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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 DependencyGraphProof (⇔)
        ↳19 TRUE
    ↳20 QDP
        ↳21 QDPOrderProof (⇔)
        ↳22 QDP
        ↳23 DependencyGraphProof (⇔)
        ↳24 TRUE
    ↳25 QDP
        ↳26 QDPOrderProof (⇔)
        ↳27 QDP
        ↳28 PisEmptyProof (⇔)
        ↳29 TRUE

(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) DependencyPairsProof (EQUIVALENT transformation)

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

(2) 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))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 5 less nodes.

(4) Complex Obligation (AND)

(5) 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))))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(7) 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(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.
We have to consider all minimal (P,Q,R)-chains.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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))))

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

(11) 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)  =  x2
s(x1)  =  s(x1)

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(12) 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(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.
We have to consider all minimal (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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))))

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(17) 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))

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

(18) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(19) TRUE

(20) 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))))

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(22) 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))

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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(24) TRUE

(25) 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))))

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUICKSORT(cons(N, L)) → QUICKSORT(high(N, L))
QUICKSORT(cons(N, L)) → QUICKSORT(low(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(x1, x2)
high(x1, x2)  =  x2
low(x1, x2)  =  x2
nil  =  nil
ifhigh(x1, x2, x3)  =  x3
le(x1, x2)  =  x2
true  =  true
iflow(x1, x2, x3)  =  x3
false  =  false
0  =  0
s(x1)  =  s

Homeomorphic Embedding Order
The following usable rules [FROCOS05] 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)
iflow(true, N, cons(M, L)) → cons(M, low(N, L))
ifhigh(false, N, cons(M, L)) → cons(M, high(N, L))

(27) 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(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.
We have to consider all minimal (P,Q,R)-chains.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE