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

(0) Obligation:

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

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

QSORT(xs) → QS(half(length(xs)), xs)
QSORT(xs) → HALF(length(xs))
QSORT(xs) → LENGTH(xs)
QS(n, cons(x, xs)) → APPEND(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
QS(n, cons(x, xs)) → QS(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs)))
QS(n, cons(x, xs)) → HALF(n)
QS(n, cons(x, xs)) → FILTERLOW(get(n, cons(x, xs)), cons(x, xs))
QS(n, cons(x, xs)) → GET(n, cons(x, xs))
QS(n, cons(x, xs)) → QS(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))
QS(n, cons(x, xs)) → FILTERHIGH(get(n, cons(x, xs)), cons(x, xs))
FILTERLOW(n, cons(x, xs)) → IF1(ge(n, x), n, x, xs)
FILTERLOW(n, cons(x, xs)) → GE(n, x)
IF1(true, n, x, xs) → FILTERLOW(n, xs)
IF1(false, n, x, xs) → FILTERLOW(n, xs)
FILTERHIGH(n, cons(x, xs)) → IF2(ge(x, n), n, x, xs)
FILTERHIGH(n, cons(x, xs)) → GE(x, n)
IF2(true, n, x, xs) → FILTERHIGH(n, xs)
IF2(false, n, x, xs) → FILTERHIGH(n, xs)
GE(s(x), s(y)) → GE(x, y)
APPEND(cons(x, xs), ys) → APPEND(xs, ys)
LENGTH(cons(x, xs)) → LENGTH(xs)
HALF(s(s(x))) → HALF(x)
GET(s(n), cons(x, cons(y, xs))) → GET(n, cons(y, xs))

The TRS R consists of the following rules:

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 10 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GET(s(n), cons(x, cons(y, xs))) → GET(n, cons(y, xs))

The TRS R consists of the following rules:

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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:

GET(s(n), cons(x, cons(y, xs))) → GET(n, cons(y, xs))

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

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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].

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

(11) Obligation:

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

GET(s(n), cons(x, cons(y, xs))) → GET(n, cons(y, xs))

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:

• GET(s(n), cons(x, cons(y, xs))) → GET(n, cons(y, xs))
  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:

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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:

HALF(s(s(x))) → HALF(x)

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

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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].

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

(18) Obligation:

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

HALF(s(s(x))) → HALF(x)

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:

• HALF(s(s(x))) → HALF(x)
  The graph contains the following edges 1 > 1

(20) YES

(21) Obligation:

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

LENGTH(cons(x, xs)) → LENGTH(xs)

The TRS R consists of the following rules:

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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:

LENGTH(cons(x, xs)) → LENGTH(xs)

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

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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].

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

(25) Obligation:

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

LENGTH(cons(x, xs)) → LENGTH(xs)

R is empty.
Q is empty.
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:

• LENGTH(cons(x, xs)) → LENGTH(xs)
  The graph contains the following edges 1 > 1

(27) YES

(28) Obligation:

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

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

The TRS R consists of the following rules:

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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:

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

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

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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].

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

(32) Obligation:

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

APPEND(cons(x, xs), ys) → APPEND(xs, ys)

R is empty.
Q is empty.
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:

• APPEND(cons(x, xs), ys) → APPEND(xs, ys)
  The graph contains the following edges 1 > 1, 2 >= 2

(34) YES

(35) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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:

GE(s(x), s(y)) → GE(x, y)

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

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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].

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

(39) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

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

(40) 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:

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

(41) YES

(42) Obligation:

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

IF2(true, n, x, xs) → FILTERHIGH(n, xs)
FILTERHIGH(n, cons(x, xs)) → IF2(ge(x, n), n, x, xs)
IF2(false, n, x, xs) → FILTERHIGH(n, xs)

The TRS R consists of the following rules:

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(43) 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.

(44) Obligation:

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

IF2(true, n, x, xs) → FILTERHIGH(n, xs)
FILTERHIGH(n, cons(x, xs)) → IF2(ge(x, n), n, x, xs)
IF2(false, n, x, xs) → FILTERHIGH(n, xs)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(45) 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].

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

(46) Obligation:

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

IF2(true, n, x, xs) → FILTERHIGH(n, xs)
FILTERHIGH(n, cons(x, xs)) → IF2(ge(x, n), n, x, xs)
IF2(false, n, x, xs) → FILTERHIGH(n, xs)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(47) 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:

• FILTERHIGH(n, cons(x, xs)) → IF2(ge(x, n), n, x, xs)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF2(true, n, x, xs) → FILTERHIGH(n, xs)
  The graph contains the following edges 2 >= 1, 4 >= 2

• IF2(false, n, x, xs) → FILTERHIGH(n, xs)
  The graph contains the following edges 2 >= 1, 4 >= 2

(48) YES

(49) Obligation:

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

IF1(true, n, x, xs) → FILTERLOW(n, xs)
FILTERLOW(n, cons(x, xs)) → IF1(ge(n, x), n, x, xs)
IF1(false, n, x, xs) → FILTERLOW(n, xs)

The TRS R consists of the following rules:

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(50) 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.

(51) Obligation:

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

IF1(true, n, x, xs) → FILTERLOW(n, xs)
FILTERLOW(n, cons(x, xs)) → IF1(ge(n, x), n, x, xs)
IF1(false, n, x, xs) → FILTERLOW(n, xs)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(52) 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].

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

(53) Obligation:

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

IF1(true, n, x, xs) → FILTERLOW(n, xs)
FILTERLOW(n, cons(x, xs)) → IF1(ge(n, x), n, x, xs)
IF1(false, n, x, xs) → FILTERLOW(n, xs)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))

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

(54) 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:

• FILTERLOW(n, cons(x, xs)) → IF1(ge(n, x), n, x, xs)
  The graph contains the following edges 1 >= 2, 2 > 3, 2 > 4

• IF1(true, n, x, xs) → FILTERLOW(n, xs)
  The graph contains the following edges 2 >= 1, 4 >= 2

• IF1(false, n, x, xs) → FILTERLOW(n, xs)
  The graph contains the following edges 2 >= 1, 4 >= 2

(55) YES

(56) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))
QS(n, cons(x, xs)) → QS(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

qsort(xs) → qs(half(length(xs)), xs)
qs(n, nil) → nil
qs(n, cons(x, xs)) → append(qs(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))), cons(get(n, cons(x, xs)), qs(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))))
filterlow(n, nil) → nil
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
append(nil, ys) → ys
append(cons(x, xs), ys) → cons(x, append(xs, ys))
length(nil) → 0
length(cons(x, xs)) → s(length(xs))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, nil) → 0
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(57) 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.

(58) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))
QS(n, cons(x, xs)) → QS(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(59) 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].

qsort(x0)
qs(x0, nil)
qs(x0, cons(x1, x2))
append(nil, ys)
append(cons(x0, x1), ys)
length(nil)
length(cons(x0, x1))

(60) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs)))
QS(n, cons(x, xs)) → QS(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(61) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(n, cons(x, xs)) → QS(half(n), filterhigh(get(n, cons(x, xs)), cons(x, xs))) at position [1] we obtained the following new rules [LPAR04]:

QS(n, cons(x, xs)) → QS(half(n), if2(ge(x, get(n, cons(x, xs))), get(n, cons(x, xs)), x, xs))

(62) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs)))
QS(n, cons(x, xs)) → QS(half(n), if2(ge(x, get(n, cons(x, xs))), get(n, cons(x, xs)), x, xs))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(63) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(n, cons(x, xs)) → QS(half(n), filterlow(get(n, cons(x, xs)), cons(x, xs))) at position [1] we obtained the following new rules [LPAR04]:

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))

(64) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if2(ge(x, get(n, cons(x, xs))), get(n, cons(x, xs)), x, xs))
QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(65) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QS(n, cons(x, xs)) → QS(half(n), if2(ge(x, get(n, cons(x, xs))), get(n, cons(x, xs)), x, xs)) at position [1] we obtained the following new rules [LPAR04]:

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), get(x0, cons(x1, nil)), x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, get(x0, cons(x1, nil))), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(s(x0), cons(x1, cons(x2, x3)))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

(66) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), get(x0, cons(x1, nil)), x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, get(x0, cons(x1, nil))), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(s(x0), cons(x1, cons(x2, x3)))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(67) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), get(x0, cons(x1, nil)), x1, nil)) at position [1,1] we obtained the following new rules [LPAR04]:

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))

(68) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, get(x0, cons(x1, nil))), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(s(x0), cons(x1, cons(x2, x3)))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(69) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2))) at position [0] we obtained the following new rules [LPAR04]:

QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))

(70) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, get(x0, cons(x1, nil))), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(s(x0), cons(x1, cons(x2, x3)))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(71) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3))) at position [1,1] we obtained the following new rules [LPAR04]:

QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

(72) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, get(x0, cons(x1, nil))), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(s(x0), cons(x1, cons(x2, x3)))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(73) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, get(x0, cons(x1, nil))), x1, x1, nil)) at position [1,0,1] we obtained the following new rules [LPAR04]:

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))

(74) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(s(x0), cons(x1, cons(x2, x3)))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(75) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2))) at position [0] we obtained the following new rules [LPAR04]:

QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))

(76) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(s(x0), cons(x1, cons(x2, x3)))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(77) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(s(x0), cons(x1, cons(x2, x3)))), get(x0, cons(x2, x3)), x1, cons(x2, x3))) at position [1,0,1] we obtained the following new rules [LPAR04]:

QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

(78) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(79) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2))) at position [1,1] we obtained the following new rules [LPAR04]:

QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))

(80) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(81) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, get(0, cons(x0, cons(x1, x2)))), x0, x0, cons(x1, x2))) at position [1,0,1] we obtained the following new rules [LPAR04]:

QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))

(82) Obligation:

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

QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs))
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(83) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule QS(n, cons(x, xs)) → QS(half(n), if1(ge(get(n, cons(x, xs)), x), get(n, cons(x, xs)), x, xs)) at position [1] we obtained the following new rules [LPAR04]:

QS(y0, cons(0, y2)) → QS(half(y0), if1(true, get(y0, cons(0, y2)), 0, y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), get(x0, cons(x1, nil)), x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(get(x0, cons(x1, nil)), x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(s(x0), cons(x1, cons(x2, x3))), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

(84) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(y0, cons(0, y2)) → QS(half(y0), if1(true, get(y0, cons(0, y2)), 0, y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), get(x0, cons(x1, nil)), x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(get(x0, cons(x1, nil)), x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(s(x0), cons(x1, cons(x2, x3))), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(85) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(y0, cons(0, y2)) → QS(half(y0), if1(true, get(y0, cons(0, y2)), 0, y2)) at position [1] we obtained the following new rules [LPAR04]:

QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))

(86) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), get(x0, cons(x1, nil)), x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(get(x0, cons(x1, nil)), x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(s(x0), cons(x1, cons(x2, x3))), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(87) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), get(x0, cons(x1, nil)), x1, nil)) at position [1,1] we obtained the following new rules [LPAR04]:

QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))

(88) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(get(x0, cons(x1, nil)), x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(s(x0), cons(x1, cons(x2, x3))), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(89) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2))) at position [0] we obtained the following new rules [LPAR04]:

QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))

(90) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(get(x0, cons(x1, nil)), x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(s(x0), cons(x1, cons(x2, x3))), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(91) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(s(x0), cons(x1, cons(x2, x3))), x1, cons(x2, x3))) at position [1,1] we obtained the following new rules [LPAR04]:

QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

(92) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(get(x0, cons(x1, nil)), x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(s(x0), cons(x1, cons(x2, x3))), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(93) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(get(x0, cons(x1, nil)), x1), x1, x1, nil)) at position [1,0,0] we obtained the following new rules [LPAR04]:

QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))

(94) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(s(x0), cons(x1, cons(x2, x3))), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(95) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(0, cons(x0, cons(x1, x2))) → QS(half(0), if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2))) at position [0] we obtained the following new rules [LPAR04]:

QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))

(96) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(s(x0), cons(x1, cons(x2, x3))), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(97) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(s(x0), cons(x1, cons(x2, x3))), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3))) at position [1,0,0] we obtained the following new rules [LPAR04]:

QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

(98) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(99) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), get(0, cons(x0, cons(x1, x2))), x0, cons(x1, x2))) at position [1,1] we obtained the following new rules [LPAR04]:

QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

(100) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(101) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(get(0, cons(x0, cons(x1, x2))), x0), x0, x0, cons(x1, x2))) at position [1,0,0] we obtained the following new rules [LPAR04]:

QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

(102) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(103) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QS(y0, cons(0, y2)) → QS(half(y0), filterlow(get(y0, cons(0, y2)), y2))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(QS(x₁, x₂)) = x₂   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(false) = 0   
POL(filterhigh(x₁, x₂)) = x₂   
POL(filterlow(x₁, x₂)) = x₂   
POL(ge(x₁, x₂)) = 0   
POL(get(x₁, x₂)) = 0   
POL(half(x₁)) = 0   
POL(if1(x₁, x₂, x₃, x₄)) = x₃ + x₄   
POL(if2(x₁, x₂, x₃, x₄)) = x₃ + x₄   
POL(nil) = 0   
POL(s(x₁)) = 0   
POL(true) = 0   

The following usable rules [FROCOS05] were oriented:

if2(true, n, x, xs) → filterhigh(n, xs)
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
if1(true, n, x, xs) → filterlow(n, xs)
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
filterlow(n, nil) → nil
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterhigh(n, nil) → nil

(104) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(105) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if2(ge(x1, get(x0, cons(x2, x3))), get(x0, cons(x2, x3)), x1, cons(x2, x3)))
QS(s(x0), cons(x1, cons(x2, x3))) → QS(half(s(x0)), if1(ge(get(x0, cons(x2, x3)), x1), get(x0, cons(x2, x3)), x1, cons(x2, x3)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( QS(x₁, x₂) ) = x₁

POL( half(x₁) ) = max{0, x₁ - 1}

POL( 0 ) = 0

POL( s(x₁) ) = x₁ + 2

POL( if1(x₁, ..., x₄) ) = max{0, 2x₃ - 2}

POL( if2(x₁, ..., x₄) ) = max{0, -1}

POL( ge(x₁, x₂) ) = max{0, -1}

POL( true ) = 0

POL( filterhigh(x₁, x₂) ) = 2x₁ + 1

POL( cons(x₁, x₂) ) = 1

POL( false ) = 0

POL( get(x₁, x₂) ) = x₂

POL( nil ) = 2

POL( filterlow(x₁, x₂) ) = max{0, 2x₁ - 2}

The following usable rules [FROCOS05] were oriented:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))

(106) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
get(n, cons(x, nil)) → x
get(0, cons(x, cons(y, xs))) → x
get(s(n), cons(x, cons(y, xs))) → get(n, cons(y, xs))
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(true, n, x, xs) → filterlow(n, xs)
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(107) 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.

(108) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

ge(x, 0) → true
ge(s(x), s(y)) → ge(x, y)
if1(true, n, x, xs) → filterlow(n, xs)
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
ge(0, s(x)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if2(true, n, x, xs) → filterhigh(n, xs)
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))
get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

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

(109) 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].

get(x0, nil)
get(x0, cons(x1, nil))
get(0, cons(x0, cons(x1, x2)))
get(s(x0), cons(x1, cons(x2, x3)))

(110) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if2(ge(x0, x0), x0, x0, cons(x1, x2)))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

ge(x, 0) → true
ge(s(x), s(y)) → ge(x, y)
if1(true, n, x, xs) → filterlow(n, xs)
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
ge(0, s(x)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if2(true, n, x, xs) → filterhigh(n, xs)
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))

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

(111) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
QS(0, cons(x0', cons(x1', x2))) → QS(0, if2(ge(x0', x0'), x0', x0', cons(x1', x2)))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(QS(x₁, x₂)) = x₂   
POL(cons(x₁, x₂)) = 1 + x₂   
POL(false) = 1   
POL(filterhigh(x₁, x₂)) = x₂   
POL(filterlow(x₁, x₂)) = x₂   
POL(ge(x₁, x₂)) = 1   
POL(half(x₁)) = 1 + x₁   
POL(if1(x₁, x₂, x₃, x₄)) = 1 + x₄   
POL(if2(x₁, x₂, x₃, x₄)) = x₁ + x₄   
POL(nil) = 0   
POL(s(x₁)) = 1   
POL(true) = 1   

At least one of these decreasing rules is always used after the deleted DP:
if2(true, n821, x1051, xs611) → filterhigh(n821, xs611)


The following formula is valid:
x0':sort[a0],x1':sort[a0],x2:sort[a0].if2'(ge(x0' , x0' ), x0' , x0' , cons(x1' , x2 ))=true


The transformed set:
if2'(true, n82, x105, xs61) → true
filterhigh'(n91, cons(x116, xs68)) → if2'(ge(x116, n91), n91, x116, xs68)
if2'(false, n100, x127, xs75) → filterhigh'(n100, xs75)
filterhigh'(n109, nil) → false
ge(x, 0) → true
ge(s(x9), s(y')) → ge(x9, y')
if1(true, n14, x20, xs10) → filterlow(n14, xs10)
filterlow(n23, cons(x31, xs17)) → if1(ge(n23, x31), n23, x31, xs17)
if1(false, n32, x42, xs24) → cons(x42, filterlow(n32, xs24))
filterlow(n41, nil) → nil
ge(0, s(x63)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x94))) → s(half(x94))
if2(true, n82, x105, xs61) → filterhigh(n82, xs61)
filterhigh(n91, cons(x116, xs68)) → if2(ge(x116, n91), n91, x116, xs68)
if2(false, n100, x127, xs75) → cons(x127, filterhigh(n100, xs75))
filterhigh(n109, nil) → nil
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a5](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a5](x0, x2), equal_sort[a5](x1, x3))
equal_sort[a5](cons(x0, x1), nil) → false
equal_sort[a5](nil, cons(x0, x1)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a67](witness_sort[a67], witness_sort[a67]) → true


The proof given by the theorem prover:
The following program was given to the internal theorem prover:

   [x, x0, x1, x2, x3, n82, x105, xs61, n100, x127, x116, xs68, n109, n91, x9, y', x63, n41, n23, x31, xs17, x94, n14, x20, n32, x42]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(x0)) -> false
   equal_sort[a0](s(x0), 0) -> false
   equal_sort[a0](s(x0), s(x1)) -> equal_sort[a0](x0, x1)
   equal_sort[a5](cons(x0, x1), cons(x2, x3)) -> equal_sort[a5](x0, x2) and equal_sort[a5](x1, x3)
   equal_sort[a5](cons(x0, x1), nil) -> false
   equal_sort[a5](nil, cons(x0, x1)) -> false
   equal_sort[a5](nil, nil) -> true
   equal_sort[a67](witness_sort[a67], witness_sort[a67]) -> true
   if2'(true, n82, x105, xs61) -> true
   if2'(false, n100, x127, cons(x116, xs68)) -> if2'(ge(x116, n100), n100, x116, xs68)
   if2'(false, n100, x127, nil) -> false
   filterhigh'(n109, nil) -> false
   equal_bool(ge(x116, n91), true) -> true | filterhigh'(n91, cons(x116, xs68)) -> true
   equal_bool(ge(x116, n91), true) -> false | filterhigh'(n91, cons(x116, xs68)) -> filterhigh'(n91, xs68)
   ge(x, 0) -> true
   ge(s(x9), s(y')) -> ge(x9, y')
   ge(0, s(x63)) -> false
   filterlow(n41, nil) -> nil
   equal_bool(ge(n23, x31), true) -> true | filterlow(n23, cons(x31, xs17)) -> filterlow(n23, xs17)
   equal_bool(ge(n23, x31), true) -> false | filterlow(n23, cons(x31, xs17)) -> cons(x31, filterlow(n23, xs17))
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x94))) -> s(half(x94))
   filterhigh(n109, nil) -> nil
   equal_bool(ge(x116, n91), true) -> true | filterhigh(n91, cons(x116, xs68)) -> filterhigh(n91, xs68)
   equal_bool(ge(x116, n91), true) -> false | filterhigh(n91, cons(x116, xs68)) -> cons(x116, filterhigh(n91, xs68))
   if1(true, n14, x20, cons(x31, xs17)) -> if1(ge(n14, x31), n14, x31, xs17)
   if1(true, n14, x20, nil) -> nil
   if1(false, n32, x42, cons(x31, xs17)) -> cons(x42, if1(ge(n32, x31), n32, x31, xs17))
   if1(false, n32, x42, nil) -> cons(x42, nil)
   if2(true, n82, x105, cons(x116, xs68)) -> if2(ge(x116, n82), n82, x116, xs68)
   if2(true, n82, x105, nil) -> nil
   if2(false, n100, x127, cons(x116, xs68)) -> cons(x127, if2(ge(x116, n100), n100, x116, xs68))
   if2(false, n100, x127, nil) -> cons(x127, nil)

The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 9a00b172b26c9abb2d4c4d5eaf341e919eb0fbf1 nowonder 20100222 unpublished dirty


Partial correctness of the following Program

   [x, x0, x1, x2, x3, n82, x105, xs61, n100, x127, x116, xs68, n109, n91, x9, y', x63, n41, n23, x31, xs17, x94, n14, x20, n32, x42]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(x0)) -> false
   equal_sort[a0](s(x0), 0) -> false
   equal_sort[a0](s(x0), s(x1)) -> equal_sort[a0](x0, x1)
   equal_sort[a5](cons(x0, x1), cons(x2, x3)) -> equal_sort[a5](x0, x2) and equal_sort[a5](x1, x3)
   equal_sort[a5](cons(x0, x1), nil) -> false
   equal_sort[a5](nil, cons(x0, x1)) -> false
   equal_sort[a5](nil, nil) -> true
   equal_sort[a67](witness_sort[a67], witness_sort[a67]) -> true
   if2'(true, n82, x105, xs61) -> true
   if2'(false, n100, x127, cons(x116, xs68)) -> if2'(ge(x116, n100), n100, x116, xs68)
   if2'(false, n100, x127, nil) -> false
   filterhigh'(n109, nil) -> false
   equal_bool(ge(x116, n91), true) -> true | filterhigh'(n91, cons(x116, xs68)) -> true
   equal_bool(ge(x116, n91), true) -> false | filterhigh'(n91, cons(x116, xs68)) -> filterhigh'(n91, xs68)
   ge(x, 0) -> true
   ge(s(x9), s(y')) -> ge(x9, y')
   ge(0, s(x63)) -> false
   filterlow(n41, nil) -> nil
   equal_bool(ge(n23, x31), true) -> true | filterlow(n23, cons(x31, xs17)) -> filterlow(n23, xs17)
   equal_bool(ge(n23, x31), true) -> false | filterlow(n23, cons(x31, xs17)) -> cons(x31, filterlow(n23, xs17))
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x94))) -> s(half(x94))
   filterhigh(n109, nil) -> nil
   equal_bool(ge(x116, n91), true) -> true | filterhigh(n91, cons(x116, xs68)) -> filterhigh(n91, xs68)
   equal_bool(ge(x116, n91), true) -> false | filterhigh(n91, cons(x116, xs68)) -> cons(x116, filterhigh(n91, xs68))
   if1(true, n14, x20, cons(x31, xs17)) -> if1(ge(n14, x31), n14, x31, xs17)
   if1(true, n14, x20, nil) -> nil
   if1(false, n32, x42, cons(x31, xs17)) -> cons(x42, if1(ge(n32, x31), n32, x31, xs17))
   if1(false, n32, x42, nil) -> cons(x42, nil)
   if2(true, n82, x105, cons(x116, xs68)) -> if2(ge(x116, n82), n82, x116, xs68)
   if2(true, n82, x105, nil) -> nil
   if2(false, n100, x127, cons(x116, xs68)) -> cons(x127, if2(ge(x116, n100), n100, x116, xs68))
   if2(false, n100, x127, nil) -> cons(x127, nil)

using the following formula:
x0':sort[a0],x1':sort[a0],x2:sort[a0].if2'(ge(x0', x0'), x0', x0', cons(x1', x2))=true

could be successfully shown:
(0) Formula
(1) Induction by data structure [EQUIVALENT]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT]
        (5) YES
    (6) Formula
        (7) Symbolic evaluation [EQUIVALENT]
        (8) Formula
        (9) Case Analysis [EQUIVALENT]
        (10) AND
            (11) Formula
                (12) Case Analysis [EQUIVALENT]
                (13) AND
                    (14) Formula
                        (15) Inverse Substitution [SOUND]
                        (16) Formula
                        (17) Induction by data structure [SOUND]
                        (18) AND
                            (19) Formula
                                (20) Symbolic evaluation [EQUIVALENT]
                                (21) YES
                            (22) Formula
                                (23) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (24) YES
                    (25) Formula
                        (26) Inverse Substitution [SOUND]
                        (27) Formula
                        (28) Induction by data structure [SOUND]
                        (29) AND
                            (30) Formula
                                (31) Symbolic evaluation [EQUIVALENT]
                                (32) YES
                            (33) Formula
                                (34) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (35) YES
            (36) Formula
                (37) Case Analysis [EQUIVALENT]
                (38) AND
                    (39) Formula
                        (40) Inverse Substitution [SOUND]
                        (41) Formula
                        (42) Induction by data structure [SOUND]
                        (43) AND
                            (44) Formula
                                (45) Symbolic evaluation [EQUIVALENT]
                                (46) YES
                            (47) Formula
                                (48) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (49) YES
                    (50) Formula
                        (51) Inverse Substitution [SOUND]
                        (52) Formula
                        (53) Induction by data structure [SOUND]
                        (54) AND
                            (55) Formula
                                (56) Symbolic evaluation [EQUIVALENT]
                                (57) YES
                            (58) Formula
                                (59) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (60) YES


----------------------------------------

(0)
Obligation:
Formula:
x0':sort[a0],x1':sort[a0],x2:sort[a0].if2'(ge(x0', x0'), x0', x0', cons(x1', x2))=true

There are no hypotheses.




