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:

qsort(nil) → nil
qsort(.(x, y)) → ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) → nil
lowers(x, .(y, z)) → if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) → nil
greaters(x, .(y, z)) → if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

qsort(nil) → nil
qsort(.(x, y)) → ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) → nil
lowers(x, .(y, z)) → if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) → nil
greaters(x, .(y, z)) → if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LOWERS(x, .(y, z)) → LOWERS(x, z)
GREATERS(x, .(y, z)) → GREATERS(x, z)
QSORT(.(x, y)) → GREATERS(x, y)
QSORT(.(x, y)) → LOWERS(x, y)
QSORT(.(x, y)) → QSORT(greaters(x, y))
QSORT(.(x, y)) → QSORT(lowers(x, y))

The TRS R consists of the following rules:

qsort(nil) → nil
qsort(.(x, y)) → ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) → nil
lowers(x, .(y, z)) → if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) → nil
greaters(x, .(y, z)) → if(<=(y, x), greaters(x, z), .(y, greaters(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:

LOWERS(x, .(y, z)) → LOWERS(x, z)
GREATERS(x, .(y, z)) → GREATERS(x, z)
QSORT(.(x, y)) → GREATERS(x, y)
QSORT(.(x, y)) → LOWERS(x, y)
QSORT(.(x, y)) → QSORT(greaters(x, y))
QSORT(.(x, y)) → QSORT(lowers(x, y))

The TRS R consists of the following rules:

qsort(nil) → nil
qsort(.(x, y)) → ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) → nil
lowers(x, .(y, z)) → if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) → nil
greaters(x, .(y, z)) → if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

GREATERS(x, .(y, z)) → GREATERS(x, z)

The TRS R consists of the following rules:

qsort(nil) → nil
qsort(.(x, y)) → ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) → nil
lowers(x, .(y, z)) → if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) → nil
greaters(x, .(y, z)) → if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


GREATERS(x, .(y, z)) → GREATERS(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(GREATERS(x1, x2)) = (2)x_2   
POL(.(x1, x2)) = 1/4 + (7/2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

qsort(nil) → nil
qsort(.(x, y)) → ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) → nil
lowers(x, .(y, z)) → if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) → nil
greaters(x, .(y, z)) → if(<=(y, x), greaters(x, z), .(y, greaters(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:

LOWERS(x, .(y, z)) → LOWERS(x, z)

The TRS R consists of the following rules:

qsort(nil) → nil
qsort(.(x, y)) → ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) → nil
lowers(x, .(y, z)) → if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) → nil
greaters(x, .(y, z)) → if(<=(y, x), greaters(x, z), .(y, greaters(x, z)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


LOWERS(x, .(y, z)) → LOWERS(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(.(x1, x2)) = 1/4 + (7/2)x_2   
POL(LOWERS(x1, x2)) = (2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

qsort(nil) → nil
qsort(.(x, y)) → ++(qsort(lowers(x, y)), .(x, qsort(greaters(x, y))))
lowers(x, nil) → nil
lowers(x, .(y, z)) → if(<=(y, x), .(y, lowers(x, z)), lowers(x, z))
greaters(x, nil) → nil
greaters(x, .(y, z)) → if(<=(y, x), greaters(x, z), .(y, greaters(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.