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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
    ↳26 QDPSizeChangeProof (⇔)
    ↳27 TRUE
  ↳28 QDP
    ↳29 UsableRulesProof (⇔)
    ↳30 QDP
    ↳31 QReductionProof (⇔)
    ↳32 QDP
    ↳33 QDPSizeChangeProof (⇔)
    ↳34 TRUE
  ↳35 QDP
    ↳36 UsableRulesProof (⇔)
    ↳37 QDP
    ↳38 QReductionProof (⇔)
    ↳39 QDP
    ↳40 QDPOrderProof (⇔)
    ↳41 QDP
    ↳42 PisEmptyProof (⇔)
    ↳43 TRUE
  ↳44 QDP
    ↳45 UsableRulesProof (⇔)
    ↳46 QDP
    ↳47 QReductionProof (⇔)
    ↳48 QDP
    ↳49 QDPSizeChangeProof (⇔)
    ↳50 TRUE
  ↳51 QDP
    ↳52 UsableRulesProof (⇔)
    ↳53 QDP
    ↳54 QReductionProof (⇔)
    ↳55 QDP
    ↳56 QDPOrderProof (⇔)
    ↳57 QDP
    ↳58 PisEmptyProof (⇔)
    ↳59 TRUE

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

(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:

MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
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(add(n, x)) → APP(quicksort(low(n, x)), add(n, quicksort(high(n, x))))
QUICKSORT(add(n, x)) → QUICKSORT(low(n, x))
QUICKSORT(add(n, x)) → LOW(n, x)
QUICKSORT(add(n, x)) → QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) → HIGH(n, x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 6 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

APP(add(n, x), y) → APP(x, y)

R is empty.
The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

(11) Obligation:

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

APP(add(n, x), y) → APP(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • APP(add(n, x), y) → APP(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

(13) TRUE

(14) 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

R is empty.
The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

(18) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LE(s(x), s(y)) → LE(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(20) TRUE

(21) 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) 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)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(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))
quicksort(nil)
quicksort(add(x0, x1))

(25) 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)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • HIGH(n, add(m, x)) → IF_HIGH(le(m, n), n, add(m, x))
    The graph contains the following edges 1 >= 2, 2 >= 3

  • IF_HIGH(true, n, add(m, x)) → HIGH(n, x)
    The graph contains the following edges 2 >= 1, 3 > 2

  • IF_HIGH(false, n, add(m, x)) → HIGH(n, x)
    The graph contains the following edges 2 >= 1, 3 > 2

(27) TRUE

(28) 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) 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)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(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))
quicksort(nil)
quicksort(add(x0, x1))

(32) 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)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))

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

(33) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • LOW(n, add(m, x)) → IF_LOW(le(m, n), n, add(m, x))
    The graph contains the following edges 1 >= 2, 2 >= 3

  • IF_LOW(true, n, add(m, x)) → LOW(n, x)
    The graph contains the following edges 2 >= 1, 3 > 2

  • IF_LOW(false, n, add(m, x)) → LOW(n, x)
    The graph contains the following edges 2 >= 1, 3 > 2

(34) TRUE

(35) Obligation:

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

QUICKSORT(add(n, x)) → QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) → QUICKSORT(low(n, x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(36) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(37) Obligation:

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

QUICKSORT(add(n, x)) → QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) → QUICKSORT(low(n, x))

The TRS R consists of the following rules:

low(n, nil) → nil
if_low(false, n, add(m, x)) → low(n, x)
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_low(true, n, add(m, x)) → add(m, low(n, x))
high(n, nil) → nil
if_high(true, n, add(m, x)) → high(n, x)
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(false, n, add(m, x)) → add(m, high(n, x))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(38) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
app(nil, x0)
app(add(x0, x1), x2)
quicksort(nil)
quicksort(add(x0, x1))

(39) Obligation:

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

QUICKSORT(add(n, x)) → QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) → QUICKSORT(low(n, x))

The TRS R consists of the following rules:

low(n, nil) → nil
if_low(false, n, add(m, x)) → low(n, x)
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_low(true, n, add(m, x)) → add(m, low(n, x))
high(n, nil) → nil
if_high(true, n, add(m, x)) → high(n, x)
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(false, n, add(m, x)) → add(m, high(n, x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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))

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUICKSORT(add(n, x)) → QUICKSORT(high(n, x))
QUICKSORT(add(n, x)) → QUICKSORT(low(n, x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(QUICKSORT(x1)) = x1   
POL(add(x1, x2)) = 1 + x2   
POL(false) = 0   
POL(high(x1, x2)) = x2   
POL(if_high(x1, x2, x3)) = x3   
POL(if_low(x1, x2, x3)) = x3   
POL(le(x1, x2)) = 0   
POL(low(x1, x2)) = x2   
POL(nil) = 0   
POL(s(x1)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

high(n, nil) → nil
if_low(true, n, add(m, x)) → add(m, low(n, x))
if_low(false, n, add(m, x)) → low(n, x)
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
low(n, nil) → nil
if_high(true, n, add(m, x)) → high(n, x)
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(false, n, add(m, x)) → add(m, high(n, x))

(41) Obligation:

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

low(n, nil) → nil
if_low(false, n, add(m, x)) → low(n, x)
low(n, add(m, x)) → if_low(le(m, n), n, add(m, x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
if_low(true, n, add(m, x)) → add(m, low(n, x))
high(n, nil) → nil
if_high(true, n, add(m, x)) → high(n, x)
high(n, add(m, x)) → if_high(le(m, n), n, add(m, x))
if_high(false, n, add(m, x)) → add(m, high(n, x))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
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))

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(43) TRUE

(44) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(45) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(46) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(47) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

(48) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

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

(49) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • MINUS(s(x), s(y)) → MINUS(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(50) TRUE

(51) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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))
quicksort(nil) → nil
quicksort(add(n, x)) → app(quicksort(low(n, x)), add(n, quicksort(high(n, x))))

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(52) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(53) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

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

(54) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

quot(0, s(x0))
quot(s(x0), s(x1))
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))
quicksort(nil)
quicksort(add(x0, x1))

(55) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

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

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(QUOT(x1, x2)) = x1   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x

(57) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

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

(58) PisEmptyProof (EQUIVALENT transformation)

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

(59) TRUE