----------------------------------------

(1) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
x1':sort[a0],x2:sort[a0].if2'(ge(0, 0), 0, 0, cons(x1', x2))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0],x1':sort[a0],x2:sort[a0].if2'(ge(s(n), s(n)), s(n), s(n), cons(x1', x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
x1':sort[a0],x2:sort[a0].if2'(ge(0, 0), 0, 0, cons(x1', x2))=true

There are no hypotheses.




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(5)
YES

----------------------------------------

(6)
Obligation:
Formula:
n:sort[a0],x1':sort[a0],x2:sort[a0].if2'(ge(s(n), s(n)), s(n), s(n), cons(x1', x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(7) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
n:sort[a0],x1':sort[a0],x2:sort[a0].if2'(ge(n, n), s(n), s(n), cons(x1', x2))=true
----------------------------------------

(8)
Obligation:
Formula:
n:sort[a0],x1':sort[a0],x2:sort[a0].if2'(ge(n, n), s(n), s(n), cons(x1', x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(9) Case Analysis (EQUIVALENT)
Case analysis leads to the following new obligations:

Formula:
n:sort[a0],x2:sort[a0].if2'(ge(n, n), s(n), s(n), cons(0, x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true





Formula:
n:sort[a0],x_1:sort[a0],x2:sort[a0].if2'(ge(n, n), s(n), s(n), cons(s(x_1), x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true






----------------------------------------

(10)
Complex Obligation (AND)

----------------------------------------

(11)
Obligation:
Formula:
n:sort[a0],x2:sort[a0].if2'(ge(n, n), s(n), s(n), cons(0, x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(12) Case Analysis (EQUIVALENT)
Case analysis leads to the following new obligations:

Formula:
n:sort[a0].if2'(ge(n, n), s(n), s(n), cons(0, 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true





Formula:
n:sort[a0],x_1:sort[a0].if2'(ge(n, n), s(n), s(n), cons(0, s(x_1)))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true






----------------------------------------

(13)
Complex Obligation (AND)

----------------------------------------

(14)
Obligation:
Formula:
n:sort[a0].if2'(ge(n, n), s(n), s(n), cons(0, 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(15) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0].if2'(ge(n, n), n', n', cons(0, 0))=true

Inverse substitution used:
[s(n)/n']


----------------------------------------

(16)
Obligation:
Formula:
n:sort[a0],n':sort[a0].if2'(ge(n, n), n', n', cons(0, 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(17) Induction by data structure (SOUND)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
n':sort[a0].if2'(ge(0, 0), n', n', cons(0, 0))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0].if2'(ge(s(n''), s(n'')), n', n', cons(0, 0))=true

Hypotheses:
n'':sort[a0],!n':sort[a0].if2'(ge(n'', n''), n', n', cons(0, 0))=true






----------------------------------------

(18)
Complex Obligation (AND)

----------------------------------------

(19)
Obligation:
Formula:
n':sort[a0].if2'(ge(0, 0), n', n', cons(0, 0))=true

There are no hypotheses.




----------------------------------------

(20) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(21)
YES

----------------------------------------

(22)
Obligation:
Formula:
n'':sort[a0],n':sort[a0].if2'(ge(s(n''), s(n'')), n', n', cons(0, 0))=true

Hypotheses:
n'':sort[a0],!n':sort[a0].if2'(ge(n'', n''), n', n', cons(0, 0))=true




----------------------------------------

(23) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n'':sort[a0],!n':sort[a0].if2'(ge(n'', n''), n', n', cons(0, 0))=true

----------------------------------------

(24)
YES

----------------------------------------

(25)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0].if2'(ge(n, n), s(n), s(n), cons(0, s(x_1)))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(26) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0],x_1:sort[a0].if2'(ge(n, n), n', n', cons(0, s(x_1)))=true

Inverse substitution used:
[s(n)/n']


----------------------------------------

(27)
Obligation:
Formula:
n:sort[a0],n':sort[a0],x_1:sort[a0].if2'(ge(n, n), n', n', cons(0, s(x_1)))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(28) Induction by data structure (SOUND)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
n':sort[a0],x_1:sort[a0].if2'(ge(0, 0), n', n', cons(0, s(x_1)))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0].if2'(ge(s(n''), s(n'')), n', n', cons(0, s(x_1)))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if2'(ge(n'', n''), n', n', cons(0, s(x_1)))=true






----------------------------------------

(29)
Complex Obligation (AND)

----------------------------------------

(30)
Obligation:
Formula:
n':sort[a0],x_1:sort[a0].if2'(ge(0, 0), n', n', cons(0, s(x_1)))=true

There are no hypotheses.




----------------------------------------

(31) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(32)
YES

----------------------------------------

(33)
Obligation:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0].if2'(ge(s(n''), s(n'')), n', n', cons(0, s(x_1)))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if2'(ge(n'', n''), n', n', cons(0, s(x_1)))=true




----------------------------------------

(34) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if2'(ge(n'', n''), n', n', cons(0, s(x_1)))=true

----------------------------------------

(35)
YES

----------------------------------------

(36)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0],x2:sort[a0].if2'(ge(n, n), s(n), s(n), cons(s(x_1), x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(37) Case Analysis (EQUIVALENT)
Case analysis leads to the following new obligations:

Formula:
n:sort[a0],x_1:sort[a0].if2'(ge(n, n), s(n), s(n), cons(s(x_1), 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true





Formula:
n:sort[a0],x_1:sort[a0],x_1':sort[a0].if2'(ge(n, n), s(n), s(n), cons(s(x_1), s(x_1')))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true






----------------------------------------

(38)
Complex Obligation (AND)

----------------------------------------

(39)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0].if2'(ge(n, n), s(n), s(n), cons(s(x_1), 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(40) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0],x_1:sort[a0].if2'(ge(n, n), n', n', cons(s(x_1), 0))=true

Inverse substitution used:
[s(n)/n']


----------------------------------------

(41)
Obligation:
Formula:
n:sort[a0],n':sort[a0],x_1:sort[a0].if2'(ge(n, n), n', n', cons(s(x_1), 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(42) Induction by data structure (SOUND)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
n':sort[a0],x_1:sort[a0].if2'(ge(0, 0), n', n', cons(s(x_1), 0))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0].if2'(ge(s(n''), s(n'')), n', n', cons(s(x_1), 0))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if2'(ge(n'', n''), n', n', cons(s(x_1), 0))=true






----------------------------------------

(43)
Complex Obligation (AND)

----------------------------------------

(44)
Obligation:
Formula:
n':sort[a0],x_1:sort[a0].if2'(ge(0, 0), n', n', cons(s(x_1), 0))=true

There are no hypotheses.




----------------------------------------

(45) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(46)
YES

----------------------------------------

(47)
Obligation:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0].if2'(ge(s(n''), s(n'')), n', n', cons(s(x_1), 0))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if2'(ge(n'', n''), n', n', cons(s(x_1), 0))=true




----------------------------------------

(48) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if2'(ge(n'', n''), n', n', cons(s(x_1), 0))=true

----------------------------------------

(49)
YES

----------------------------------------

(50)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0],x_1':sort[a0].if2'(ge(n, n), s(n), s(n), cons(s(x_1), s(x_1')))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(51) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0],x_1:sort[a0],x_1':sort[a0].if2'(ge(n, n), n', n', cons(s(x_1), s(x_1')))=true

Inverse substitution used:
[s(n)/n']


----------------------------------------

(52)
Obligation:
Formula:
n:sort[a0],n':sort[a0],x_1:sort[a0],x_1':sort[a0].if2'(ge(n, n), n', n', cons(s(x_1), s(x_1')))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if2'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(53) Induction by data structure (SOUND)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
n':sort[a0],x_1:sort[a0],x_1':sort[a0].if2'(ge(0, 0), n', n', cons(s(x_1), s(x_1')))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0],x_1':sort[a0].if2'(ge(s(n''), s(n'')), n', n', cons(s(x_1), s(x_1')))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_1':sort[a0].if2'(ge(n'', n''), n', n', cons(s(x_1), s(x_1')))=true






----------------------------------------

(54)
Complex Obligation (AND)

----------------------------------------

(55)
Obligation:
Formula:
n':sort[a0],x_1:sort[a0],x_1':sort[a0].if2'(ge(0, 0), n', n', cons(s(x_1), s(x_1')))=true

There are no hypotheses.




----------------------------------------

(56) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(57)
YES

----------------------------------------

(58)
Obligation:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0],x_1':sort[a0].if2'(ge(s(n''), s(n'')), n', n', cons(s(x_1), s(x_1')))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_1':sort[a0].if2'(ge(n'', n''), n', n', cons(s(x_1), s(x_1')))=true




----------------------------------------

(59) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_1':sort[a0].if2'(ge(n'', n''), n', n', cons(s(x_1), s(x_1')))=true

----------------------------------------

(60)
YES

(112) Complex Obligation (AND)

(113) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(x0, cons(x1, nil)) → QS(half(x0), if1(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

ge(x, 0) → true
ge(s(x), s(y)) → ge(x, y)
if1(true, n, x, xs) → filterlow(n, xs)
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
ge(0, s(x)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if2(true, n, x, xs) → filterhigh(n, xs)
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))

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

(114) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
QS(x0', cons(x1', nil)) → QS(half(x0'), if1(ge(x1', x1'), x1', x1', nil))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(QS(x₁, x₂)) = x₂   
POL(cons(x₁, x₂)) = 1 + x₂   
POL(false) = 0   
POL(filterhigh(x₁, x₂)) = x₂   
POL(filterlow(x₁, x₂)) = x₂   
POL(ge(x₁, x₂)) = 1   
POL(half(x₁)) = x₁   
POL(if1(x₁, x₂, x₃, x₄)) = 1 + x₄   
POL(if2(x₁, x₂, x₃, x₄)) = 1 + x₄   
POL(nil) = 1   
POL(s(x₁)) = 1 + x₁   
POL(true) = 1   

At least one of these decreasing rules is always used after the deleted DP:
if1(true, n14'', x20'', xs10'') → filterlow(n14'', xs10'')


The following formula is valid:
x1':sort[a0].if1'(ge(x1' , x1' ), x1' , x1' , nil)=true


The transformed set:
if1'(true, n14, x20, xs10) → true
filterlow'(n23, cons(x31, xs17)) → if1'(ge(n23, x31), n23, x31, xs17)
if1'(false, n32, x42, xs24) → filterlow'(n32, xs24)
filterlow'(n41, nil) → false
ge(x, 0) → true
ge(s(x9), s(y')) → ge(x9, y')
if1(true, n14, x20, xs10) → filterlow(n14, xs10)
filterlow(n23, cons(x31, xs17)) → if1(ge(n23, x31), n23, x31, xs17)
if1(false, n32, x42, xs24) → cons(x42, filterlow(n32, xs24))
filterlow(n41, nil) → nil
ge(0, s(x63)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x94))) → s(half(x94))
if2(true, n82, x105, xs61) → filterhigh(n82, xs61)
filterhigh(n91, cons(x116, xs68)) → if2(ge(x116, n91), n91, x116, xs68)
if2(false, n100, x127, xs75) → cons(x127, filterhigh(n100, xs75))
filterhigh(n109, nil) → nil
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a5](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a5](x0, x2), equal_sort[a5](x1, x3))
equal_sort[a5](cons(x0, x1), nil) → false
equal_sort[a5](nil, cons(x0, x1)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → true


The proof given by the theorem prover:
The following program was given to the internal theorem prover:

