(true, n, add(m, x)) → HIGH(n, x)

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(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

IF_HIGH(false, n, add(m, x)) → HIGH(n, x)
HIGH(n, add(m, x)) → IF_HIGH(le(m, n), n, add(m, x))
IF_HIGH(true, n, add(m, x)) → HIGH(n, x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(add(x₁, x₂)) = 3 + (4)x_1 + (4)x_2
POL(le(x₁, x₂)) = 3 + x_1 + (2)x_2
POL(true) = 4
POL(IF_HIGH(x₁, x₂, x₃)) = 2 + (4)x_2 + x_3
POL(false) = 2
POL(s(x₁)) = x_1
POL(HIGH(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(0) = 3

The value of delta used in the strict ordering is 1.
The following usable rules [17] 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(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:

LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))
IF_LOW(false, n, add(m, x)) → LOW(n, x)
IF_LOW(true, n, add(m, x)) → LOW(n, x)

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(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))

The remaining pairs can at least be oriented weakly.

IF_LOW(false, n, add(m, x)) → LOW(n, x)
IF_LOW(true, n, add(m, x)) → LOW(n, x)

Used ordering: Polynomial interpretation [25,35]:

POL(add(x₁, x₂)) = 3 + x_1 + (2)x_2
POL(le(x₁, x₂)) = 3 + (3)x_1 + x_2
POL(true) = 2
POL(false) = 3
POL(s(x₁)) = 3 + (3)x_1
POL(LOW(x₁, x₂)) = 3 + x_2
POL(IF_LOW(x₁, x₂, x₃)) = x_3
POL(0) = 2

The value of delta used in the strict ordering is 3.
The following usable rules [17] 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:

IF_LOW(false, n, add(m, x)) → LOW(n, x)
IF_LOW(true, n, add(m, x)) → LOW(n, x)

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(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF_QS(false, x, n, y) → QUICKSORT(x)
QUICKSORT(x) → IF_QS(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
IF_QS(false, x, n, y) → QUICKSORT(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(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

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