YES Termination w.r.t. Q proof of Rubio_04_quick.ari

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

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

(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(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

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(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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:

LE(s(X), s(Y)) → LE(X, Y)
APP(cons(N, L), Y) → APP(L, Y)
LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
LOW(N, cons(M, L)) → LE(M, N)
IFLOW(true, N, cons(M, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)
HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
HIGH(N, cons(M, L)) → LE(M, N)
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
QUICKSORT(cons(N, L)) → APP(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
QUICKSORT(cons(N, L)) → LOW(N, L)
QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → HIGH(N, L)

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(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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 5 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

APP(cons(N, L), Y) → APP(L, 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(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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(cons(N, L), Y) → APP(L, Y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

(11) Obligation:

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

APP(cons(N, L), Y) → APP(L, 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(cons(N, L), Y) → APP(L, Y)
    The graph contains the following edges 1 > 1, 2 >= 2

(13) YES

(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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
app(nil, Y) → Y
app(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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].

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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) YES

(21) Obligation:

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

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)

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(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)

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))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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].

app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

(25) Obligation:

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

HIGH(N, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
IFHIGH(false, N, cons(M, L)) → HIGH(N, L)

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, cons(M, L)) → IFHIGH(le(M, N), N, cons(M, L))
    The graph contains the following edges 1 >= 2, 2 >= 3

  • IFHIGH(true, N, cons(M, L)) → HIGH(N, L)
    The graph contains the following edges 2 >= 1, 3 > 2

  • IFHIGH(false, N, cons(M, L)) → HIGH(N, L)
    The graph contains the following edges 2 >= 1, 3 > 2

(27) YES

(28) Obligation:

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

LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
IFLOW(true, N, cons(M, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)

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(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
IFLOW(true, N, cons(M, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)

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))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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].

app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(x0, x1))

(32) Obligation:

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

LOW(N, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
IFLOW(true, N, cons(M, L)) → LOW(N, L)
IFLOW(false, N, cons(M, L)) → LOW(N, L)

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, cons(M, L)) → IFLOW(le(M, N), N, cons(M, L))
    The graph contains the following edges 1 >= 2, 2 >= 3

  • IFLOW(true, N, cons(M, L)) → LOW(N, L)
    The graph contains the following edges 2 >= 1, 3 > 2

  • IFLOW(false, N, cons(M, L)) → LOW(N, L)
    The graph contains the following edges 2 >= 1, 3 > 2

(34) YES

(35) Obligation:

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

QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))

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(cons(N, L), Y) → cons(N, app(L, Y))
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
iflow(false, N, cons(M, L)) → low(N, L)
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))
quicksort(nil) → nil
quicksort(cons(N, L)) → app(quicksort(low(N, L)), cons(N, quicksort(high(N, L))))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))

The TRS R consists of the following rules:

low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(false, N, cons(M, L)) → low(N, L)
le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
app(nil, x0)
app(cons(x0, x1), x2)
low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))
quicksort(nil)
quicksort(cons(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].

app(nil, x0)
app(cons(x0, x1), x2)
quicksort(nil)
quicksort(cons(x0, x1))

(39) Obligation:

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

QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))

The TRS R consists of the following rules:

low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(false, N, cons(M, L)) → low(N, L)
le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))

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, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))

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

(40) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


QUICKSORT(cons(N, L)) → QUICKSORT(low(N, L))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( QUICKSORT(x1) ) = max{0, 2x1 - 1}

POL( high(x1, x2) ) = x2 + 2

POL( nil ) = 0

POL( cons(x1, x2) ) = x2 + 2

POL( ifhigh(x1, ..., x3) ) = x3 + 2

POL( le(x1, x2) ) = 1

POL( true ) = 1

POL( low(x1, x2) ) = x2

POL( iflow(x1, ..., x3) ) = max{0, 2x1 + x3 - 2}

POL( false ) = 0

POL( 0 ) = 2

POL( s(x1) ) = 2x1


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(false, N, cons(M, L)) → low(N, L)
le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))

(41) Obligation:

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

QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))

The TRS R consists of the following rules:

low(N, nil) → nil
low(N, cons(M, L)) → iflow(le(M, N), N, cons(M, L))
iflow(false, N, cons(M, L)) → low(N, L)
le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))

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, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))

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

(42) 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.

(43) Obligation:

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

QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))

The TRS R consists of the following rules:

high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))

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, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))

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

(44) 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].

low(x0, nil)
low(x0, cons(x1, x2))
iflow(true, x0, cons(x1, x2))
iflow(false, x0, cons(x1, x2))

(45) Obligation:

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

QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))

The TRS R consists of the following rules:

high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))

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

(46) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUICKSORT(x1)  =  x1
cons(x1, x2)  =  cons(x2)
high(x1, x2)  =  x2
nil  =  nil
ifhigh(x1, x2, x3)  =  x3

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

cons_1=1
nil=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))

(47) Obligation:

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

high(N, nil) → nil
high(N, cons(M, L)) → ifhigh(le(M, N), N, cons(M, L))
ifhigh(true, N, cons(M, L)) → high(N, L)
le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
high(x0, nil)
high(x0, cons(x1, x2))
ifhigh(true, x0, cons(x1, x2))
ifhigh(false, x0, cons(x1, x2))

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

(48) PisEmptyProof (EQUIVALENT transformation)

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

(49) YES