   [x, x0, x1, x2, x3, n14, x20, xs10, n32, x42, x31, xs17, n41, n23, x9, y', x63, x94, n109, n91, x116, xs68, n82, x105, n100, x127]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(x0)) -> false
   equal_sort[a0](s(x0), 0) -> false
   equal_sort[a0](s(x0), s(x1)) -> equal_sort[a0](x0, x1)
   equal_sort[a5](cons(x0, x1), cons(x2, x3)) -> equal_sort[a5](x0, x2) and equal_sort[a5](x1, x3)
   equal_sort[a5](cons(x0, x1), nil) -> false
   equal_sort[a5](nil, cons(x0, x1)) -> false
   equal_sort[a5](nil, nil) -> true
   equal_sort[a64](witness_sort[a64], witness_sort[a64]) -> true
   if1'(true, n14, x20, xs10) -> true
   if1'(false, n32, x42, cons(x31, xs17)) -> if1'(ge(n32, x31), n32, x31, xs17)
   if1'(false, n32, x42, nil) -> false
   filterlow'(n41, nil) -> false
   equal_bool(ge(n23, x31), true) -> true | filterlow'(n23, cons(x31, xs17)) -> true
   equal_bool(ge(n23, x31), true) -> false | filterlow'(n23, cons(x31, xs17)) -> filterlow'(n23, xs17)
   ge(x, 0) -> true
   ge(s(x9), s(y')) -> ge(x9, y')
   ge(0, s(x63)) -> false
   filterlow(n41, nil) -> nil
   equal_bool(ge(n23, x31), true) -> true | filterlow(n23, cons(x31, xs17)) -> filterlow(n23, xs17)
   equal_bool(ge(n23, x31), true) -> false | filterlow(n23, cons(x31, xs17)) -> cons(x31, filterlow(n23, xs17))
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x94))) -> s(half(x94))
   filterhigh(n109, nil) -> nil
   equal_bool(ge(x116, n91), true) -> true | filterhigh(n91, cons(x116, xs68)) -> filterhigh(n91, xs68)
   equal_bool(ge(x116, n91), true) -> false | filterhigh(n91, cons(x116, xs68)) -> cons(x116, filterhigh(n91, xs68))
   if1(true, n14, x20, cons(x31, xs17)) -> if1(ge(n14, x31), n14, x31, xs17)
   if1(true, n14, x20, nil) -> nil
   if1(false, n32, x42, cons(x31, xs17)) -> cons(x42, if1(ge(n32, x31), n32, x31, xs17))
   if1(false, n32, x42, nil) -> cons(x42, nil)
   if2(true, n82, x105, cons(x116, xs68)) -> if2(ge(x116, n82), n82, x116, xs68)
   if2(true, n82, x105, nil) -> nil
   if2(false, n100, x127, cons(x116, xs68)) -> cons(x127, if2(ge(x116, n100), n100, x116, xs68))
   if2(false, n100, x127, nil) -> cons(x127, nil)

The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 9a00b172b26c9abb2d4c4d5eaf341e919eb0fbf1 nowonder 20100222 unpublished dirty


Partial correctness of the following Program

   [x, x0, x1, x2, x3, n14, x20, xs10, n32, x42, x31, xs17, n41, n23, x9, y', x63, x94, n109, n91, x116, xs68, n82, x105, n100, x127]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(x0)) -> false
   equal_sort[a0](s(x0), 0) -> false
   equal_sort[a0](s(x0), s(x1)) -> equal_sort[a0](x0, x1)
   equal_sort[a5](cons(x0, x1), cons(x2, x3)) -> equal_sort[a5](x0, x2) and equal_sort[a5](x1, x3)
   equal_sort[a5](cons(x0, x1), nil) -> false
   equal_sort[a5](nil, cons(x0, x1)) -> false
   equal_sort[a5](nil, nil) -> true
   equal_sort[a64](witness_sort[a64], witness_sort[a64]) -> true
   if1'(true, n14, x20, xs10) -> true
   if1'(false, n32, x42, cons(x31, xs17)) -> if1'(ge(n32, x31), n32, x31, xs17)
   if1'(false, n32, x42, nil) -> false
   filterlow'(n41, nil) -> false
   equal_bool(ge(n23, x31), true) -> true | filterlow'(n23, cons(x31, xs17)) -> true
   equal_bool(ge(n23, x31), true) -> false | filterlow'(n23, cons(x31, xs17)) -> filterlow'(n23, xs17)
   ge(x, 0) -> true
   ge(s(x9), s(y')) -> ge(x9, y')
   ge(0, s(x63)) -> false
   filterlow(n41, nil) -> nil
   equal_bool(ge(n23, x31), true) -> true | filterlow(n23, cons(x31, xs17)) -> filterlow(n23, xs17)
   equal_bool(ge(n23, x31), true) -> false | filterlow(n23, cons(x31, xs17)) -> cons(x31, filterlow(n23, xs17))
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x94))) -> s(half(x94))
   filterhigh(n109, nil) -> nil
   equal_bool(ge(x116, n91), true) -> true | filterhigh(n91, cons(x116, xs68)) -> filterhigh(n91, xs68)
   equal_bool(ge(x116, n91), true) -> false | filterhigh(n91, cons(x116, xs68)) -> cons(x116, filterhigh(n91, xs68))
   if1(true, n14, x20, cons(x31, xs17)) -> if1(ge(n14, x31), n14, x31, xs17)
   if1(true, n14, x20, nil) -> nil
   if1(false, n32, x42, cons(x31, xs17)) -> cons(x42, if1(ge(n32, x31), n32, x31, xs17))
   if1(false, n32, x42, nil) -> cons(x42, nil)
   if2(true, n82, x105, cons(x116, xs68)) -> if2(ge(x116, n82), n82, x116, xs68)
   if2(true, n82, x105, nil) -> nil
   if2(false, n100, x127, cons(x116, xs68)) -> cons(x127, if2(ge(x116, n100), n100, x116, xs68))
   if2(false, n100, x127, nil) -> cons(x127, nil)

using the following formula:
x1':sort[a0].if1'(ge(x1', x1'), x1', x1', nil)=true

could be successfully shown:
(0) Formula
(1) Induction by data structure [EQUIVALENT]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT]
        (5) YES
    (6) Formula
        (7) Symbolic evaluation [EQUIVALENT]
        (8) Formula
        (9) Hypothesis Lifting [EQUIVALENT]
        (10) Formula
        (11) Inverse Substitution [SOUND]
        (12) Formula
        (13) Inverse Substitution [SOUND]
        (14) Formula
        (15) Induction by data structure [EQUIVALENT]
        (16) AND
            (17) Formula
                (18) Symbolic evaluation [EQUIVALENT]
                (19) YES
            (20) Formula
                (21) Symbolic evaluation [EQUIVALENT]
                (22) YES


----------------------------------------

(0)
Obligation:
Formula:
x1':sort[a0].if1'(ge(x1', x1'), x1', x1', nil)=true

There are no hypotheses.




----------------------------------------

(1) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
if1'(ge(0, 0), 0, 0, nil)=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0].if1'(ge(s(n), s(n)), s(n), s(n), nil)=true

Hypotheses:
n:sort[a0].if1'(ge(n, n), n, n, nil)=true






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
if1'(ge(0, 0), 0, 0, nil)=true

There are no hypotheses.




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(5)
YES

----------------------------------------

(6)
Obligation:
Formula:
n:sort[a0].if1'(ge(s(n), s(n)), s(n), s(n), nil)=true

Hypotheses:
n:sort[a0].if1'(ge(n, n), n, n, nil)=true




----------------------------------------

(7) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
n:sort[a0].if1'(ge(n, n), s(n), s(n), nil)=true
----------------------------------------

(8)
Obligation:
Formula:
n:sort[a0].if1'(ge(n, n), s(n), s(n), nil)=true

Hypotheses:
n:sort[a0].if1'(ge(n, n), n, n, nil)=true




----------------------------------------

(9) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(if1'(ge(n, n), n, n, nil)=true->if1'(ge(n, n), s(n), s(n), nil)=true)

There are no hypotheses.




----------------------------------------

(10)
Obligation:
Formula:
n:sort[a0].(if1'(ge(n, n), n, n, nil)=true->if1'(ge(n, n), s(n), s(n), nil)=true)

There are no hypotheses.




----------------------------------------

(11) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n':bool,n:sort[a0].(if1'(n', n, n, nil)=true->if1'(n', s(n), s(n), nil)=true)

Inverse substitution used:
[ge(n, n)/n']


----------------------------------------

(12)
Obligation:
Formula:
n':bool,n:sort[a0].(if1'(n', n, n, nil)=true->if1'(n', s(n), s(n), nil)=true)

There are no hypotheses.




----------------------------------------

(13) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n':bool,n:sort[a0],n'':sort[a0].(if1'(n', n, n, nil)=true->if1'(n', n'', n'', nil)=true)

Inverse substitution used:
[s(n)/n'']


----------------------------------------

(14)
Obligation:
Formula:
n':bool,n:sort[a0],n'':sort[a0].(if1'(n', n, n, nil)=true->if1'(n', n'', n'', nil)=true)

There are no hypotheses.




----------------------------------------

(15) Induction by data structure (EQUIVALENT)
Induction by data structure bool generates the following cases:



1. Base Case:
Formula:
n:sort[a0],n'':sort[a0].(if1'(true, n, n, nil)=true->if1'(true, n'', n'', nil)=true)

There are no hypotheses.





1. Base Case:
Formula:
n:sort[a0],n'':sort[a0].(if1'(false, n, n, nil)=true->if1'(false, n'', n'', nil)=true)

There are no hypotheses.






----------------------------------------

(16)
Complex Obligation (AND)

----------------------------------------

(17)
Obligation:
Formula:
n:sort[a0],n'':sort[a0].(if1'(true, n, n, nil)=true->if1'(true, n'', n'', nil)=true)

There are no hypotheses.




----------------------------------------

(18) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(19)
YES

----------------------------------------

(20)
Obligation:
Formula:
n:sort[a0],n'':sort[a0].(if1'(false, n, n, nil)=true->if1'(false, n'', n'', nil)=true)

There are no hypotheses.




----------------------------------------

(21) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(22)
YES

(115) Complex Obligation (AND)

(116) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))
QS(0, cons(x0, cons(x1, x2))) → QS(0, if1(ge(x0, x0), x0, x0, cons(x1, x2)))

The TRS R consists of the following rules:

ge(x, 0) → true
ge(s(x), s(y)) → ge(x, y)
if1(true, n, x, xs) → filterlow(n, xs)
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
ge(0, s(x)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if2(true, n, x, xs) → filterhigh(n, xs)
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))

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

(117) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
QS(0, cons(x0', cons(x1', x2))) → QS(0, if1(ge(x0', x0'), x0', x0', cons(x1', x2)))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(QS(x₁, x₂)) = x₂   
POL(cons(x₁, x₂)) = 1 + x₂   
POL(false) = 0   
POL(filterhigh(x₁, x₂)) = x₂   
POL(filterlow(x₁, x₂)) = x₂   
POL(ge(x₁, x₂)) = 0   
POL(half(x₁)) = x₁   
POL(if1(x₁, x₂, x₃, x₄)) = 1 + x₄   
POL(if2(x₁, x₂, x₃, x₄)) = 1 + x₄   
POL(nil) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

At least one of these decreasing rules is always used after the deleted DP:
if1(true, n14'', x20'', xs10'') → filterlow(n14'', xs10'')


The following formula is valid:
x0':sort[a0],x1':sort[a0],x2:sort[a0].if1'(ge(x0' , x0' ), x0' , x0' , cons(x1' , x2 ))=true


The transformed set:
if1'(true, n14, x20, xs10) → true
filterlow'(n23, cons(x31, xs17)) → if1'(ge(n23, x31), n23, x31, xs17)
if1'(false, n32, x42, xs24) → filterlow'(n32, xs24)
filterlow'(n41, nil) → false
ge(x, 0) → true
ge(s(x9), s(y')) → ge(x9, y')
if1(true, n14, x20, xs10) → filterlow(n14, xs10)
filterlow(n23, cons(x31, xs17)) → if1(ge(n23, x31), n23, x31, xs17)
if1(false, n32, x42, xs24) → cons(x42, filterlow(n32, xs24))
filterlow(n41, nil) → nil
ge(0, s(x63)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x94))) → s(half(x94))
if2(true, n82, x105, xs61) → filterhigh(n82, xs61)
filterhigh(n91, cons(x116, xs68)) → if2(ge(x116, n91), n91, x116, xs68)
if2(false, n100, x127, xs75) → cons(x127, filterhigh(n100, xs75))
filterhigh(n109, nil) → nil
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a5](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a5](x0, x2), equal_sort[a5](x1, x3))
equal_sort[a5](cons(x0, x1), nil) → false
equal_sort[a5](nil, cons(x0, x1)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true


The proof given by the theorem prover:
The following program was given to the internal theorem prover:

