Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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)))

The set Q consists of the following terms:

qsort(nil)
qsort(.(x0, x1))
lowers(x0, nil)
lowers(x0, .(x1, x2))
greaters(x0, nil)
greaters(x0, .(x1, x2))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

The set Q consists of the following terms:

qsort(nil)
qsort(.(x0, x1))
lowers(x0, nil)
lowers(x0, .(x1, x2))
greaters(x0, nil)
greaters(x0, .(x1, x2))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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)))

The set Q consists of the following terms:

qsort(nil)
qsort(.(x0, x1))
lowers(x0, nil)
lowers(x0, .(x1, x2))
greaters(x0, nil)
greaters(x0, .(x1, x2))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
GREATERS(x1, x2)  =  GREATERS(x1, x2)
.(x1, x2)  =  .(x1, x2)
qsort(x1)  =  qsort(x1)
nil  =  nil
++(x1, x2)  =  ++(x1)
lowers(x1, x2)  =  lowers(x1, x2)
greaters(x1, x2)  =  greaters(x1, x2)
if(x1, x2, x3)  =  if(x1)
<=(x1, x2)  =  x2

Lexicographic path order with status [LPO].
Quasi-Precedence:

GREATERS₂ > [nil, ++₁, greaters₂, if₁]
[.₂, qsort₁] > lowers₂ > [nil, ++₁, greaters₂, if₁]

Status:

GREATERS₂: [2,1]
.₂: [2,1]
qsort₁: [1]
nil: []
++₁: [1]
lowers₂: [2,1]
greaters₂: [1,2]
if₁: [1]

The following usable rules [FROCOS05] were oriented:

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)))

(9) Obligation:

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)))

The set Q consists of the following terms:

qsort(nil)
qsort(.(x0, x1))
lowers(x0, nil)
lowers(x0, .(x1, x2))
greaters(x0, nil)
greaters(x0, .(x1, x2))

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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)))

The set Q consists of the following terms:

qsort(nil)
qsort(.(x0, x1))
lowers(x0, nil)
lowers(x0, .(x1, x2))
greaters(x0, nil)
greaters(x0, .(x1, x2))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
LOWERS(x1, x2)  =  LOWERS(x1, x2)
.(x1, x2)  =  .(x1, x2)
qsort(x1)  =  qsort(x1)
nil  =  nil
++(x1, x2)  =  ++(x1)
lowers(x1, x2)  =  lowers(x1, x2)
greaters(x1, x2)  =  greaters(x1, x2)
if(x1, x2, x3)  =  if(x1)
<=(x1, x2)  =  x2

Lexicographic path order with status [LPO].
Quasi-Precedence:

LOWERS₂ > [nil, ++₁, greaters₂, if₁]
[.₂, qsort₁] > lowers₂ > [nil, ++₁, greaters₂, if₁]

Status:

LOWERS₂: [2,1]
.₂: [2,1]
qsort₁: [1]
nil: []
++₁: [1]
lowers₂: [2,1]
greaters₂: [1,2]
if₁: [1]

The following usable rules [FROCOS05] were oriented:

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)))

(14) Obligation:

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)))

The set Q consists of the following terms:

qsort(nil)
qsort(.(x0, x1))
lowers(x0, nil)
lowers(x0, .(x1, x2))
greaters(x0, nil)
greaters(x0, .(x1, x2))

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

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) TRUE