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:

qsort1(nil) -> nil
qsort1(.2(x, y)) -> ++2(qsort1(lowers2(x, y)), .2(x, qsort1(greaters2(x, y))))
lowers2(x, nil) -> nil
lowers2(x, .2(y, z)) -> if3(<=2(y, x), .2(y, lowers2(x, z)), lowers2(x, z))
greaters2(x, nil) -> nil
greaters2(x, .2(y, z)) -> if3(<=2(y, x), greaters2(x, z), .2(y, greaters2(x, z)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

qsort1(nil) -> nil
qsort1(.2(x, y)) -> ++2(qsort1(lowers2(x, y)), .2(x, qsort1(greaters2(x, y))))
lowers2(x, nil) -> nil
lowers2(x, .2(y, z)) -> if3(<=2(y, x), .2(y, lowers2(x, z)), lowers2(x, z))
greaters2(x, nil) -> nil
greaters2(x, .2(y, z)) -> if3(<=2(y, x), greaters2(x, z), .2(y, greaters2(x, z)))

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:

QSORT1(.2(x, y)) -> LOWERS2(x, y)
GREATERS2(x, .2(y, z)) -> GREATERS2(x, z)
LOWERS2(x, .2(y, z)) -> LOWERS2(x, z)
QSORT1(.2(x, y)) -> GREATERS2(x, y)
QSORT1(.2(x, y)) -> QSORT1(lowers2(x, y))
QSORT1(.2(x, y)) -> QSORT1(greaters2(x, y))

The TRS R consists of the following rules:

qsort1(nil) -> nil
qsort1(.2(x, y)) -> ++2(qsort1(lowers2(x, y)), .2(x, qsort1(greaters2(x, y))))
lowers2(x, nil) -> nil
lowers2(x, .2(y, z)) -> if3(<=2(y, x), .2(y, lowers2(x, z)), lowers2(x, z))
greaters2(x, nil) -> nil
greaters2(x, .2(y, z)) -> if3(<=2(y, x), greaters2(x, z), .2(y, greaters2(x, z)))

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:

QSORT1(.2(x, y)) -> LOWERS2(x, y)
GREATERS2(x, .2(y, z)) -> GREATERS2(x, z)
LOWERS2(x, .2(y, z)) -> LOWERS2(x, z)
QSORT1(.2(x, y)) -> GREATERS2(x, y)
QSORT1(.2(x, y)) -> QSORT1(lowers2(x, y))
QSORT1(.2(x, y)) -> QSORT1(greaters2(x, y))

The TRS R consists of the following rules:

qsort1(nil) -> nil
qsort1(.2(x, y)) -> ++2(qsort1(lowers2(x, y)), .2(x, qsort1(greaters2(x, y))))
lowers2(x, nil) -> nil
lowers2(x, .2(y, z)) -> if3(<=2(y, x), .2(y, lowers2(x, z)), lowers2(x, z))
greaters2(x, nil) -> nil
greaters2(x, .2(y, z)) -> if3(<=2(y, x), greaters2(x, z), .2(y, greaters2(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 less nodes.

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

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

GREATERS2(x, .2(y, z)) -> GREATERS2(x, z)

The TRS R consists of the following rules:

qsort1(nil) -> nil
qsort1(.2(x, y)) -> ++2(qsort1(lowers2(x, y)), .2(x, qsort1(greaters2(x, y))))
lowers2(x, nil) -> nil
lowers2(x, .2(y, z)) -> if3(<=2(y, x), .2(y, lowers2(x, z)), lowers2(x, z))
greaters2(x, nil) -> nil
greaters2(x, .2(y, z)) -> if3(<=2(y, x), greaters2(x, z), .2(y, greaters2(x, z)))

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.


GREATERS2(x, .2(y, z)) -> GREATERS2(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( .2(x1, x2) ) = 3x1 + x2 + 1


POL( GREATERS2(x1, x2) ) = x1 + 2x2 + 3



The following usable rules [14] were oriented: none



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

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

qsort1(nil) -> nil
qsort1(.2(x, y)) -> ++2(qsort1(lowers2(x, y)), .2(x, qsort1(greaters2(x, y))))
lowers2(x, nil) -> nil
lowers2(x, .2(y, z)) -> if3(<=2(y, x), .2(y, lowers2(x, z)), lowers2(x, z))
greaters2(x, nil) -> nil
greaters2(x, .2(y, z)) -> if3(<=2(y, x), greaters2(x, z), .2(y, greaters2(x, z)))

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

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

LOWERS2(x, .2(y, z)) -> LOWERS2(x, z)

The TRS R consists of the following rules:

qsort1(nil) -> nil
qsort1(.2(x, y)) -> ++2(qsort1(lowers2(x, y)), .2(x, qsort1(greaters2(x, y))))
lowers2(x, nil) -> nil
lowers2(x, .2(y, z)) -> if3(<=2(y, x), .2(y, lowers2(x, z)), lowers2(x, z))
greaters2(x, nil) -> nil
greaters2(x, .2(y, z)) -> if3(<=2(y, x), greaters2(x, z), .2(y, greaters2(x, z)))

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.


LOWERS2(x, .2(y, z)) -> LOWERS2(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( .2(x1, x2) ) = 3x1 + x2 + 1


POL( LOWERS2(x1, x2) ) = x1 + 2x2 + 3



The following usable rules [14] were oriented: none



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

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

qsort1(nil) -> nil
qsort1(.2(x, y)) -> ++2(qsort1(lowers2(x, y)), .2(x, qsort1(greaters2(x, y))))
lowers2(x, nil) -> nil
lowers2(x, .2(y, z)) -> if3(<=2(y, x), .2(y, lowers2(x, z)), lowers2(x, z))
greaters2(x, nil) -> nil
greaters2(x, .2(y, z)) -> if3(<=2(y, x), greaters2(x, z), .2(y, greaters2(x, z)))

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.