Termination w.r.t. Q proof of TRS/Rubioquick.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(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

  ↳ 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))))

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:

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))))

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:

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))))

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))))

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(cons₂(N, L), Y) -> APP₂(L, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( APP₂(x₁, x₂) ) = x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(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.
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

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))))

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 Order [17,21] with Interpretation:

POL( LE₂(x₁, x₂) ) = x₁

POL( s₁(x₁) ) = x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(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.
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

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))))

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.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = 1

POL( IFHIGH₃(x₁, ..., x₃) ) = max{0, x₂ + x₃ - 1}

POL( false ) = 0

POL( 0 ) = 1

POL( s₁(x₁) ) = x₁ + 1

POL( le₂(x₁, x₂) ) = 0

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( HIGH₂(x₁, x₂) ) = x₁ + x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

          ↳ QDP

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

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))))

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 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ 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))))

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.

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = 1

POL( false ) = max{0, -1}

POL( IFLOW₃(x₁, ..., x₃) ) = x₃

POL( 0 ) = 1

POL( s₁(x₁) ) = max{0, x₁ - 1}

POL( le₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( LOW₂(x₁, x₂) ) = x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ 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))))

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

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))))

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

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = 1

POL( false ) = 1

POL( iflow₃(x₁, ..., x₃) ) = x₃ + 1

POL( s₁(x₁) ) = 1

POL( 0 ) = max{0, -1}

POL( ifhigh₃(x₁, ..., x₃) ) = x₃

POL( le₂(x₁, x₂) ) = 1

POL( high₂(x₁, x₂) ) = x₂

POL( QUICKSORT₁(x₁) ) = x₁

POL( nil ) = max{0, -1}

POL( cons₂(x₁, x₂) ) = x₂ + 1

POL( low₂(x₁, x₂) ) = x₂ + 1

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

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

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

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = 1

POL( false ) = max{0, -1}

POL( iflow₃(x₁, ..., x₃) ) = x₂ + x₃

POL( s₁(x₁) ) = max{0, x₁ - 1}

POL( 0 ) = 1

POL( le₂(x₁, x₂) ) = max{0, x₁ - 1}

POL( QUICKSORT₁(x₁) ) = x₁

POL( nil ) = max{0, -1}

POL( cons₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( low₂(x₁, x₂) ) = x₁ + x₂

The following usable rules [14] were oriented:

low₂(N, nil) -> nil
low₂(N, cons₂(M, L)) -> iflow₃(le₂(M, N), N, cons₂(M, L))
iflow₃(false, N, cons₂(M, L)) -> low₂(N, L)
iflow₃(true, N, cons₂(M, L)) -> cons₂(M, low₂(N, L))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

le₂(0, Y) -> true
le₂(s₁(X), 0) -> false
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
app₂(nil, Y) -> Y
app₂(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.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.