Termination w.r.t. Q proof of /tmp/tmpCVlf77/quicksort.trs

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

    ↳23 QDPOrderProof (⇔)

    ↳24 QDP

    ↳25 PisEmptyProof (⇔)

    ↳26 TRUE

  ↳27 QDP

(0) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, 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:

LE(s(x), s(y)) → LE(x, y)
APP(add(n, x), y) → APP(x, y)
LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))
LOW(n, add(m, x)) → LE(m, n)
IF_LOW(true, n, add(m, x)) → LOW(n, x)
IF_LOW(false, n, add(m, x)) → LOW(n, x)
HIGH(n, add(m, x)) → IF_HIGH(le(m, n), n, add(m, x))
HIGH(n, add(m, x)) → LE(m, n)
IF_HIGH(true, n, add(m, x)) → HIGH(n, x)
IF_HIGH(false, n, add(m, x)) → HIGH(n, x)
QUICKSORT(x) → IF_QS(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
QUICKSORT(x) → ISEMPTY(x)
QUICKSORT(x) → LOW(head(x), tail(x))
QUICKSORT(x) → HEAD(x)
QUICKSORT(x) → TAIL(x)
QUICKSORT(x) → HIGH(head(x), tail(x))
IF_QS(false, x, n, y) → APP(quicksort(x), add(n, quicksort(y)))
IF_QS(false, x, n, y) → QUICKSORT(x)
IF_QS(false, x, n, y) → QUICKSORT(y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, 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 5 SCCs with 8 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(add(n, x), y) → APP(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, 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.

APP(add(n, x), y) → APP(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2) = APP(x1)
add(x1, x2) = add(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

add₂ > APP₁

Status:

APP₁: [1]
add₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, 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:

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, 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.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1, x2) = x1
s(x1) = s(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, 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

(17) Obligation:

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

HIGH(n, add(m, x)) → IF_HIGH(le(m, n), n, add(m, x))
IF_HIGH(true, n, add(m, x)) → HIGH(n, x)
IF_HIGH(false, n, add(m, x)) → HIGH(n, x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, x0, x1, x2)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

HIGH(n, add(m, x)) → IF_HIGH(le(m, n), n, add(m, x))
IF_HIGH(true, n, add(m, x)) → HIGH(n, x)
IF_HIGH(false, n, add(m, x)) → HIGH(n, x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HIGH(x1, x2) = HIGH(x2)
add(x1, x2) = add(x2)
IF_HIGH(x1, x2, x3) = x3
le(x1, x2) = le(x1, x2)
true = true
false = false
s(x1) = x1
0 = 0

Lexicographic path order with status [LPO].
Precedence:

add₁ > HIGH₁
le₂ > HIGH₁
true > HIGH₁
false > HIGH₁
0 > HIGH₁

Status:

add₁: [1]
le₂: [1,2]
true: []
false: []
HIGH₁: [1]
0: []

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, x0, x1, x2)

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))
IF_LOW(true, n, add(m, x)) → LOW(n, x)
IF_LOW(false, n, add(m, x)) → LOW(n, x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, x0, x1, x2)

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

(23) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))
IF_LOW(true, n, add(m, x)) → LOW(n, x)
IF_LOW(false, n, add(m, x)) → LOW(n, x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LOW(x1, x2) = LOW(x2)
add(x1, x2) = add(x2)
IF_LOW(x1, x2, x3) = x3
le(x1, x2) = le(x1, x2)
true = true
false = false
s(x1) = x1
0 = 0

Lexicographic path order with status [LPO].
Precedence:

add₁ > LOW₁
le₂ > LOW₁
true > LOW₁
false > LOW₁
0 > LOW₁

Status:

add₁: [1]
le₂: [1,2]
true: []
false: []
LOW₁: [1]
0: []

The following usable rules [FROCOS05] were oriented: none

(24) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, x0, x1, x2)

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) TRUE

(27) Obligation:

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

QUICKSORT(x) → IF_QS(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
IF_QS(false, x, n, y) → QUICKSORT(x)
IF_QS(false, x, n, y) → QUICKSORT(y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
app(nil, y) → y
app(add(n, x), y) → add(n, app(x, y))
low(n, nil) → nil
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
high(n, nil) → nil
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(true, n, add(m, x)) → high(n, x)
if_high(false, n, add(m, x)) → add(m, high(n, x))
head(add(n, x)) → n
tail(add(n, x)) → x
isempty(nil) → true
isempty(add(n, x)) → false
quicksort(x) → if_qs(isempty(x), low(head(x), tail(x)), head(x), high(head(x), tail(x)))
if_qs(true, x, n, y) → nil
if_qs(false, x, n, y) → app(quicksort(x), add(n, quicksort(y)))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
low(x0, nil)
low(x0, add(x1, x2))
if_low(true, x0, add(x1, x2))
if_low(false, x0, add(x1, x2))
high(x0, nil)
high(x0, add(x1, x2))
if_high(true, x0, add(x1, x2))
if_high(false, x0, add(x1, x2))
head(add(x0, x1))
tail(add(x0, x1))
isempty(nil)
isempty(add(x0, x1))
quicksort(x0)
if_qs(true, x0, x1, x2)
if_qs(false, x0, x1, x2)

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