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(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

Q is empty.


QTRS
  ↳ 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(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

Q is empty.

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(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))


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

HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))
QUICKSORT1(cons2(N, L)) -> LOW2(N, L)
IFLOW3(false, N, cons2(M, L)) -> LOW2(N, L)
QUICKSORT1(cons2(N, L)) -> APP2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))
IFLOW3(true, N, cons2(M, L)) -> LOW2(N, L)
IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(low2(N, L))
LOW2(N, cons2(M, L)) -> LE2(M, N)
LE2(s1(X), s1(Y)) -> LE2(X, Y)
IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
HIGH2(N, cons2(M, L)) -> LE2(M, N)
LOW2(N, cons2(M, L)) -> IFLOW3(le2(M, N), N, cons2(M, L))
QUICKSORT1(cons2(N, L)) -> HIGH2(N, L)
APP2(cons2(N, L), Y) -> APP2(L, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(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:

HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))
QUICKSORT1(cons2(N, L)) -> LOW2(N, L)
IFLOW3(false, N, cons2(M, L)) -> LOW2(N, L)
QUICKSORT1(cons2(N, L)) -> APP2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))
IFLOW3(true, N, cons2(M, L)) -> LOW2(N, L)
IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(low2(N, L))
LOW2(N, cons2(M, L)) -> LE2(M, N)
LE2(s1(X), s1(Y)) -> LE2(X, Y)
IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
HIGH2(N, cons2(M, L)) -> LE2(M, N)
LOW2(N, cons2(M, L)) -> IFLOW3(le2(M, N), N, cons2(M, L))
QUICKSORT1(cons2(N, L)) -> HIGH2(N, L)
APP2(cons2(N, L), Y) -> APP2(L, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 5 less nodes.

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

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

APP2(cons2(N, L), Y) -> APP2(L, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(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(cons2(N, L), Y) -> APP2(L, Y)
Used argument filtering: APP2(x1, x2)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial


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

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

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(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

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

LE2(s1(X), s1(Y)) -> LE2(X, Y)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(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

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

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(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

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

HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))
IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(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.

IFHIGH3(false, N, cons2(M, L)) -> HIGH2(N, L)
IFHIGH3(true, N, cons2(M, L)) -> HIGH2(N, L)
Used argument filtering: HIGH2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
IFHIGH3(x1, x2, x3)  =  x3
le2(x1, x2)  =  le
0  =  0
true  =  true
s1(x1)  =  s
false  =  false
Used ordering: Quasi Precedence: [le, true] > false


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

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

HIGH2(N, cons2(M, L)) -> IFHIGH3(le2(M, N), N, cons2(M, L))

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(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

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

IFLOW3(false, N, cons2(M, L)) -> LOW2(N, L)
LOW2(N, cons2(M, L)) -> IFLOW3(le2(M, N), N, cons2(M, L))
IFLOW3(true, N, cons2(M, L)) -> LOW2(N, L)

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(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.

IFLOW3(false, N, cons2(M, L)) -> LOW2(N, L)
IFLOW3(true, N, cons2(M, L)) -> LOW2(N, L)
Used argument filtering: IFLOW3(x1, x2, x3)  =  x3
cons2(x1, x2)  =  cons1(x2)
LOW2(x1, x2)  =  x2
le2(x1, x2)  =  le
0  =  0
true  =  true
s1(x1)  =  s
false  =  false
Used ordering: Quasi Precedence: [le, true] > false


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

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

LOW2(N, cons2(M, L)) -> IFLOW3(le2(M, N), N, cons2(M, L))

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(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

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

QUICKSORT1(cons2(N, L)) -> QUICKSORT1(low2(N, L))
QUICKSORT1(cons2(N, L)) -> QUICKSORT1(high2(N, L))

The TRS R consists of the following rules:

le2(0, Y) -> true
le2(s1(X), 0) -> false
le2(s1(X), s1(Y)) -> le2(X, Y)
app2(nil, Y) -> Y
app2(cons2(N, L), Y) -> cons2(N, app2(L, Y))
low2(N, nil) -> nil
low2(N, cons2(M, L)) -> iflow3(le2(M, N), N, cons2(M, L))
iflow3(true, N, cons2(M, L)) -> cons2(M, low2(N, L))
iflow3(false, N, cons2(M, L)) -> low2(N, L)
high2(N, nil) -> nil
high2(N, cons2(M, L)) -> ifhigh3(le2(M, N), N, cons2(M, L))
ifhigh3(true, N, cons2(M, L)) -> high2(N, L)
ifhigh3(false, N, cons2(M, L)) -> cons2(M, high2(N, L))
quicksort1(nil) -> nil
quicksort1(cons2(N, L)) -> app2(quicksort1(low2(N, L)), cons2(N, quicksort1(high2(N, L))))

The set Q consists of the following terms:

le2(0, x0)
le2(s1(x0), 0)
le2(s1(x0), s1(x1))
app2(nil, x0)
app2(cons2(x0, x1), x2)
low2(x0, nil)
low2(x0, cons2(x1, x2))
iflow3(true, x0, cons2(x1, x2))
iflow3(false, x0, cons2(x1, x2))
high2(x0, nil)
high2(x0, cons2(x1, x2))
ifhigh3(true, x0, cons2(x1, x2))
ifhigh3(false, x0, cons2(x1, x2))
quicksort1(nil)
quicksort1(cons2(x0, x1))

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