Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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:

QUICKSORT1(x) -> LOW2(head1(x), tail1(x))
IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
QUICKSORT1(x) -> HEAD1(x)
IF_QS4(false, x, n, y) -> APP2(quicksort1(x), add2(n, quicksort1(y)))
QUICKSORT1(x) -> TAIL1(x)
LOW2(n, add2(m, x)) -> LE2(m, n)
QUICKSORT1(x) -> HIGH2(head1(x), tail1(x))
LOW2(n, add2(m, x)) -> IF_LOW3(le2(m, n), n, add2(m, x))
HIGH2(n, add2(m, x)) -> IF_HIGH3(le2(m, n), n, add2(m, x))
IF_QS4(false, x, n, y) -> QUICKSORT1(y)
LE2(s1(x), s1(y)) -> LE2(x, y)
IF_QS4(false, x, n, y) -> QUICKSORT1(x)
HIGH2(n, add2(m, x)) -> LE2(m, n)
QUICKSORT1(x) -> IF_QS4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
IF_LOW3(true, n, add2(m, x)) -> LOW2(n, x)
IF_LOW3(false, n, add2(m, x)) -> LOW2(n, x)
IF_HIGH3(true, n, add2(m, x)) -> HIGH2(n, x)
APP2(add2(n, x), y) -> APP2(x, y)
QUICKSORT1(x) -> ISEMPTY1(x)

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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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:

QUICKSORT1(x) -> LOW2(head1(x), tail1(x))
IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
QUICKSORT1(x) -> HEAD1(x)
IF_QS4(false, x, n, y) -> APP2(quicksort1(x), add2(n, quicksort1(y)))
QUICKSORT1(x) -> TAIL1(x)
LOW2(n, add2(m, x)) -> LE2(m, n)
QUICKSORT1(x) -> HIGH2(head1(x), tail1(x))
LOW2(n, add2(m, x)) -> IF_LOW3(le2(m, n), n, add2(m, x))
HIGH2(n, add2(m, x)) -> IF_HIGH3(le2(m, n), n, add2(m, x))
IF_QS4(false, x, n, y) -> QUICKSORT1(y)
LE2(s1(x), s1(y)) -> LE2(x, y)
IF_QS4(false, x, n, y) -> QUICKSORT1(x)
HIGH2(n, add2(m, x)) -> LE2(m, n)
QUICKSORT1(x) -> IF_QS4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
IF_LOW3(true, n, add2(m, x)) -> LOW2(n, x)
IF_LOW3(false, n, add2(m, x)) -> LOW2(n, x)
IF_HIGH3(true, n, add2(m, x)) -> HIGH2(n, x)
APP2(add2(n, x), y) -> APP2(x, y)
QUICKSORT1(x) -> ISEMPTY1(x)

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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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 8 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(add2(n, x), y) -> APP2(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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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(add2(n, x), y) -> APP2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(APP2(x1, x2)) = 3·x1   
POL(add2(x1, x2)) = 1 + 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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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)) = 3·x1 + 3·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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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:

IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
HIGH2(n, add2(m, x)) -> IF_HIGH3(le2(m, n), n, add2(m, x))
IF_HIGH3(true, n, add2(m, x)) -> HIGH2(n, x)

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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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.


IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
HIGH2(n, add2(m, x)) -> IF_HIGH3(le2(m, n), n, add2(m, x))
IF_HIGH3(true, n, add2(m, x)) -> HIGH2(n, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(HIGH2(x1, x2)) = 3 + 3·x1 + 3·x2   
POL(IF_HIGH3(x1, x2, x3)) = 2 + 3·x2 + 3·x3   
POL(add2(x1, x2)) = 3 + 3·x1 + 3·x2   
POL(false) = 0   
POL(le2(x1, x2)) = 3 + x2   
POL(s1(x1)) = 0   
POL(true) = 2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ 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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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
            ↳ QDPOrderProof
          ↳ QDP

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

LOW2(n, add2(m, x)) -> IF_LOW3(le2(m, n), n, add2(m, x))
IF_LOW3(true, n, add2(m, x)) -> LOW2(n, x)
IF_LOW3(false, n, add2(m, x)) -> LOW2(n, x)

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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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.


LOW2(n, add2(m, x)) -> IF_LOW3(le2(m, n), n, add2(m, x))
IF_LOW3(true, n, add2(m, x)) -> LOW2(n, x)
IF_LOW3(false, n, add2(m, x)) -> LOW2(n, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(IF_LOW3(x1, x2, x3)) = 1 + 3·x2 + 3·x3   
POL(LOW2(x1, x2)) = 3 + 3·x1 + 3·x2   
POL(add2(x1, x2)) = 3 + x1 + x2   
POL(false) = 2   
POL(le2(x1, x2)) = 0   
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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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

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

IF_QS4(false, x, n, y) -> QUICKSORT1(y)
IF_QS4(false, x, n, y) -> QUICKSORT1(x)
QUICKSORT1(x) -> IF_QS4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))

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(add2(n, x), y) -> add2(n, app2(x, y))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
high2(n, nil) -> nil
high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> high2(n, x)
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
head1(add2(n, x)) -> n
tail1(add2(n, x)) -> x
isempty1(nil) -> true
isempty1(add2(n, x)) -> false
quicksort1(x) -> if_qs4(isempty1(x), low2(head1(x), tail1(x)), head1(x), high2(head1(x), tail1(x)))
if_qs4(true, x, n, y) -> nil
if_qs4(false, x, n, y) -> app2(quicksort1(x), add2(n, quicksort1(y)))

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