Term Rewriting System R:
[Y, X, N, L, M]
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
app(nil, Y) -> Y
app(cons(N, L), Y) -> cons(N, app(L, Y))
low(N, nil) -> nil
low(N, cons(M, L)) -> iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) -> cons(M, low(N, L))
iflow(false, N, cons(M, L)) -> low(N, L)
high(N, nil) -> nil
high(N, cons(M, L)) -> ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) -> high(N, L)
ifhigh(false, N, cons(M, L)) -> cons(M, high(N, L))
quicksort(nil) -> nil
quicksort(cons(N, L)) -> app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

LE(s(X), s(Y)) -> LE(X, Y)
APP(cons(N, L), Y) -> APP(L, Y)
LOW(N, cons(M, L)) -> IFLOW(le(M, N), N, cons(M, L))
LOW(N, cons(M, L)) -> LE(M, N)
IFLOW(true, N, cons(M, L)) -> LOW(N, L)
IFLOW(false, N, cons(M, L)) -> LOW(N, L)
HIGH(N, cons(M, L)) -> IFHIGH(le(M, N), N, cons(M, L))
HIGH(N, cons(M, L)) -> LE(M, N)
IFHIGH(true, N, cons(M, L)) -> HIGH(N, L)
IFHIGH(false, N, cons(M, L)) -> HIGH(N, L)
QUICKSORT(cons(N, L)) -> APP(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))
QUICKSORT(cons(N, L)) -> QUICKSORT(low(N, L))
QUICKSORT(cons(N, L)) -> LOW(N, L)
QUICKSORT(cons(N, L)) -> QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) -> HIGH(N, L)

Furthermore, R contains five SCCs. 


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


Dependency Pair:

LE(s(X), s(Y)) -> LE(X, Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
app(nil, Y) -> Y
app(cons(N, L), Y) -> cons(N, app(L, Y))
low(N, nil) -> nil
low(N, cons(M, L)) -> iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) -> cons(M, low(N, L))
iflow(false, N, cons(M, L)) -> low(N, L)
high(N, nil) -> nil
high(N, cons(M, L)) -> ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) -> high(N, L)
ifhigh(false, N, cons(M, L)) -> cons(M, high(N, L))
quicksort(nil) -> nil
quicksort(cons(N, L)) -> app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))


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 6
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining


Dependency Pair:


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
app(nil, Y) -> Y
app(cons(N, L), Y) -> cons(N, app(L, Y))
low(N, nil) -> nil
low(N, cons(M, L)) -> iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) -> cons(M, low(N, L))
iflow(false, N, cons(M, L)) -> low(N, L)
high(N, nil) -> nil
high(N, cons(M, L)) -> ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) -> high(N, L)
ifhigh(false, N, cons(M, L)) -> cons(M, high(N, L))
quicksort(nil) -> nil
quicksort(cons(N, L)) -> app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))


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
         ↳Remaining
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining


Dependency Pair:

APP(cons(N, L), Y) -> APP(L, Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
app(nil, Y) -> Y
app(cons(N, L), Y) -> cons(N, app(L, Y))
low(N, nil) -> nil
low(N, cons(M, L)) -> iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) -> cons(M, low(N, L))
iflow(false, N, cons(M, L)) -> low(N, L)
high(N, nil) -> nil
high(N, cons(M, L)) -> ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) -> high(N, L)
ifhigh(false, N, cons(M, L)) -> cons(M, high(N, L))
quicksort(nil) -> nil
quicksort(cons(N, L)) -> app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

APP(cons(N, L), Y) -> APP(L, 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)
cons(x1, x2) -> cons(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 7
             ↳Dependency Graph
       →DP Problem 3
         ↳Remaining
       →DP Problem 4
         ↳Remaining
       →DP Problem 5
         ↳Remaining


Dependency Pair:


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
app(nil, Y) -> Y
app(cons(N, L), Y) -> cons(N, app(L, Y))
low(N, nil) -> nil
low(N, cons(M, L)) -> iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) -> cons(M, low(N, L))
iflow(false, N, cons(M, L)) -> low(N, L)
high(N, nil) -> nil
high(N, cons(M, L)) -> ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) -> high(N, L)
ifhigh(false, N, cons(M, L)) -> cons(M, high(N, L))
quicksort(nil) -> nil
quicksort(cons(N, L)) -> app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))


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
         ↳Remaining Obligation(s)
       →DP Problem 4
         ↳Remaining Obligation(s)
       →DP Problem 5
         ↳Remaining Obligation(s)




The following remains to be proven:


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




The following remains to be proven:


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




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:00 minutes