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
  ↳ 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))))

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:

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))))

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:

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))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 7 SCCs with 6 less nodes.

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

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)) = 2·x1   
POL(add2(x1, x2)) = 1 + 2·x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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))))

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
          ↳ 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))))

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


The following pairs can be oriented strictly and are deleted.


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

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

The following usable rules [14] were oriented: none



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

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
          ↳ 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))))

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) = 2   
POL(HIGH2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(IF_HIGH3(x1, x2, x3)) = 2·x1 + 2·x2 + x3   
POL(add2(x1, x2)) = 2·x1 + 2·x2   
POL(false) = 2   
POL(le2(x1, x2)) = x1   
POL(s1(x1)) = 2 + 2·x1   
POL(true) = 2   

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
                ↳ PisEmptyProof
          ↳ 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))))

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
          ↳ 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))))

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_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.

LOW2(n, add2(m, x)) -> IF_LOW3(le2(m, n), n, add2(m, x))
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(IF_LOW3(x1, x2, x3)) = x1 + x3   
POL(LOW2(x1, x2)) = 2·x2   
POL(add2(x1, x2)) = 2·x1 + 2·x2   
POL(false) = 2   
POL(le2(x1, x2)) = 2·x1   
POL(s1(x1)) = 2 + 2·x1   
POL(true) = 2   

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
          ↳ QDP
            ↳ QDPOrderProof
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))))

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
          ↳ 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))))

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(add2(n, x)) -> QUICKSORT1(low2(n, x))
QUICKSORT1(add2(n, x)) -> QUICKSORT1(high2(n, x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(QUICKSORT1(x1)) = 2·x1   
POL(add2(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(false) = 0   
POL(high2(x1, x2)) = x2   
POL(if_high3(x1, x2, x3)) = x3   
POL(if_low3(x1, x2, x3)) = 2·x3   
POL(le2(x1, x2)) = 0   
POL(low2(x1, x2)) = 2·x2   
POL(nil) = 0   
POL(s1(x1)) = 0   
POL(true) = 0   

The following usable rules [14] were oriented:

high2(n, add2(m, x)) -> if_high3(le2(m, n), n, add2(m, x))
if_high3(true, n, add2(m, x)) -> 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))
if_high3(false, n, add2(m, x)) -> add2(m, high2(n, x))
high2(n, nil) -> nil



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

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
QDP
            ↳ QDPOrderProof
          ↳ 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))))

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.


MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MINUS2(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



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

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
          ↳ QDP
QDP
            ↳ QDPOrderProof

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))))

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.


QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(QUOT2(x1, x2)) = 2·x1   
POL(minus2(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented:

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



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

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.