   [x, x0, x1, x2, x3, n14, x20, xs10, n32, x42, x31, xs17, n41, n23, x9, y', x63, x94, n109, n91, x116, xs68, n82, x105, n100, x127]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(x0)) -> false
   equal_sort[a0](s(x0), 0) -> false
   equal_sort[a0](s(x0), s(x1)) -> equal_sort[a0](x0, x1)
   equal_sort[a5](cons(x0, x1), cons(x2, x3)) -> equal_sort[a5](x0, x2) and equal_sort[a5](x1, x3)
   equal_sort[a5](cons(x0, x1), nil) -> false
   equal_sort[a5](nil, cons(x0, x1)) -> false
   equal_sort[a5](nil, nil) -> true
   equal_sort[a62](witness_sort[a62], witness_sort[a62]) -> true
   if1'(true, n14, x20, xs10) -> true
   if1'(false, n32, x42, cons(x31, xs17)) -> if1'(ge(n32, x31), n32, x31, xs17)
   if1'(false, n32, x42, nil) -> false
   filterlow'(n41, nil) -> false
   equal_bool(ge(n23, x31), true) -> true | filterlow'(n23, cons(x31, xs17)) -> true
   equal_bool(ge(n23, x31), true) -> false | filterlow'(n23, cons(x31, xs17)) -> filterlow'(n23, xs17)
   ge(x, 0) -> true
   ge(s(x9), s(y')) -> ge(x9, y')
   ge(0, s(x63)) -> false
   filterlow(n41, nil) -> nil
   equal_bool(ge(n23, x31), true) -> true | filterlow(n23, cons(x31, xs17)) -> filterlow(n23, xs17)
   equal_bool(ge(n23, x31), true) -> false | filterlow(n23, cons(x31, xs17)) -> cons(x31, filterlow(n23, xs17))
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x94))) -> s(half(x94))
   filterhigh(n109, nil) -> nil
   equal_bool(ge(x116, n91), true) -> true | filterhigh(n91, cons(x116, xs68)) -> filterhigh(n91, xs68)
   equal_bool(ge(x116, n91), true) -> false | filterhigh(n91, cons(x116, xs68)) -> cons(x116, filterhigh(n91, xs68))
   if1(true, n14, x20, cons(x31, xs17)) -> if1(ge(n14, x31), n14, x31, xs17)
   if1(true, n14, x20, nil) -> nil
   if1(false, n32, x42, cons(x31, xs17)) -> cons(x42, if1(ge(n32, x31), n32, x31, xs17))
   if1(false, n32, x42, nil) -> cons(x42, nil)
   if2(true, n82, x105, cons(x116, xs68)) -> if2(ge(x116, n82), n82, x116, xs68)
   if2(true, n82, x105, nil) -> nil
   if2(false, n100, x127, cons(x116, xs68)) -> cons(x127, if2(ge(x116, n100), n100, x116, xs68))
   if2(false, n100, x127, nil) -> cons(x127, nil)

The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 9a00b172b26c9abb2d4c4d5eaf341e919eb0fbf1 nowonder 20100222 unpublished dirty


Partial correctness of the following Program

   [x, x0, x1, x2, x3, n14, x20, xs10, n32, x42, x31, xs17, n41, n23, x9, y', x63, x94, n109, n91, x116, xs68, n82, x105, n100, x127]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a0](0, 0) -> true
   equal_sort[a0](0, s(x0)) -> false
   equal_sort[a0](s(x0), 0) -> false
   equal_sort[a0](s(x0), s(x1)) -> equal_sort[a0](x0, x1)
   equal_sort[a5](cons(x0, x1), cons(x2, x3)) -> equal_sort[a5](x0, x2) and equal_sort[a5](x1, x3)
   equal_sort[a5](cons(x0, x1), nil) -> false
   equal_sort[a5](nil, cons(x0, x1)) -> false
   equal_sort[a5](nil, nil) -> true
   equal_sort[a62](witness_sort[a62], witness_sort[a62]) -> true
   if1'(true, n14, x20, xs10) -> true
   if1'(false, n32, x42, cons(x31, xs17)) -> if1'(ge(n32, x31), n32, x31, xs17)
   if1'(false, n32, x42, nil) -> false
   filterlow'(n41, nil) -> false
   equal_bool(ge(n23, x31), true) -> true | filterlow'(n23, cons(x31, xs17)) -> true
   equal_bool(ge(n23, x31), true) -> false | filterlow'(n23, cons(x31, xs17)) -> filterlow'(n23, xs17)
   ge(x, 0) -> true
   ge(s(x9), s(y')) -> ge(x9, y')
   ge(0, s(x63)) -> false
   filterlow(n41, nil) -> nil
   equal_bool(ge(n23, x31), true) -> true | filterlow(n23, cons(x31, xs17)) -> filterlow(n23, xs17)
   equal_bool(ge(n23, x31), true) -> false | filterlow(n23, cons(x31, xs17)) -> cons(x31, filterlow(n23, xs17))
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x94))) -> s(half(x94))
   filterhigh(n109, nil) -> nil
   equal_bool(ge(x116, n91), true) -> true | filterhigh(n91, cons(x116, xs68)) -> filterhigh(n91, xs68)
   equal_bool(ge(x116, n91), true) -> false | filterhigh(n91, cons(x116, xs68)) -> cons(x116, filterhigh(n91, xs68))
   if1(true, n14, x20, cons(x31, xs17)) -> if1(ge(n14, x31), n14, x31, xs17)
   if1(true, n14, x20, nil) -> nil
   if1(false, n32, x42, cons(x31, xs17)) -> cons(x42, if1(ge(n32, x31), n32, x31, xs17))
   if1(false, n32, x42, nil) -> cons(x42, nil)
   if2(true, n82, x105, cons(x116, xs68)) -> if2(ge(x116, n82), n82, x116, xs68)
   if2(true, n82, x105, nil) -> nil
   if2(false, n100, x127, cons(x116, xs68)) -> cons(x127, if2(ge(x116, n100), n100, x116, xs68))
   if2(false, n100, x127, nil) -> cons(x127, nil)

using the following formula:
x0':sort[a0],x1':sort[a0],x2:sort[a0].if1'(ge(x0', x0'), x0', x0', cons(x1', x2))=true

could be successfully shown:
(0) Formula
(1) Induction by data structure [EQUIVALENT]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT]
        (5) YES
    (6) Formula
        (7) Symbolic evaluation [EQUIVALENT]
        (8) Formula
        (9) Case Analysis [EQUIVALENT]
        (10) AND
            (11) Formula
                (12) Case Analysis [EQUIVALENT]
                (13) AND
                    (14) Formula
                        (15) Inverse Substitution [SOUND]
                        (16) Formula
                        (17) Induction by data structure [SOUND]
                        (18) AND
                            (19) Formula
                                (20) Symbolic evaluation [EQUIVALENT]
                                (21) YES
                            (22) Formula
                                (23) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (24) YES
                    (25) Formula
                        (26) Inverse Substitution [SOUND]
                        (27) Formula
                        (28) Induction by data structure [SOUND]
                        (29) AND
                            (30) Formula
                                (31) Symbolic evaluation [EQUIVALENT]
                                (32) YES
                            (33) Formula
                                (34) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (35) YES
            (36) Formula
                (37) Case Analysis [EQUIVALENT]
                (38) AND
                    (39) Formula
                        (40) Inverse Substitution [SOUND]
                        (41) Formula
                        (42) Induction by data structure [SOUND]
                        (43) AND
                            (44) Formula
                                (45) Symbolic evaluation [EQUIVALENT]
                                (46) YES
                            (47) Formula
                                (48) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (49) YES
                    (50) Formula
                        (51) Inverse Substitution [SOUND]
                        (52) Formula
                        (53) Induction by data structure [SOUND]
                        (54) AND
                            (55) Formula
                                (56) Symbolic evaluation [EQUIVALENT]
                                (57) YES
                            (58) Formula
                                (59) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (60) YES


----------------------------------------

(0)
Obligation:
Formula:
x0':sort[a0],x1':sort[a0],x2:sort[a0].if1'(ge(x0', x0'), x0', x0', cons(x1', x2))=true

There are no hypotheses.




----------------------------------------

(1) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
x1':sort[a0],x2:sort[a0].if1'(ge(0, 0), 0, 0, cons(x1', x2))=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0],x1':sort[a0],x2:sort[a0].if1'(ge(s(n), s(n)), s(n), s(n), cons(x1', x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true






----------------------------------------

(2)
Complex Obligation (AND)

----------------------------------------

(3)
Obligation:
Formula:
x1':sort[a0],x2:sort[a0].if1'(ge(0, 0), 0, 0, cons(x1', x2))=true

There are no hypotheses.




----------------------------------------

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(5)
YES

----------------------------------------

(6)
Obligation:
Formula:
n:sort[a0],x1':sort[a0],x2:sort[a0].if1'(ge(s(n), s(n)), s(n), s(n), cons(x1', x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(7) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
n:sort[a0],x1':sort[a0],x2:sort[a0].if1'(ge(n, n), s(n), s(n), cons(x1', x2))=true
----------------------------------------

(8)
Obligation:
Formula:
n:sort[a0],x1':sort[a0],x2:sort[a0].if1'(ge(n, n), s(n), s(n), cons(x1', x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(9) Case Analysis (EQUIVALENT)
Case analysis leads to the following new obligations:

Formula:
n:sort[a0],x2:sort[a0].if1'(ge(n, n), s(n), s(n), cons(0, x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true





Formula:
n:sort[a0],x_1:sort[a0],x2:sort[a0].if1'(ge(n, n), s(n), s(n), cons(s(x_1), x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true






----------------------------------------

(10)
Complex Obligation (AND)

----------------------------------------

(11)
Obligation:
Formula:
n:sort[a0],x2:sort[a0].if1'(ge(n, n), s(n), s(n), cons(0, x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(12) Case Analysis (EQUIVALENT)
Case analysis leads to the following new obligations:

Formula:
n:sort[a0].if1'(ge(n, n), s(n), s(n), cons(0, 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true





Formula:
n:sort[a0],x_1:sort[a0].if1'(ge(n, n), s(n), s(n), cons(0, s(x_1)))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true






----------------------------------------

(13)
Complex Obligation (AND)

----------------------------------------

(14)
Obligation:
Formula:
n:sort[a0].if1'(ge(n, n), s(n), s(n), cons(0, 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(15) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0].if1'(ge(n, n), n', n', cons(0, 0))=true

Inverse substitution used:
[s(n)/n']


----------------------------------------

(16)
Obligation:
Formula:
n:sort[a0],n':sort[a0].if1'(ge(n, n), n', n', cons(0, 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(17) Induction by data structure (SOUND)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
n':sort[a0].if1'(ge(0, 0), n', n', cons(0, 0))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0].if1'(ge(s(n''), s(n'')), n', n', cons(0, 0))=true

Hypotheses:
n'':sort[a0],!n':sort[a0].if1'(ge(n'', n''), n', n', cons(0, 0))=true






----------------------------------------

(18)
Complex Obligation (AND)

----------------------------------------

(19)
Obligation:
Formula:
n':sort[a0].if1'(ge(0, 0), n', n', cons(0, 0))=true

There are no hypotheses.




----------------------------------------

(20) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(21)
YES

----------------------------------------

(22)
Obligation:
Formula:
n'':sort[a0],n':sort[a0].if1'(ge(s(n''), s(n'')), n', n', cons(0, 0))=true

Hypotheses:
n'':sort[a0],!n':sort[a0].if1'(ge(n'', n''), n', n', cons(0, 0))=true




----------------------------------------

(23) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n'':sort[a0],!n':sort[a0].if1'(ge(n'', n''), n', n', cons(0, 0))=true

----------------------------------------

(24)
YES

----------------------------------------

(25)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0].if1'(ge(n, n), s(n), s(n), cons(0, s(x_1)))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(26) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0],x_1:sort[a0].if1'(ge(n, n), n', n', cons(0, s(x_1)))=true

Inverse substitution used:
[s(n)/n']


----------------------------------------

(27)
Obligation:
Formula:
n:sort[a0],n':sort[a0],x_1:sort[a0].if1'(ge(n, n), n', n', cons(0, s(x_1)))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(28) Induction by data structure (SOUND)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
n':sort[a0],x_1:sort[a0].if1'(ge(0, 0), n', n', cons(0, s(x_1)))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0].if1'(ge(s(n''), s(n'')), n', n', cons(0, s(x_1)))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if1'(ge(n'', n''), n', n', cons(0, s(x_1)))=true






----------------------------------------

(29)
Complex Obligation (AND)

----------------------------------------

(30)
Obligation:
Formula:
n':sort[a0],x_1:sort[a0].if1'(ge(0, 0), n', n', cons(0, s(x_1)))=true

There are no hypotheses.




