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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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))


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

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
QUICKSORT1(add2(n, x)) -> APP2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
IF_HIGH3(false, n, add2(m, x)) -> 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)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
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)
QUICKSORT1(add2(n, x)) -> LOW2(n, x)
HIGH2(n, add2(m, x)) -> LE2(m, n)
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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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:

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
QUICKSORT1(add2(n, x)) -> APP2(quicksort1(low2(n, x)), add2(n, quicksort1(high2(n, x))))
IF_HIGH3(false, n, add2(m, x)) -> 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)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)
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)
QUICKSORT1(add2(n, x)) -> LOW2(n, x)
HIGH2(n, add2(m, x)) -> LE2(m, n)
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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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 contains 7 SCCs with 6 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP
              ↳ 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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

APP2(add2(n, x), y) -> APP2(x, y)
Used argument filtering: APP2(x1, x2)  =  x1
add2(x1, x2)  =  add1(x2)
Used ordering: Quasi Precedence: trivial 


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

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

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP
              ↳ 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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LE2(s1(x), s1(y)) -> LE2(x, y)
Used argument filtering: LE2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


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

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

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP
              ↳ 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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IF_HIGH3(false, n, add2(m, x)) -> HIGH2(n, x)
IF_HIGH3(true, n, add2(m, x)) -> HIGH2(n, x)
Used argument filtering: IF_HIGH3(x1, x2, x3)  =  x3
add2(x1, x2)  =  add1(x2)
HIGH2(x1, x2)  =  x2
Used ordering: Quasi Precedence: trivial 


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

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

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

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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 contains 0 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP
              ↳ QDP
              ↳ 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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

IF_LOW3(true, n, add2(m, x)) -> LOW2(n, x)
IF_LOW3(false, n, add2(m, x)) -> LOW2(n, x)
Used argument filtering: LOW2(x1, x2)  =  x2
add2(x1, x2)  =  add1(x2)
IF_LOW3(x1, x2, x3)  =  x3
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ DependencyGraphProof
              ↳ QDP
              ↳ QDP
              ↳ 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))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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 contains 0 SCCs with 1 less node.

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

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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

QUICKSORT1(add2(n, x)) -> QUICKSORT1(low2(n, x))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(high2(n, x))
Used argument filtering: QUICKSORT1(x1)  =  x1
add2(x1, x2)  =  add1(x2)
low2(x1, x2)  =  x2
high2(x1, x2)  =  x2
nil  =  nil
if_high3(x1, x2, x3)  =  x3
if_low3(x1, x2, x3)  =  x3
Used ordering: Quasi Precedence: trivial 


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

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

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP

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

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


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

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

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof

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

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
Used argument filtering: QUOT2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  x1
Used ordering: Quasi Precedence: trivial 


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

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

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
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:

minus2(x0, 0)
minus2(s1(x0), s1(x1))
quot2(0, s1(x0))
quot2(s1(x0), s1(x1))
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.