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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

Q is empty.


QTRS
  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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:

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)
QUICKSORT(cons(N, L)) → APP(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))
IFLOW(true, N, cons(M, L)) → LOW(N, L)
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
LOW(N, cons(M, L)) → LE(M, N)
LE(s(X), s(Y)) → LE(X, Y)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
HIGH(N, cons(M, L)) → LE(M, N)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → HIGH(N, L)
APP(cons(N, L), Y) → APP(L, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ EdgeDeletionProof

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

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)
QUICKSORT(cons(N, L)) → APP(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))
IFLOW(true, N, cons(M, L)) → LOW(N, L)
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
LOW(N, cons(M, L)) → LE(M, N)
LE(s(X), s(Y)) → LE(X, Y)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
HIGH(N, cons(M, L)) → LE(M, N)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → HIGH(N, L)
APP(cons(N, L), Y) → APP(L, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
QDP
              ↳ DependencyGraphProof

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

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)
QUICKSORT(cons(N, L)) → APP(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
IFLOW(true, N, cons(M, L)) → LOW(N, L)
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
LOW(N, cons(M, L)) → LE(M, N)
LE(s(X), s(Y)) → LE(X, Y)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
HIGH(N, cons(M, L)) → LE(M, N)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
QUICKSORT(cons(N, L)) → HIGH(N, L)
APP(cons(N, L), Y) → APP(L, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

APP(cons(N, L), Y) → APP(L, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.


APP(cons(N, L), Y) → APP(L, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

LE(s(X), s(Y)) → LE(X, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.


LE(s(X), s(Y)) → LE(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LE(x1, x2)  =  LE(x2)
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
s1 > LE1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.


HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
HIGH(x1, x2)  =  HIGH(x2)
cons(x1, x2)  =  cons(x2)
IFHIGH(x1, x2, x3)  =  IFHIGH(x1, x3)
le(x1, x2)  =  le
false  =  false
true  =  true
s(x1)  =  s
0  =  0

Lexicographic Path Order [19].
Precedence:
cons1 > le > false > HIGH1 > IFHIGH2
cons1 > le > true > IFHIGH2
s > IFHIGH2
0 > false > HIGH1 > IFHIGH2
0 > true > IFHIGH2

The following usable rules [14] were oriented:

le(s(X), 0) → false
le(0, Y) → true
le(s(X), s(Y)) → le(X, Y)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

IFLOW(false, N, cons(M, L)) → LOW(N, L)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
IFLOW(true, N, cons(M, L)) → LOW(N, L)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.


IFLOW(false, N, cons(M, L)) → LOW(N, L)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
IFLOW(true, N, cons(M, L)) → LOW(N, L)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
IFLOW(x1, x2, x3)  =  IFLOW(x1, x3)
false  =  false
cons(x1, x2)  =  cons(x1, x2)
LOW(x1, x2)  =  LOW(x2)
le(x1, x2)  =  le(x1)
true  =  true
s(x1)  =  s(x1)
0  =  0

Lexicographic Path Order [19].
Precedence:
cons2 > LOW1 > IFLOW2 > false
cons2 > LOW1 > le1 > false
s1 > le1 > false
0 > true > LOW1 > IFLOW2 > false
0 > true > LOW1 > le1 > false

The following usable rules [14] were oriented:

le(s(X), 0) → false
le(0, Y) → true
le(s(X), s(Y)) → le(X, Y)



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ QDPOrderProof

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

QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

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.


QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
QUICKSORT(x1)  =  QUICKSORT(x1)
cons(x1, x2)  =  cons(x1, x2)
low(x1, x2)  =  x2
high(x1, x2)  =  x2
le(x1, x2)  =  le(x1, x2)
s(x1)  =  s
0  =  0
false  =  false
nil  =  nil
iflow(x1, x2, x3)  =  x3
true  =  true
ifhigh(x1, x2, x3)  =  x3

Lexicographic Path Order [19].
Precedence:
cons2 > le2 > false
cons2 > le2 > true
0 > false

The following usable rules [14] were oriented:

low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(false, N, cons(M, L)) → low(N, L)
ifhigh(true, N, cons(M, L)) → high(N, L)
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
high(N, nil) → nil
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ 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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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.