----------------------------------------

(31) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(32)
YES

----------------------------------------

(33)
Obligation:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0].if1'(ge(s(n''), s(n'')), n', n', cons(0, s(x_1)))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if1'(ge(n'', n''), n', n', cons(0, s(x_1)))=true




----------------------------------------

(34) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if1'(ge(n'', n''), n', n', cons(0, s(x_1)))=true

----------------------------------------

(35)
YES

----------------------------------------

(36)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0],x2:sort[a0].if1'(ge(n, n), s(n), s(n), cons(s(x_1), x2))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(37) Case Analysis (EQUIVALENT)
Case analysis leads to the following new obligations:

Formula:
n:sort[a0],x_1:sort[a0].if1'(ge(n, n), s(n), s(n), cons(s(x_1), 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true





Formula:
n:sort[a0],x_1:sort[a0],x_1':sort[a0].if1'(ge(n, n), s(n), s(n), cons(s(x_1), s(x_1')))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true






----------------------------------------

(38)
Complex Obligation (AND)

----------------------------------------

(39)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0].if1'(ge(n, n), s(n), s(n), cons(s(x_1), 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(40) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0],x_1:sort[a0].if1'(ge(n, n), n', n', cons(s(x_1), 0))=true

Inverse substitution used:
[s(n)/n']


----------------------------------------

(41)
Obligation:
Formula:
n:sort[a0],n':sort[a0],x_1:sort[a0].if1'(ge(n, n), n', n', cons(s(x_1), 0))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(42) Induction by data structure (SOUND)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
n':sort[a0],x_1:sort[a0].if1'(ge(0, 0), n', n', cons(s(x_1), 0))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0].if1'(ge(s(n''), s(n'')), n', n', cons(s(x_1), 0))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if1'(ge(n'', n''), n', n', cons(s(x_1), 0))=true






----------------------------------------

(43)
Complex Obligation (AND)

----------------------------------------

(44)
Obligation:
Formula:
n':sort[a0],x_1:sort[a0].if1'(ge(0, 0), n', n', cons(s(x_1), 0))=true

There are no hypotheses.




----------------------------------------

(45) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(46)
YES

----------------------------------------

(47)
Obligation:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0].if1'(ge(s(n''), s(n'')), n', n', cons(s(x_1), 0))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if1'(ge(n'', n''), n', n', cons(s(x_1), 0))=true




----------------------------------------

(48) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n'':sort[a0],!n':sort[a0],!x_1:sort[a0].if1'(ge(n'', n''), n', n', cons(s(x_1), 0))=true

----------------------------------------

(49)
YES

----------------------------------------

(50)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0],x_1':sort[a0].if1'(ge(n, n), s(n), s(n), cons(s(x_1), s(x_1')))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(51) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0],x_1:sort[a0],x_1':sort[a0].if1'(ge(n, n), n', n', cons(s(x_1), s(x_1')))=true

Inverse substitution used:
[s(n)/n']


----------------------------------------

(52)
Obligation:
Formula:
n:sort[a0],n':sort[a0],x_1:sort[a0],x_1':sort[a0].if1'(ge(n, n), n', n', cons(s(x_1), s(x_1')))=true

Hypotheses:
n:sort[a0],!x1':sort[a0],!x2:sort[a0].if1'(ge(n, n), n, n, cons(x1', x2))=true




----------------------------------------

(53) Induction by data structure (SOUND)
Induction by data structure sort[a0] generates the following cases:



1. Base Case:
Formula:
n':sort[a0],x_1:sort[a0],x_1':sort[a0].if1'(ge(0, 0), n', n', cons(s(x_1), s(x_1')))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0],x_1':sort[a0].if1'(ge(s(n''), s(n'')), n', n', cons(s(x_1), s(x_1')))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_1':sort[a0].if1'(ge(n'', n''), n', n', cons(s(x_1), s(x_1')))=true






----------------------------------------

(54)
Complex Obligation (AND)

----------------------------------------

(55)
Obligation:
Formula:
n':sort[a0],x_1:sort[a0],x_1':sort[a0].if1'(ge(0, 0), n', n', cons(s(x_1), s(x_1')))=true

There are no hypotheses.




----------------------------------------

(56) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(57)
YES

----------------------------------------

(58)
Obligation:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0],x_1':sort[a0].if1'(ge(s(n''), s(n'')), n', n', cons(s(x_1), s(x_1')))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_1':sort[a0].if1'(ge(n'', n''), n', n', cons(s(x_1), s(x_1')))=true




----------------------------------------

(59) Symbolic evaluation under hypothesis (EQUIVALENT)
Could be shown using symbolic evaluation under hypothesis, by using the following hypotheses:

n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_1':sort[a0].if1'(ge(n'', n''), n', n', cons(s(x_1), s(x_1')))=true

----------------------------------------

(60)
YES

(118) Complex Obligation (AND)

(119) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))

The TRS R consists of the following rules:

ge(x, 0) → true
ge(s(x), s(y)) → ge(x, y)
if1(true, n, x, xs) → filterlow(n, xs)
filterlow(n, cons(x, xs)) → if1(ge(n, x), n, x, xs)
if1(false, n, x, xs) → cons(x, filterlow(n, xs))
filterlow(n, nil) → nil
ge(0, s(x)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
if2(true, n, x, xs) → filterhigh(n, xs)
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))

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

(120) 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.

(121) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
ge(x, 0) → true
ge(s(x), s(y)) → ge(x, y)
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil
ge(0, s(x)) → false

The set Q consists of the following terms:

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)
filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))

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

(122) 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].

filterlow(x0, nil)
filterlow(x0, cons(x1, x2))
if1(true, x0, x1, x2)
if1(false, x0, x1, x2)

(123) Obligation:

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

QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
ge(x, 0) → true
ge(s(x), s(y)) → ge(x, y)
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil
ge(0, s(x)) → false

The set Q consists of the following terms:

filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))

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

(124) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
QS(x0, cons(x1, nil)) → QS(half(x0), if2(ge(x1, x1), x1, x1, nil))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(QS(x₁, x₂)) = x₁ + x₂   
POL(cons(x₁, x₂)) = 1 + x₂   
POL(false) = 0   
POL(filterhigh(x₁, x₂)) = x₂   
POL(ge(x₁, x₂)) = 0   
POL(half(x₁)) = x₁   
POL(if2(x₁, x₂, x₃, x₄)) = 1 + x₄   
POL(nil) = 1   
POL(s(x₁)) = 0   
POL(true) = 0   

At least one of these decreasing rules is always used after the deleted DP:
if2(true, n224, x434, xs164) → filterhigh(n224, xs164)


The following formula is valid:
x1:sort[a19].(false=true∨if2'(ge(x1 , x1 ), x1 , x1 , nil)=true)


The transformed set:
filterhigh'(n17, cons(x35, xs12)) → if2'(ge(x35, n17), n17, x35, xs12)
if2'(true, n22, x43, xs16) → true
if2'(false, n27, x51, xs20) → filterhigh'(n27, xs20)
filterhigh'(n32, nil) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x11))) → s(half(x11))
ge(x19, 0) → true
ge(s(x27), s(y2)) → ge(x27, y2)
filterhigh(n17, cons(x35, xs12)) → if2(ge(x35, n17), n17, x35, xs12)
if2(true, n22, x43, xs16) → filterhigh(n22, xs16)
if2(false, n27, x51, xs20) → cons(x51, filterhigh(n27, xs20))
filterhigh(n32, nil) → nil
ge(0, s(x66)) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a6](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a6](x0, x2), equal_sort[a6](x1, x3))
equal_sort[a6](cons(x0, x1), nil) → false
equal_sort[a6](nil, cons(x0, x1)) → false
equal_sort[a6](nil, nil) → true
equal_sort[a41](witness_sort[a41], witness_sort[a41]) → true


The proof given by the theorem prover:
The following program was given to the internal theorem prover:

   [x, x0, x1, x2, x3, n22, x43, xs16, n27, x51, x35, xs12, n32, n17, x11, x19, x27, y2, x66]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a19](0, 0) -> true
   equal_sort[a19](0, s(x0)) -> false
   equal_sort[a19](s(x0), 0) -> false
   equal_sort[a19](s(x0), s(x1)) -> equal_sort[a19](x0, x1)
   equal_sort[a6](cons(x0, x1), cons(x2, x3)) -> equal_sort[a6](x0, x2) and equal_sort[a6](x1, x3)
   equal_sort[a6](cons(x0, x1), nil) -> false
   equal_sort[a6](nil, cons(x0, x1)) -> false
   equal_sort[a6](nil, nil) -> true
   equal_sort[a41](witness_sort[a41], witness_sort[a41]) -> true
   if2'(true, n22, x43, xs16) -> true
   if2'(false, n27, x51, cons(x35, xs12)) -> if2'(ge(x35, n27), n27, x35, xs12)
   if2'(false, n27, x51, nil) -> false
   filterhigh'(n32, nil) -> false
   equal_bool(ge(x35, n17), true) -> true | filterhigh'(n17, cons(x35, xs12)) -> true
   equal_bool(ge(x35, n17), true) -> false | filterhigh'(n17, cons(x35, xs12)) -> filterhigh'(n17, xs12)
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x11))) -> s(half(x11))
   ge(x19, 0) -> true
   ge(s(x27), s(y2)) -> ge(x27, y2)
   ge(0, s(x66)) -> false
   filterhigh(n32, nil) -> nil
   equal_bool(ge(x35, n17), true) -> true | filterhigh(n17, cons(x35, xs12)) -> filterhigh(n17, xs12)
   equal_bool(ge(x35, n17), true) -> false | filterhigh(n17, cons(x35, xs12)) -> cons(x35, filterhigh(n17, xs12))
   if2(true, n22, x43, cons(x35, xs12)) -> if2(ge(x35, n22), n22, x35, xs12)
   if2(true, n22, x43, nil) -> nil
   if2(false, n27, x51, cons(x35, xs12)) -> cons(x51, if2(ge(x35, n27), n27, x35, xs12))
   if2(false, n27, x51, nil) -> cons(x51, nil)

The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 9a00b172b26c9abb2d4c4d5eaf341e919eb0fbf1 nowonder 20100222 unpublished dirty


Partial correctness of the following Program

   [x, x0, x1, x2, x3, n22, x43, xs16, n27, x51, x35, xs12, n32, n17, x11, x19, x27, y2, x66]
   equal_bool(true, false) -> false
   equal_bool(false, true) -> false
   equal_bool(true, true) -> true
   equal_bool(false, false) -> true
   true and x -> x
   false and x -> false
   true or x -> true
   false or x -> x
   not(false) -> true
   not(true) -> false
   isa_true(true) -> true
   isa_true(false) -> false
   isa_false(true) -> false
   isa_false(false) -> true
   equal_sort[a19](0, 0) -> true
   equal_sort[a19](0, s(x0)) -> false
   equal_sort[a19](s(x0), 0) -> false
   equal_sort[a19](s(x0), s(x1)) -> equal_sort[a19](x0, x1)
   equal_sort[a6](cons(x0, x1), cons(x2, x3)) -> equal_sort[a6](x0, x2) and equal_sort[a6](x1, x3)
   equal_sort[a6](cons(x0, x1), nil) -> false
   equal_sort[a6](nil, cons(x0, x1)) -> false
   equal_sort[a6](nil, nil) -> true
   equal_sort[a41](witness_sort[a41], witness_sort[a41]) -> true
   if2'(true, n22, x43, xs16) -> true
   if2'(false, n27, x51, cons(x35, xs12)) -> if2'(ge(x35, n27), n27, x35, xs12)
   if2'(false, n27, x51, nil) -> false
   filterhigh'(n32, nil) -> false
   equal_bool(ge(x35, n17), true) -> true | filterhigh'(n17, cons(x35, xs12)) -> true
   equal_bool(ge(x35, n17), true) -> false | filterhigh'(n17, cons(x35, xs12)) -> filterhigh'(n17, xs12)
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(x11))) -> s(half(x11))
   ge(x19, 0) -> true
   ge(s(x27), s(y2)) -> ge(x27, y2)
   ge(0, s(x66)) -> false
   filterhigh(n32, nil) -> nil
   equal_bool(ge(x35, n17), true) -> true | filterhigh(n17, cons(x35, xs12)) -> filterhigh(n17, xs12)
   equal_bool(ge(x35, n17), true) -> false | filterhigh(n17, cons(x35, xs12)) -> cons(x35, filterhigh(n17, xs12))
   if2(true, n22, x43, cons(x35, xs12)) -> if2(ge(x35, n22), n22, x35, xs12)
   if2(true, n22, x43, nil) -> nil
   if2(false, n27, x51, cons(x35, xs12)) -> cons(x51, if2(ge(x35, n27), n27, x35, xs12))
   if2(false, n27, x51, nil) -> cons(x51, nil)

using the following formula:
x1:sort[a19].(false=true\/if2'(ge(x1, x1), x1, x1, nil)=true)

could be successfully shown:
(0) Formula
(1) Symbolic evaluation [EQUIVALENT]
(2) Formula
(3) Induction by data structure [EQUIVALENT]
(4) AND
    (5) Formula
        (6) Symbolic evaluation [EQUIVALENT]
        (7) YES
    (8) Formula
        (9) Symbolic evaluation [EQUIVALENT]
        (10) Formula
        (11) Hypothesis Lifting [EQUIVALENT]
        (12) Formula
        (13) Inverse Substitution [SOUND]
        (14) Formula
        (15) Inverse Substitution [SOUND]
        (16) Formula
        (17) Induction by data structure [EQUIVALENT]
        (18) AND
            (19) Formula
                (20) Symbolic evaluation [EQUIVALENT]
                (21) YES
            (22) Formula
                (23) Symbolic evaluation [EQUIVALENT]
                (24) YES


----------------------------------------

(0)
Obligation:
Formula:
x1:sort[a19].(false=true\/if2'(ge(x1, x1), x1, x1, nil)=true)

There are no hypotheses.




