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 Order [17,21] with Interpretation:

POL( APP2(x1, x2) ) = max{0, x1 - 2}


POL( cons2(x1, x2) ) = x2 + 3



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 Order [17,21] with Interpretation:

POL( LE2(x1, x2) ) = max{0, x2 - 2}


POL( s1(x1) ) = x1 + 3



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))
The remaining pairs can at least be oriented weakly.

IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( HIGH2(x1, x2) ) = max{0, x2 - 2}


POL( cons2(x1, x2) ) = x2 + 3


POL( IFHIGH3(x1, ..., x3) ) = max{0, x1 + x3 - 3}


POL( le2(x1, x2) ) = 0


POL( false ) = 0


POL( true ) = 0



The following usable rules [14] were oriented:

le2(s1(X), 0) -> false
le2(0, Y) -> true
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(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.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.

↳ 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)
IFLOW3(true, N, cons2(M, L)) -> LOW2(N, L)
The remaining pairs can at least be oriented weakly.

LOW2(N, cons2(M, L)) -> IFLOW3(le2(M, N), N, cons2(M, L))
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( IFLOW3(x1, ..., x3) ) = max{0, x3 - 2}


POL( cons2(x1, x2) ) = x2 + 3


POL( LOW2(x1, x2) ) = max{0, x2 - 2}



The following usable rules [14] were oriented: none



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

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

LOW2(N, cons2(M, L)) -> IFLOW3(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))))

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
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 Order [17,21] with Interpretation:

POL( QUICKSORT1(x1) ) = max{0, x1 - 2}


POL( cons2(x1, x2) ) = x2 + 3


POL( low2(x1, x2) ) = x2


POL( high2(x1, x2) ) = x2


POL( iflow3(x1, ..., x3) ) = x3


POL( nil ) = 0


POL( ifhigh3(x1, ..., x3) ) = x3



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



↳ 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.