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
  ↳ 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))))

Q is empty.

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))))

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

↳ 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))))

Q is empty.
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
  ↳ 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))))

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


The following pairs can be oriented strictly and are deleted.


APP2(cons2(N, L), Y) -> APP2(L, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = 2·x1   
POL(cons2(x1, x2)) = 1 + 2·x2   

The following usable rules [14] were oriented: none



↳ 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))))

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

↳ 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))))

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


The following pairs can be oriented strictly and are deleted.


LE2(s1(X), s1(Y)) -> LE2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE2(x1, x2)) = 2·x1 + x2   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ 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))))

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

↳ 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))))

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


The following pairs can be oriented strictly and are deleted.


HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))
IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
The remaining pairs can at least be oriented weakly.

IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(HIGH2(x1, x2)) = 1 + 2·x2   
POL(IFHIGH3(x1, x2, x3)) = x1 + x3   
POL(cons2(x1, x2)) = 1 + x1 + 2·x2   
POL(false) = 1   
POL(le2(x1, x2)) = x1   
POL(s1(x1)) = 2 + 2·x1   
POL(true) = 0   

The following usable rules [14] were oriented:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ DependencyGraphProof
          ↳ QDP
          ↳ QDP

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

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))))

Q is empty.
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
  ↳ 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))))

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


The following pairs can be oriented strictly 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 be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(IFLOW3(x1, x2, x3)) = 1 + 2·x2 + x3   
POL(LOW2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(cons2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(false) = 2   
POL(le2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(s1(x1)) = 0   
POL(true) = 0   

The following usable rules [14] were oriented: none



↳ 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))))

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

↳ 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))))

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


The following pairs can be oriented strictly and are deleted.


QUICKSORT1(cons2(N, L)) -> QUICKSORT1(low2(N, L))
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(QUICKSORT1(x1)) = 2·x1   
POL(cons2(x1, x2)) = 1 + 2·x2   
POL(false) = 0   
POL(high2(x1, x2)) = x2   
POL(ifhigh3(x1, x2, x3)) = x3   
POL(iflow3(x1, x2, x3)) = x3   
POL(le2(x1, x2)) = 0   
POL(low2(x1, x2)) = x2   
POL(nil) = 2   
POL(s1(x1)) = 1 + 2·x1   
POL(true) = 0   

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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))))

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