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
  ↳ Non-Overlap Check

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.

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:

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)))

The set Q consists of the following terms:

qsort1(nil)
qsort1(.2(x0, x1))
lowers2(x0, nil)
lowers2(x0, .2(x1, x2))
greaters2(x0, nil)
greaters2(x0, .2(x1, x2))


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)))

The set Q consists of the following terms:

qsort1(nil)
qsort1(.2(x0, x1))
lowers2(x0, nil)
lowers2(x0, .2(x1, x2))
greaters2(x0, nil)
greaters2(x0, .2(x1, x2))

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:

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)))

The set Q consists of the following terms:

qsort1(nil)
qsort1(.2(x0, x1))
lowers2(x0, nil)
lowers2(x0, .2(x1, x2))
greaters2(x0, nil)
greaters2(x0, .2(x1, x2))

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
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

qsort1(nil)
qsort1(.2(x0, x1))
lowers2(x0, nil)
lowers2(x0, .2(x1, x2))
greaters2(x0, nil)
greaters2(x0, .2(x1, x2))

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


The following pairs can be strictly oriented and are deleted.


GREATERS2(x, .2(y, z)) -> GREATERS2(x, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
GREATERS2(x1, x2)  =  GREATERS1(x2)
.2(x1, x2)  =  .1(x2)

Lexicographic Path Order [19].
Precedence:
[GREATERS1, .1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

qsort1(nil)
qsort1(.2(x0, x1))
lowers2(x0, nil)
lowers2(x0, .2(x1, x2))
greaters2(x0, nil)
greaters2(x0, .2(x1, x2))

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
                ↳ 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)))

The set Q consists of the following terms:

qsort1(nil)
qsort1(.2(x0, x1))
lowers2(x0, nil)
lowers2(x0, .2(x1, x2))
greaters2(x0, nil)
greaters2(x0, .2(x1, x2))

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


The following pairs can be strictly oriented and are deleted.


LOWERS2(x, .2(y, z)) -> LOWERS2(x, z)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LOWERS2(x1, x2)  =  LOWERS1(x2)
.2(x1, x2)  =  .1(x2)

Lexicographic Path Order [19].
Precedence:
[LOWERS1, .1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

qsort1(nil)
qsort1(.2(x0, x1))
lowers2(x0, nil)
lowers2(x0, .2(x1, x2))
greaters2(x0, nil)
greaters2(x0, .2(x1, x2))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.