Termination w.r.t. Q proof of /tmp/tmpL1xTYC/#3.55.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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))
IF_LOW(false, n, add(m, x)) → LOW(n, x)
QUICKSORT(add(n, x)) → QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) → HIGH(n, 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(add(n, x)) → APP(quicksort(low(n, x)), add(n, quicksort(high(n, x))))
QUOT(s(x), s(y)) → MINUS(x, y)
APP(add(n, x), y) → APP(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
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_LOW(true, n, add(m, x)) → LOW(n, x)
IF_HIGH(true, n, add(m, x)) → HIGH(n, x)
QUICKSORT(add(n, x)) → QUICKSORT(low(n, x))
QUICKSORT(add(n, x)) → LOW(n, x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))
IF_LOW(false, n, add(m, x)) → LOW(n, x)
QUICKSORT(add(n, x)) → QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) → HIGH(n, 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(add(n, x)) → APP(quicksort(low(n, x)), add(n, quicksort(high(n, x))))
QUOT(s(x), s(y)) → MINUS(x, y)
APP(add(n, x), y) → APP(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
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_LOW(true, n, add(m, x)) → LOW(n, x)
IF_HIGH(true, n, add(m, x)) → HIGH(n, x)
QUICKSORT(add(n, x)) → QUICKSORT(low(n, x))
QUICKSORT(add(n, x)) → LOW(n, x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 7 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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

          ↳ QDP

          ↳ QDP

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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

          ↳ 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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)) = 1 + (4)x_1
POL(LE(x₁, x₂)) = (3)x_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

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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

          ↳ 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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)) = 4 + (2)x_1 + x_2
POL(le(x₁, x₂)) = 2 + (2)x_2
POL(true) = 2
POL(IF_HIGH(x₁, x₂, x₃)) = (2)x_3
POL(false) = 2
POL(s(x₁)) = 2 + (3)x_1
POL(HIGH(x₁, x₂)) = 4 + (2)x_2
POL(0) = 1

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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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

          ↳ QDP

          ↳ 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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)) = (4)x_1 + (4)x_2
POL(le(x₁, x₂)) = x_1
POL(true) = 1
POL(false) = 1
POL(s(x₁)) = 4 + (2)x_1
POL(LOW(x₁, x₂)) = 1 + (4)x_2
POL(IF_LOW(x₁, x₂, x₃)) = x_1 + (2)x_3
POL(0) = 2

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

le(0, y) → true
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

          ↳ 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

QUICKSORT(add(n, x)) → QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) → QUICKSORT(low(n, x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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.

QUICKSORT(add(n, x)) → QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) → QUICKSORT(low(n, x))

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

POL(add(x₁, x₂)) = 4 + (4)x_2
POL(if_high(x₁, x₂, x₃)) = (4)x_3
POL(high(x₁, x₂)) = 3 + (4)x_2
POL(low(x₁, x₂)) = (4)x_2
POL(le(x₁, x₂)) = 0
POL(true) = 0
POL(false) = 0
POL(s(x₁)) = 1 + (4)x_1
POL(QUICKSORT(x₁)) = (4)x_1
POL(0) = 4
POL(nil) = 4
POL(if_low(x₁, x₂, x₃)) = (4)x_3

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

if_high(false, n, add(m, x)) → add(m, high(n, x))
high(n, nil) → nil
if_high(true, n, add(m, x)) → high(n, x)
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_low(false, n, add(m, x)) → low(n, x)
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
low(n, nil) → nil
if_low(true, n, add(m, x)) → add(m, low(n, x))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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.

MINUS(s(x), s(y)) → MINUS(x, y)

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

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

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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.

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

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

POL(minus(x₁, x₂)) = 3 + x_1
POL(QUOT(x₁, x₂)) = (2)x_1
POL(s(x₁)) = 4 + (4)x_1
POL(0) = 0

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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.