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



   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)) -> IFLOW(le(m, n), n, add(m, x))
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)) -> IFHIGH(le(m, n), n, add(m, 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)
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.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Nar
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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)) -> iflow(le(m, n), n, add(m, x))
iflow(true, n, add(m, x)) -> add(m, low(n, x))
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> ifhigh(le(m, n), n, add(m, x))
ifhigh(true, n, add(m, x)) -> high(n, x)
ifhigh(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: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 8
Dependency Graph
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Nar
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Argument Filtering and Ordering
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Nar
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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)) -> iflow(le(m, n), n, add(m, x))
iflow(true, n, add(m, x)) -> add(m, low(n, x))
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> ifhigh(le(m, n), n, add(m, x))
ifhigh(true, n, add(m, x)) -> high(n, x)
ifhigh(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: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
LE(x1, x2) -> LE(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 9
Dependency Graph
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Nar
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Argument Filtering and Ordering
       →DP Problem 4
AFS
       →DP Problem 5
Nar
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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)) -> iflow(le(m, n), n, add(m, x))
iflow(true, n, add(m, x)) -> add(m, low(n, x))
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> ifhigh(le(m, n), n, add(m, x))
ifhigh(true, n, add(m, x)) -> high(n, x)
ifhigh(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: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> APP(x1, x2)
add(x1, x2) -> add(x1, x2)


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
           →DP Problem 10
Dependency Graph
       →DP Problem 4
AFS
       →DP Problem 5
Nar
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
Argument Filtering and Ordering
       →DP Problem 5
Nar
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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


The following usable rules for innermost can be oriented:

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


Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
minus(x1, x2) -> x1


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
           →DP Problem 11
Dependency Graph
       →DP Problem 5
Nar
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


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


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Narrowing Transformation
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


Dependency Pairs:

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




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

LOW(n, add(m, x)) -> IFLOW(le(m, n), n, add(m, x))
three new Dependency Pairs are created:

LOW(n', add(0, x)) -> IFLOW(true, n', add(0, x))
LOW(0, add(s(x''), x)) -> IFLOW(false, 0, add(s(x''), x))
LOW(s(y'), add(s(x''), x)) -> IFLOW(le(x'', y'), s(y'), add(s(x''), x))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Nar
           →DP Problem 12
Instantiation Transformation
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


Dependency Pairs:

LOW(s(y'), add(s(x''), x)) -> IFLOW(le(x'', y'), s(y'), add(s(x''), x))
LOW(0, add(s(x''), x)) -> IFLOW(false, 0, add(s(x''), x))
IFLOW(true, n, add(m, x)) -> LOW(n, x)
LOW(n', add(0, x)) -> IFLOW(true, n', add(0, x))
IFLOW(false, 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
app(add(n, x), y) -> add(n, app(x, y))
low(n, nil) -> nil
low(n, add(m, x)) -> iflow(le(m, n), n, add(m, x))
iflow(true, n, add(m, x)) -> add(m, low(n, x))
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> ifhigh(le(m, n), n, add(m, x))
ifhigh(true, n, add(m, x)) -> high(n, x)
ifhigh(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




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

IFLOW(true, n, add(m, x)) -> LOW(n, x)
two new Dependency Pairs are created:

IFLOW(true, n', add(0, x'')) -> LOW(n', x'')
IFLOW(true, s(y'''), add(s(x'''''), x')) -> LOW(s(y'''), x')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Nar
           →DP Problem 12
Inst
             ...
               →DP Problem 13
Instantiation Transformation
       →DP Problem 6
Remaining
       →DP Problem 7
Remaining


Dependency Pairs:

IFLOW(true, s(y'''), add(s(x'''''), x')) -> LOW(s(y'''), x')
LOW(0, add(s(x''), x)) -> IFLOW(false, 0, add(s(x''), x))
IFLOW(true, n', add(0, x'')) -> LOW(n', x'')
LOW(n', add(0, x)) -> IFLOW(true, n', add(0, x))
IFLOW(false, n, add(m, x)) -> LOW(n, x)
LOW(s(y'), add(s(x''), x)) -> IFLOW(le(x'', y'), s(y'), add(s(x''), 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)) -> iflow(le(m, n), n, add(m, x))
iflow(true, n, add(m, x)) -> add(m, low(n, x))
iflow(false, n, add(m, x)) -> low(n, x)
high(n, nil) -> nil
high(n, add(m, x)) -> ifhigh(le(m, n), n, add(m, x))
ifhigh(true, n, add(m, x)) -> high(n, x)
ifhigh(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




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

IFLOW(false, n, add(m, x)) -> LOW(n, x)
two new Dependency Pairs are created:

IFLOW(false, 0, add(s(x''''), x'')) -> LOW(0, x'')
IFLOW(false, s(y'''), add(s(x'''''), x')) -> LOW(s(y'''), x')

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Nar
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Nar
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
AFS
       →DP Problem 4
AFS
       →DP Problem 5
Nar
       →DP Problem 6
Remaining Obligation(s)
       →DP Problem 7
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:01 minutes