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))))

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
Polynomial Ordering
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 5
Polo

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))))

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 6
Dependency Graph
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 5
Polo

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))))

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

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))))

The following dependency pair can be strictly oriented:

APP(cons(N, L), Y) -> APP(L, 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(cons(x1, x2)) =  1 + x2 POL(APP(x1, x2)) =  x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 7
Dependency Graph
→DP Problem 3
Polo
→DP Problem 4
Polo
→DP Problem 5
Polo

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))))

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

Dependency Pairs:

IFLOW(false, N, cons(M, L)) -> LOW(N, L)
IFLOW(true, N, cons(M, L)) -> LOW(N, L)
LOW(N, cons(M, L)) -> IFLOW(le(M, N), N, cons(M, L))

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))))

The following dependency pairs can be strictly oriented:

IFLOW(false, N, cons(M, L)) -> LOW(N, L)
IFLOW(true, N, cons(M, L)) -> LOW(N, L)

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(cons(x1, x2)) =  1 + x2 POL(true) =  0 POL(s(x1)) =  0 POL(le(x1, x2)) =  0 POL(IFLOW(x1, x2, x3)) =  x3

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 8
Dependency Graph
→DP Problem 4
Polo
→DP Problem 5
Polo

Dependency Pair:

LOW(N, cons(M, L)) -> IFLOW(le(M, N), N, cons(M, L))

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))))

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

Dependency Pairs:

IFHIGH(false, N, cons(M, L)) -> HIGH(N, L)
IFHIGH(true, N, cons(M, L)) -> HIGH(N, L)
HIGH(N, cons(M, L)) -> IFHIGH(le(M, N), N, cons(M, L))

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))))

The following dependency pairs can be strictly oriented:

IFHIGH(false, N, cons(M, L)) -> HIGH(N, L)
IFHIGH(true, N, cons(M, L)) -> HIGH(N, L)

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

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 9
Dependency Graph
→DP Problem 5
Polo

Dependency Pair:

HIGH(N, cons(M, L)) -> IFHIGH(le(M, N), N, cons(M, L))

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))))

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

Dependency Pairs:

QUICKSORT(cons(N, L)) -> QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) -> QUICKSORT(low(N, L))

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))))

The following dependency pairs can be strictly oriented:

QUICKSORT(cons(N, L)) -> QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) -> QUICKSORT(low(N, L))

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

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))
iflow(true, N, cons(M, L)) -> cons(M, low(N, L))
iflow(false, N, cons(M, L)) -> low(N, L)
low(N, nil) -> nil
low(N, cons(M, L)) -> iflow(le(M, N), N, cons(M, L))

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(QUICKSORT(x1)) =  x1 POL(iflow(x1, x2, x3)) =  x3 POL(0) =  0 POL(cons(x1, x2)) =  1 + x2 POL(false) =  0 POL(high(x1, x2)) =  x2 POL(low(x1, x2)) =  x2 POL(nil) =  0 POL(true) =  0 POL(s(x1)) =  0 POL(ifhigh(x1, x2, x3)) =  x3 POL(le(x1, x2)) =  0

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 10
Dependency Graph

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))))

Using the Dependency Graph resulted in no new DP problems.

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