----------------------------------------

(1) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x1:sort[a19].if2'(ge(x1, x1), x1, x1, nil)=true
----------------------------------------

(2)
Obligation:
Formula:
x1:sort[a19].if2'(ge(x1, x1), x1, x1, nil)=true

There are no hypotheses.




----------------------------------------

(3) Induction by data structure (EQUIVALENT)
Induction by data structure sort[a19] generates the following cases:



1. Base Case:
Formula:
if2'(ge(0, 0), 0, 0, nil)=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a19].if2'(ge(s(n), s(n)), s(n), s(n), nil)=true

Hypotheses:
n:sort[a19].if2'(ge(n, n), n, n, nil)=true






----------------------------------------

(4)
Complex Obligation (AND)

----------------------------------------

(5)
Obligation:
Formula:
if2'(ge(0, 0), 0, 0, nil)=true

There are no hypotheses.




----------------------------------------

(6) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(7)
YES

----------------------------------------

(8)
Obligation:
Formula:
n:sort[a19].if2'(ge(s(n), s(n)), s(n), s(n), nil)=true

Hypotheses:
n:sort[a19].if2'(ge(n, n), n, n, nil)=true




----------------------------------------

(9) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
n:sort[a19].if2'(ge(n, n), s(n), s(n), nil)=true
----------------------------------------

(10)
Obligation:
Formula:
n:sort[a19].if2'(ge(n, n), s(n), s(n), nil)=true

Hypotheses:
n:sort[a19].if2'(ge(n, n), n, n, nil)=true




----------------------------------------

(11) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a19].(if2'(ge(n, n), n, n, nil)=true->if2'(ge(n, n), s(n), s(n), nil)=true)

There are no hypotheses.




----------------------------------------

(12)
Obligation:
Formula:
n:sort[a19].(if2'(ge(n, n), n, n, nil)=true->if2'(ge(n, n), s(n), s(n), nil)=true)

There are no hypotheses.




----------------------------------------

(13) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n':bool,n:sort[a19].(if2'(n', n, n, nil)=true->if2'(n', s(n), s(n), nil)=true)

Inverse substitution used:
[ge(n, n)/n']


----------------------------------------

(14)
Obligation:
Formula:
n':bool,n:sort[a19].(if2'(n', n, n, nil)=true->if2'(n', s(n), s(n), nil)=true)

There are no hypotheses.




----------------------------------------

(15) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n':bool,n:sort[a19],n'':sort[a19].(if2'(n', n, n, nil)=true->if2'(n', n'', n'', nil)=true)

Inverse substitution used:
[s(n)/n'']


----------------------------------------

(16)
Obligation:
Formula:
n':bool,n:sort[a19],n'':sort[a19].(if2'(n', n, n, nil)=true->if2'(n', n'', n'', nil)=true)

There are no hypotheses.




----------------------------------------

(17) Induction by data structure (EQUIVALENT)
Induction by data structure bool generates the following cases:



1. Base Case:
Formula:
n:sort[a19],n'':sort[a19].(if2'(true, n, n, nil)=true->if2'(true, n'', n'', nil)=true)

There are no hypotheses.





1. Base Case:
Formula:
n:sort[a19],n'':sort[a19].(if2'(false, n, n, nil)=true->if2'(false, n'', n'', nil)=true)

There are no hypotheses.






----------------------------------------

(18)
Complex Obligation (AND)

----------------------------------------

(19)
Obligation:
Formula:
n:sort[a19],n'':sort[a19].(if2'(true, n, n, nil)=true->if2'(true, n'', n'', nil)=true)

There are no hypotheses.




----------------------------------------

(20) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(21)
YES

----------------------------------------

(22)
Obligation:
Formula:
n:sort[a19],n'':sort[a19].(if2'(false, n, n, nil)=true->if2'(false, n'', n'', nil)=true)

There are no hypotheses.




----------------------------------------

(23) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
True
----------------------------------------

(24)
YES

(125) Complex Obligation (AND)

(126) Obligation:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
ge(x, 0) → true
ge(s(x), s(y)) → ge(x, y)
filterhigh(n, cons(x, xs)) → if2(ge(x, n), n, x, xs)
if2(true, n, x, xs) → filterhigh(n, xs)
if2(false, n, x, xs) → cons(x, filterhigh(n, xs))
filterhigh(n, nil) → nil
ge(0, s(x)) → false

The set Q consists of the following terms:

filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
half(0)
half(s(0))
half(s(s(x0)))

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

(127) PisEmptyProof (EQUIVALENT transformation)

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

(128) YES

(129) Obligation:

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

filterhigh'(n17, cons(x35, xs12)) → if2'(ge(x35, n17), n17, x35, xs12)
if2'(true, n22, x43, xs16) → true
if2'(false, n27, x51, xs20) → filterhigh'(n27, xs20)
filterhigh'(n32, nil) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x11))) → s(half(x11))
ge(x19, 0) → true
ge(s(x27), s(y2)) → ge(x27, y2)
filterhigh(n17, cons(x35, xs12)) → if2(ge(x35, n17), n17, x35, xs12)
if2(true, n22, x43, xs16) → filterhigh(n22, xs16)
if2(false, n27, x51, xs20) → cons(x51, filterhigh(n27, xs20))
filterhigh(n32, nil) → nil
ge(0, s(x66)) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a6](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a6](x0, x2), equal_sort[a6](x1, x3))
equal_sort[a6](cons(x0, x1), nil) → false
equal_sort[a6](nil, cons(x0, x1)) → false
equal_sort[a6](nil, nil) → true
equal_sort[a41](witness_sort[a41], witness_sort[a41]) → true

Q is empty.

(130) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:

half₁ > [false, nil, 0] > [filterhigh'₂, if2'₄, ge₂, true, filterhigh₂, if2₄] > cons₂
half₁ > s₁ > equal_sort[a19]₂
not₁ > [false, nil, 0] > [filterhigh'₂, if2'₄, ge₂, true, filterhigh₂, if2₄] > cons₂
isa_true1 > [false, nil, 0] > [filterhigh'₂, if2'₄, ge₂, true, filterhigh₂, if2₄] > cons₂
isa_false1 > [false, nil, 0] > [filterhigh'₂, if2'₄, ge₂, true, filterhigh₂, if2₄] > cons₂
equal_sort[a6]₂ > [false, nil, 0] > [filterhigh'₂, if2'₄, ge₂, true, filterhigh₂, if2₄] > cons₂
equal_sort[a6]₂ > and₂
equal_sort[a41]₂ > [filterhigh'₂, if2'₄, ge₂, true, filterhigh₂, if2₄] > cons₂

Status:

filterhigh'₂: [2,1]
cons₂: [1,2]
if2'₄: [4,2,1,3]
ge₂: [1,2]
true: multiset
false: multiset
nil: multiset
half₁: multiset
0: multiset
s₁: multiset
filterhigh₂: [2,1]
if2₄: [4,2,1,3]
equal_bool2: [2,1]
and₂: multiset
or₂: [2,1]
not₁: multiset
isa_true1: multiset
isa_false1: multiset
equal_sort[a19]₂: [1,2]
equal_sort[a6]₂: [2,1]
equal_sort[a41]₂: [2,1]
witness_sort[a41]: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

filterhigh'(n17, cons(x35, xs12)) → if2'(ge(x35, n17), n17, x35, xs12)
if2'(true, n22, x43, xs16) → true
if2'(false, n27, x51, xs20) → filterhigh'(n27, xs20)
filterhigh'(n32, nil) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x11))) → s(half(x11))
ge(x19, 0) → true
ge(s(x27), s(y2)) → ge(x27, y2)
filterhigh(n17, cons(x35, xs12)) → if2(ge(x35, n17), n17, x35, xs12)
if2(true, n22, x43, xs16) → filterhigh(n22, xs16)
if2(false, n27, x51, xs20) → cons(x51, filterhigh(n27, xs20))
filterhigh(n32, nil) → nil
ge(0, s(x66)) → false
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a19](0, 0) → true
equal_sort[a19](0, s(x0)) → false
equal_sort[a19](s(x0), 0) → false
equal_sort[a19](s(x0), s(x1)) → equal_sort[a19](x0, x1)
equal_sort[a6](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a6](x0, x2), equal_sort[a6](x1, x3))
equal_sort[a6](cons(x0, x1), nil) → false
equal_sort[a6](nil, cons(x0, x1)) → false
equal_sort[a6](nil, nil) → true
equal_sort[a41](witness_sort[a41], witness_sort[a41]) → true

(131) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(132) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(133) YES

(134) Obligation:

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

if1'(true, n14, x20, xs10) → true
filterlow'(n23, cons(x31, xs17)) → if1'(ge(n23, x31), n23, x31, xs17)
if1'(false, n32, x42, xs24) → filterlow'(n32, xs24)
filterlow'(n41, nil) → false
ge(x, 0) → true
ge(s(x9), s(y')) → ge(x9, y')
if1(true, n14, x20, xs10) → filterlow(n14, xs10)
filterlow(n23, cons(x31, xs17)) → if1(ge(n23, x31), n23, x31, xs17)
if1(false, n32, x42, xs24) → cons(x42, filterlow(n32, xs24))
filterlow(n41, nil) → nil
ge(0, s(x63)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x94))) → s(half(x94))
if2(true, n82, x105, xs61) → filterhigh(n82, xs61)
filterhigh(n91, cons(x116, xs68)) → if2(ge(x116, n91), n91, x116, xs68)
if2(false, n100, x127, xs75) → cons(x127, filterhigh(n100, xs75))
filterhigh(n109, nil) → nil
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a5](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a5](x0, x2), equal_sort[a5](x1, x3))
equal_sort[a5](cons(x0, x1), nil) → false
equal_sort[a5](nil, cons(x0, x1)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

Q is empty.

