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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

Termination of R to be shown.

`   R`
`     ↳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)) -> LE(m, n)
IFLOW(true, n, add(m, x)) -> LOW(n, x)
IFLOW(false, n, add(m, x)) -> LOW(n, x)
HIGH(n, add(m, x)) -> LE(m, n)
IFHIGH(true, n, add(m, x)) -> HIGH(n, x)
IFHIGH(false, n, add(m, x)) -> HIGH(n, x)

Furthermore, R contains seven SCCs.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS 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.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS 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.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS 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.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules w.r.t. to the implicit AFS 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) =  1 POL(minus(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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`
`         ↳Nar`

Dependency Pairs:

IFLOW(false, n, add(m, x)) -> LOW(n, x)
IFLOW(true, n, add(m, x)) -> 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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

The following dependency pairs can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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`
`         ↳Nar`

Dependency Pairs:

IFHIGH(false, n, add(m, x)) -> HIGH(n, x)
IFHIGH(true, n, add(m, x)) -> HIGH(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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

The following dependency pairs can be strictly oriented:

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

There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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.

`   R`
`     ↳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`
`         ↳Nar`

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳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`
`         ↳Narrowing Transformation`

Dependency Pairs:

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

two new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Narrowing Transformation`

Dependency Pairs:

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

two new Dependency Pairs are created:

The transformation is resulting in one new DP problem:

`   R`
`     ↳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`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 15`
`                 ↳Polynomial Ordering`

Dependency Pairs:

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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
quicksort(nil) -> nil

The following dependency pairs can be strictly oriented:

Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)
iflow(false, n, add(m, x)) -> low(n, x)
low(n, nil) -> nil

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.

`   R`
`     ↳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`
`         ↳Nar`
`           →DP Problem 14`
`             ↳Nar`
`             ...`
`               →DP Problem 16`
`                 ↳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
low(n, nil) -> nil
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
ifhigh(true, n, add(m, x)) -> high(n, x)