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(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

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(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

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(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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:

IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
QUICKSORT1(add2(n, x)) -> APP2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(low2(n, x))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(high2(n, x))
QUICKSORT1(add2(n, x)) -> HIGH2(n, x)
LOW2(n, add2(m, x)) -> LE2(m, n)
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))
LE2(s1(x), s1(y)) -> LE2(x, y)
HIGH2(n, add2(m, x)) -> LE2(m, n)
QUICKSORT1(add2(n, x)) -> LOW2(n, 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)

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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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:

IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
QUICKSORT1(add2(n, x)) -> APP2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(low2(n, x))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(high2(n, x))
QUICKSORT1(add2(n, x)) -> HIGH2(n, x)
LOW2(n, add2(m, x)) -> LE2(m, n)
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))
LE2(s1(x), s1(y)) -> LE2(x, y)
HIGH2(n, add2(m, x)) -> LE2(m, n)
QUICKSORT1(add2(n, x)) -> LOW2(n, 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)

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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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(add2(n, x), y) -> APP2(x, 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)
add2(x1, x2)  =  add1(x2)

Lexicographic Path Order [19].
Precedence:
[APP1, add1]


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(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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:

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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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.


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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
IF_HIGH3(x1, x2, x3)  =  x3
false  =  false
add2(x1, x2)  =  add2(x1, x2)
HIGH2(x1, x2)  =  HIGH1(x2)
le2(x1, x2)  =  le1(x1)
true  =  true
0  =  0
s1(x1)  =  s

Lexicographic Path Order [19].
Precedence:
add2 > [false, le1] > HIGH1
true > HIGH1
0 > HIGH1
s > HIGH1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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
                ↳ 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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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.


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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LOW2(x1, x2)  =  LOW1(x2)
add2(x1, x2)  =  add2(x1, x2)
IF_LOW3(x1, x2, x3)  =  x3
le2(x1, x2)  =  le
true  =  true
false  =  false
0  =  0
s1(x1)  =  s

Lexicographic Path Order [19].
Precedence:
add2 > LOW1
add2 > le > true
add2 > le > false
0 > true
0 > false
s > le > true
s > le > false


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(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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(add2(n, x)) -> QUICKSORT1(low2(n, x))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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(add2(n, x)) -> QUICKSORT1(low2(n, x))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(high2(n, x))
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)
add2(x1, x2)  =  add1(x2)
low2(x1, x2)  =  x2
high2(x1, x2)  =  x2
if_high3(x1, x2, x3)  =  x3
le2(x1, x2)  =  x2
true  =  true
0  =  0
s1(x1)  =  s
false  =  false
nil  =  nil
if_low3(x1, x2, x3)  =  x3

Lexicographic Path Order [19].
Precedence:
0 > true > add1
0 > false > add1


The following usable rules [14] were oriented:

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))
low2(n, nil) -> nil
low2(n, add2(m, x)) -> if_low3(le2(m, n), n, add2(m, x))
if_low3(false, n, add2(m, x)) -> low2(n, x)
if_low3(true, n, add2(m, x)) -> add2(m, low2(n, x))



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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(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))
quicksort1(nil) -> nil
quicksort1(add2(n, x)) -> app2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(add2(x0, x1), x2)
low2(x0, nil)
low2(x0, add2(x1, x2))
if_low3(true, x0, add2(x1, x2))
if_low3(false, x0, add2(x1, x2))
high2(x0, nil)
high2(x0, add2(x1, x2))
if_high3(true, x0, add2(x1, x2))
if_high3(false, x0, add2(x1, x2))
quicksort1(nil)
quicksort1(add2(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.