Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

Q is empty.


QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))
QUICKSORT1(cons2(N, L)) -> LOW2(N, L)
IFLOW3(false, N, cons2(M, L)) -> LOW2(N, L)
QUICKSORT1(cons2(N, L)) -> APP2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))
IFLOW3(true, N, cons2(M, L)) -> LOW2(N, L)
IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(low2(N, L))
LOW2(N, cons2(M, L)) -> LE2(M, N)
LE2(s1(X), s1(Y)) -> LE2(X, Y)
IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
HIGH2(N, cons2(M, L)) -> LE2(M, N)
LOW2(N, cons2(M, L)) -> IFLOW3(le2(M, N), N, cons2(M, L))
QUICKSORT1(cons2(N, L)) -> HIGH2(N, L)
APP2(cons2(N, L), Y) -> APP2(L, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))
QUICKSORT1(cons2(N, L)) -> LOW2(N, L)
IFLOW3(false, N, cons2(M, L)) -> LOW2(N, L)
QUICKSORT1(cons2(N, L)) -> APP2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))
IFLOW3(true, N, cons2(M, L)) -> LOW2(N, L)
IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(low2(N, L))
LOW2(N, cons2(M, L)) -> LE2(M, N)
LE2(s1(X), s1(Y)) -> LE2(X, Y)
IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
HIGH2(N, cons2(M, L)) -> LE2(M, N)
LOW2(N, cons2(M, L)) -> IFLOW3(le2(M, N), N, cons2(M, L))
QUICKSORT1(cons2(N, L)) -> HIGH2(N, L)
APP2(cons2(N, L), Y) -> APP2(L, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 5 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

APP2(cons2(N, L), Y) -> APP2(L, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


APP2(cons2(N, L), Y) -> APP2(L, Y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
APP2(x1, x2)  =  APP1(x1)
cons2(x1, x2)  =  cons1(x2)

Lexicographic Path Order [19].
Precedence:
[APP1, cons1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

LE2(s1(X), s1(Y)) -> LE2(X, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


LE2(s1(X), s1(Y)) -> LE2(X, Y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LE2(x1, x2)  =  LE1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP
              ↳ QDP
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))
IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)
The remaining pairs can at least by weakly be oriented.

HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))
Used ordering: Combined order from the following AFS and order.
HIGH2(x1, x2)  =  HIGH2(x1, x2)
cons2(x1, x2)  =  cons2(x1, x2)
IFHIGH3(x1, x2, x3)  =  IFHIGH2(x2, x3)
le2(x1, x2)  =  le
false  =  false
true  =  true
s1(x1)  =  s
0  =  0

Lexicographic Path Order [19].
Precedence:
false > [HIGH2, cons2, IFHIGH2] > [le, true, 0]
s > [le, true, 0]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ DependencyGraphProof
              ↳ QDP
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

IFLOW3(false, N, cons2(M, L)) -> LOW2(N, L)
LOW2(N, cons2(M, L)) -> IFLOW3(le2(M, N), N, cons2(M, L))
IFLOW3(true, N, cons2(M, L)) -> LOW2(N, L)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


IFLOW3(false, N, cons2(M, L)) -> LOW2(N, L)
LOW2(N, cons2(M, L)) -> IFLOW3(le2(M, N), N, cons2(M, L))
IFLOW3(true, N, cons2(M, L)) -> LOW2(N, L)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
IFLOW3(x1, x2, x3)  =  x3
false  =  false
cons2(x1, x2)  =  cons1(x2)
LOW2(x1, x2)  =  LOW1(x2)
le2(x1, x2)  =  le
true  =  true
s1(x1)  =  x1
0  =  0

Lexicographic Path Order [19].
Precedence:
cons1 > LOW1 > le
[true, 0] > false > le
[true, 0] > LOW1 > le


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
QDP
                ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

QUICKSORT1(cons2(N, L)) -> QUICKSORT1(low2(N, L))
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


QUICKSORT1(cons2(N, L)) -> QUICKSORT1(low2(N, L))
The remaining pairs can at least by weakly be oriented.

QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))
Used ordering: Combined order from the following AFS and order.
QUICKSORT1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
low2(x1, x2)  =  x2
high2(x1, x2)  =  high1(x2)
le2(x1, x2)  =  le
s1(x1)  =  x1
0  =  0
false  =  false
nil  =  nil
iflow3(x1, x2, x3)  =  x3
true  =  true
ifhigh3(x1, x2, x3)  =  ifhigh1(x3)

Lexicographic Path Order [19].
Precedence:
[cons1, high1, le, nil, true, ifhigh1] > [0, false]


The following usable rules [14] were oriented:

low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
high2(N, nil) -> nil
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
QDP
                    ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUICKSORT1(x1)  =  QUICKSORT1(x1)
cons2(x1, x2)  =  cons1(x2)
high2(x1, x2)  =  x2
le2(x1, x2)  =  le2(x1, x2)
s1(x1)  =  s
0  =  0
false  =  false
ifhigh3(x1, x2, x3)  =  x3
true  =  true
nil  =  nil

Lexicographic Path Order [19].
Precedence:
[cons1, le2] > false


The following usable rules [14] were oriented:

high2(N, nil) -> nil
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.