Termination Proof using AProVE

Term Rewriting System R:
[x, y, n, m]
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))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains seven SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

MINUS(s(x), s(y)) -> MINUS(x, y)

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))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(x, y)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

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))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)

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))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LE(s(x), s(y)) -> LE(x, y)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(LE(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

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))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

APP(add(n, x), y) -> APP(x, y)

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))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(add(n, x), y) -> APP(x, y)

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(APP(x₁, x₂)) = x₁

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 10

             ↳Dependency Graph

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

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))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polynomial Ordering

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

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))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Additionally, the following usable rules for innermost can be oriented:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = x₁

POL(0) = 0

POL(minus(x₁, x₂)) = x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

           →DP Problem 11

             ↳Dependency Graph

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

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))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polynomial Ordering

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pairs:

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

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))))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_LOW(false, n, add(m, x)) -> LOW(n, x)
IF_LOW(true, n, add(m, x)) -> LOW(n, x)

Additionally, the following usable rules for innermost can be oriented:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(LOW(x₁, x₂)) = x₂

POL(false) = 0

POL(IF_LOW(x₁, x₂, x₃)) = x₃

POL(true) = 0

POL(s(x₁)) = 0

POL(le(x₁, x₂)) = 0

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

           →DP Problem 12

             ↳Dependency Graph

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

Dependency Pair:

LOW(n, add(m, x)) -> IF_LOW(le(m, n), n, add(m, x))

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))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polynomial Ordering

       →DP Problem 7

         ↳Polo

Dependency Pairs:

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

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))))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_HIGH(false, n, add(m, x)) -> HIGH(n, x)
IF_HIGH(true, n, add(m, x)) -> HIGH(n, x)

Additionally, the following usable rules for innermost can be oriented:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(false) = 0

POL(IF_HIGH(x₁, x₂, x₃)) = x₃

POL(HIGH(x₁, x₂)) = x₂

POL(true) = 0

POL(s(x₁)) = 0

POL(le(x₁, x₂)) = 0

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

           →DP Problem 13

             ↳Dependency Graph

       →DP Problem 7

         ↳Polo

Dependency Pair:

HIGH(n, add(m, x)) -> IF_HIGH(le(m, n), n, add(m, x))

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))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polynomial Ordering

Dependency Pairs:

QUICKSORT(add(n, x)) -> QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) -> QUICKSORT(low(n, x))

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))))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

QUICKSORT(add(n, x)) -> QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) -> QUICKSORT(low(n, x))

Additionally, the following usable rules for innermost can be oriented:

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))
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)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUICKSORT(x₁)) = 1 + x₁

POL(0) = 0

POL(if_low(x₁, x₂, x₃)) = x₃

POL(false) = 0

POL(high(x₁, x₂)) = x₂

POL(if_high(x₁, x₂, x₃)) = x₃

POL(low(x₁, x₂)) = x₂

POL(nil) = 0

POL(true) = 0

POL(s(x₁)) = 0

POL(le(x₁, x₂)) = 0

POL(add(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Polo

       →DP Problem 5

         ↳Polo

       →DP Problem 6

         ↳Polo

       →DP Problem 7

         ↳Polo

           →DP Problem 14

             ↳Dependency Graph

Dependency Pair:

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))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes

POL(APP(x₁, x₂))	= x₁
POL(add(x₁, x₂))	= 1 + x₂

POL(QUOT(x₁, x₂))	= x₁
POL(0)	= 0
POL(minus(x₁, x₂))	= x₁
POL(s(x₁))	= 1 + x₁

POL(0)	= 0
POL(LOW(x₁, x₂))	= x₂
POL(false)	= 0
POL(IF_LOW(x₁, x₂, x₃))	= x₃
POL(true)	= 0
POL(s(x₁))	= 0
POL(le(x₁, x₂))	= 0
POL(add(x₁, x₂))	= 1 + x₂

POL(0)	= 0
POL(false)	= 0
POL(IF_HIGH(x₁, x₂, x₃))	= x₃
POL(HIGH(x₁, x₂))	= x₂
POL(true)	= 0
POL(s(x₁))	= 0
POL(le(x₁, x₂))	= 0
POL(add(x₁, x₂))	= 1 + x₂

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