(135) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
if1'(x1, x2, x3, x4)  =  if1'(x1, x2, x3, x4)
true  =  true
filterlow'(x1, x2)  =  filterlow'(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
ge(x1, x2)  =  ge(x1, x2)
false  =  false
nil  =  nil
0  =  0
s(x1)  =  x1
if1(x1, x2, x3, x4)  =  if1(x1, x2, x3, x4)
filterlow(x1, x2)  =  filterlow(x1, x2)
half(x1)  =  half(x1)
if2(x1, x2, x3, x4)  =  if2(x1, x2, x3, x4)
filterhigh(x1, x2)  =  filterhigh(x1, x2)
equal_bool(x1, x2)  =  equal_bool(x1, x2)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
not(x1)  =  not(x1)
isa_true(x1)  =  isa_true(x1)
isa_false(x1)  =  isa_false(x1)
equal_sort[a0](x1, x2)  =  equal_sort[a0](x1, x2)
equal_sort[a5](x1, x2)  =  equal_sort[a5](x1, x2)
equal_sort[a62](x1, x2)  =  equal_sort[a62](x1, x2)
witness_sort[a62]  =  witness_sort[a62]

Recursive path order with status [RPO].
Quasi-Precedence:

[if1'₄, filterlow'₂] > [ge₂, false, isa_true1] > [true, 0, witness_sort[a62]]
[if1'₄, filterlow'₂] > [ge₂, false, isa_true1] > cons₂
[if1₄, filterlow₂] > [ge₂, false, isa_true1] > [true, 0, witness_sort[a62]]
[if1₄, filterlow₂] > [ge₂, false, isa_true1] > cons₂
half₁ > [true, 0, witness_sort[a62]]
[if2₄, filterhigh₂] > nil > [ge₂, false, isa_true1] > [true, 0, witness_sort[a62]]
[if2₄, filterhigh₂] > nil > [ge₂, false, isa_true1] > cons₂
or₂ > [true, 0, witness_sort[a62]]
not₁ > [ge₂, false, isa_true1] > [true, 0, witness_sort[a62]]
not₁ > [ge₂, false, isa_true1] > cons₂
isa_false1 > [ge₂, false, isa_true1] > [true, 0, witness_sort[a62]]
isa_false1 > [ge₂, false, isa_true1] > cons₂
equal_sort[a0]₂ > [ge₂, false, isa_true1] > [true, 0, witness_sort[a62]]
equal_sort[a0]₂ > [ge₂, false, isa_true1] > cons₂
equal_sort[a5]₂ > and₂ > [ge₂, false, isa_true1] > [true, 0, witness_sort[a62]]
equal_sort[a5]₂ > and₂ > [ge₂, false, isa_true1] > cons₂
equal_sort[a62]₂ > [true, 0, witness_sort[a62]]

Status:

if1'₄: [2,4,3,1]
true: multiset
filterlow'₂: [1,2]
cons₂: [1,2]
ge₂: multiset
false: multiset
nil: multiset
0: multiset
if1₄: [2,4,3,1]
filterlow₂: [1,2]
half₁: [1]
if2₄: [4,2,3,1]
filterhigh₂: [2,1]
equal_bool2: multiset
and₂: [2,1]
or₂: [1,2]
not₁: multiset
isa_true1: multiset
isa_false1: multiset
equal_sort[a0]₂: [1,2]
equal_sort[a5]₂: [2,1]
equal_sort[a62]₂: multiset
witness_sort[a62]: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

if1'(true, n14, x20, xs10) → true
filterlow'(n23, cons(x31, xs17)) → if1'(ge(n23, x31), n23, x31, xs17)
if1'(false, n32, x42, xs24) → filterlow'(n32, xs24)
filterlow'(n41, nil) → false
ge(x, 0) → true
if1(true, n14, x20, xs10) → filterlow(n14, xs10)
filterlow(n23, cons(x31, xs17)) → if1(ge(n23, x31), n23, x31, xs17)
if1(false, n32, x42, xs24) → cons(x42, filterlow(n32, xs24))
filterlow(n41, nil) → nil
ge(0, s(x63)) → false
half(0) → 0
half(s(0)) → 0
if2(true, n82, x105, xs61) → filterhigh(n82, xs61)
filterhigh(n91, cons(x116, xs68)) → if2(ge(x116, n91), n91, x116, xs68)
if2(false, n100, x127, xs75) → cons(x127, filterhigh(n100, xs75))
filterhigh(n109, nil) → nil
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a5](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a5](x0, x2), equal_sort[a5](x1, x3))
equal_sort[a5](cons(x0, x1), nil) → false
equal_sort[a5](nil, cons(x0, x1)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

(136) Obligation:

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

ge(s(x9), s(y')) → ge(x9, y')
half(s(s(x94))) → s(half(x94))
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

Q is empty.

(137) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(equal_sort[a0](x₁, x₂)) = x₁ + x₂   
POL(ge(x₁, x₂)) = 2·x₁ + x₂   
POL(half(x₁)) = 2 + 2·x₁   
POL(s(x₁)) = 1 + 2·x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

ge(s(x9), s(y')) → ge(x9, y')
half(s(s(x94))) → s(half(x94))
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

(138) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(139) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(140) YES

(141) Obligation:

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

if1'(true, n14, x20, xs10) → true
filterlow'(n23, cons(x31, xs17)) → if1'(ge(n23, x31), n23, x31, xs17)
if1'(false, n32, x42, xs24) → filterlow'(n32, xs24)
filterlow'(n41, nil) → false
ge(x, 0) → true
ge(s(x9), s(y')) → ge(x9, y')
if1(true, n14, x20, xs10) → filterlow(n14, xs10)
filterlow(n23, cons(x31, xs17)) → if1(ge(n23, x31), n23, x31, xs17)
if1(false, n32, x42, xs24) → cons(x42, filterlow(n32, xs24))
filterlow(n41, nil) → nil
ge(0, s(x63)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x94))) → s(half(x94))
if2(true, n82, x105, xs61) → filterhigh(n82, xs61)
filterhigh(n91, cons(x116, xs68)) → if2(ge(x116, n91), n91, x116, xs68)
if2(false, n100, x127, xs75) → cons(x127, filterhigh(n100, xs75))
filterhigh(n109, nil) → nil
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a5](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a5](x0, x2), equal_sort[a5](x1, x3))
equal_sort[a5](cons(x0, x1), nil) → false
equal_sort[a5](nil, cons(x0, x1)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → true

Q is empty.

(142) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
if1'(x1, x2, x3, x4)  =  if1'(x1, x2, x3, x4)
true  =  true
filterlow'(x1, x2)  =  filterlow'(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
ge(x1, x2)  =  ge(x1, x2)
false  =  false
nil  =  nil
0  =  0
s(x1)  =  x1
if1(x1, x2, x3, x4)  =  if1(x1, x2, x3, x4)
filterlow(x1, x2)  =  filterlow(x1, x2)
half(x1)  =  half(x1)
if2(x1, x2, x3, x4)  =  if2(x1, x2, x3, x4)
filterhigh(x1, x2)  =  filterhigh(x1, x2)
equal_bool(x1, x2)  =  equal_bool(x1, x2)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
not(x1)  =  not(x1)
isa_true(x1)  =  isa_true(x1)
isa_false(x1)  =  isa_false(x1)
equal_sort[a0](x1, x2)  =  equal_sort[a0](x1, x2)
equal_sort[a5](x1, x2)  =  equal_sort[a5](x1, x2)
equal_sort[a64](x1, x2)  =  equal_sort[a64](x1, x2)
witness_sort[a64]  =  witness_sort[a64]

Recursive path order with status [RPO].
Quasi-Precedence:

[if1'₄, filterlow'₂] > [ge₂, false, isa_true1] > [true, 0, witness_sort[a64]]
[if1'₄, filterlow'₂] > [ge₂, false, isa_true1] > cons₂
[if1₄, filterlow₂] > [ge₂, false, isa_true1] > [true, 0, witness_sort[a64]]
[if1₄, filterlow₂] > [ge₂, false, isa_true1] > cons₂
half₁ > [true, 0, witness_sort[a64]]
[if2₄, filterhigh₂] > nil > [ge₂, false, isa_true1] > [true, 0, witness_sort[a64]]
[if2₄, filterhigh₂] > nil > [ge₂, false, isa_true1] > cons₂
or₂ > [true, 0, witness_sort[a64]]
not₁ > [ge₂, false, isa_true1] > [true, 0, witness_sort[a64]]
not₁ > [ge₂, false, isa_true1] > cons₂
isa_false1 > [ge₂, false, isa_true1] > [true, 0, witness_sort[a64]]
isa_false1 > [ge₂, false, isa_true1] > cons₂
equal_sort[a0]₂ > [ge₂, false, isa_true1] > [true, 0, witness_sort[a64]]
equal_sort[a0]₂ > [ge₂, false, isa_true1] > cons₂
equal_sort[a5]₂ > and₂ > [ge₂, false, isa_true1] > [true, 0, witness_sort[a64]]
equal_sort[a5]₂ > and₂ > [ge₂, false, isa_true1] > cons₂
equal_sort[a64]₂ > [true, 0, witness_sort[a64]]

Status:

if1'₄: [2,4,3,1]
true: multiset
filterlow'₂: [1,2]
cons₂: [1,2]
ge₂: multiset
false: multiset
nil: multiset
0: multiset
if1₄: [2,4,3,1]
filterlow₂: [1,2]
half₁: [1]
if2₄: [4,2,3,1]
filterhigh₂: [2,1]
equal_bool2: multiset
and₂: [2,1]
or₂: [1,2]
not₁: multiset
isa_true1: multiset
isa_false1: multiset
equal_sort[a0]₂: [1,2]
equal_sort[a5]₂: [2,1]
equal_sort[a64]₂: multiset
witness_sort[a64]: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

if1'(true, n14, x20, xs10) → true
filterlow'(n23, cons(x31, xs17)) → if1'(ge(n23, x31), n23, x31, xs17)
if1'(false, n32, x42, xs24) → filterlow'(n32, xs24)
filterlow'(n41, nil) → false
ge(x, 0) → true
if1(true, n14, x20, xs10) → filterlow(n14, xs10)
filterlow(n23, cons(x31, xs17)) → if1(ge(n23, x31), n23, x31, xs17)
if1(false, n32, x42, xs24) → cons(x42, filterlow(n32, xs24))
filterlow(n41, nil) → nil
ge(0, s(x63)) → false
half(0) → 0
half(s(0)) → 0
if2(true, n82, x105, xs61) → filterhigh(n82, xs61)
filterhigh(n91, cons(x116, xs68)) → if2(ge(x116, n91), n91, x116, xs68)
if2(false, n100, x127, xs75) → cons(x127, filterhigh(n100, xs75))
filterhigh(n109, nil) → nil
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a5](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a5](x0, x2), equal_sort[a5](x1, x3))
equal_sort[a5](cons(x0, x1), nil) → false
equal_sort[a5](nil, cons(x0, x1)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a64](witness_sort[a64], witness_sort[a64]) → true

(143) Obligation:

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

ge(s(x9), s(y')) → ge(x9, y')
half(s(s(x94))) → s(half(x94))
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

Q is empty.

(144) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(equal_sort[a0](x₁, x₂)) = x₁ + x₂   
POL(ge(x₁, x₂)) = 2·x₁ + x₂   
POL(half(x₁)) = 2 + 2·x₁   
POL(s(x₁)) = 1 + 2·x₁   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

ge(s(x9), s(y')) → ge(x9, y')
half(s(s(x94))) → s(half(x94))
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

(145) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(146) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(147) YES

(148) Obligation:

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

if2'(true, n82, x105, xs61) → true
filterhigh'(n91, cons(x116, xs68)) → if2'(ge(x116, n91), n91, x116, xs68)
if2'(false, n100, x127, xs75) → filterhigh'(n100, xs75)
filterhigh'(n109, nil) → false
ge(x, 0) → true
ge(s(x9), s(y')) → ge(x9, y')
if1(true, n14, x20, xs10) → filterlow(n14, xs10)
filterlow(n23, cons(x31, xs17)) → if1(ge(n23, x31), n23, x31, xs17)
if1(false, n32, x42, xs24) → cons(x42, filterlow(n32, xs24))
filterlow(n41, nil) → nil
ge(0, s(x63)) → false
half(0) → 0
half(s(0)) → 0
half(s(s(x94))) → s(half(x94))
if2(true, n82, x105, xs61) → filterhigh(n82, xs61)
filterhigh(n91, cons(x116, xs68)) → if2(ge(x116, n91), n91, x116, xs68)
if2(false, n100, x127, xs75) → cons(x127, filterhigh(n100, xs75))
filterhigh(n109, nil) → nil
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a5](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a5](x0, x2), equal_sort[a5](x1, x3))
equal_sort[a5](cons(x0, x1), nil) → false
equal_sort[a5](nil, cons(x0, x1)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a67](witness_sort[a67], witness_sort[a67]) → true

Q is empty.

(149) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
if2'(x1, x2, x3, x4)  =  if2'(x1, x2, x3, x4)
true  =  true
filterhigh'(x1, x2)  =  filterhigh'(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
ge(x1, x2)  =  ge(x1, x2)
false  =  false
nil  =  nil
0  =  0
s(x1)  =  s(x1)
if1(x1, x2, x3, x4)  =  if1(x1, x2, x3, x4)
filterlow(x1, x2)  =  filterlow(x1, x2)
half(x1)  =  x1
if2(x1, x2, x3, x4)  =  if2(x1, x2, x3, x4)
filterhigh(x1, x2)  =  filterhigh(x1, x2)
equal_bool(x1, x2)  =  equal_bool(x1, x2)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
not(x1)  =  not(x1)
isa_true(x1)  =  x1
isa_false(x1)  =  x1
equal_sort[a0](x1, x2)  =  equal_sort[a0](x1, x2)
equal_sort[a5](x1, x2)  =  equal_sort[a5](x1, x2)
equal_sort[a67](x1, x2)  =  equal_sort[a67](x1, x2)
witness_sort[a67]  =  witness_sort[a67]

Recursive path order with status [RPO].
Quasi-Precedence:

nil > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if2'₄, filterhigh'₂] > ge₂
nil > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > cons₂
nil > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > ge₂
nil > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > and₂
0 > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if2'₄, filterhigh'₂] > ge₂
0 > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > cons₂
0 > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > ge₂
0 > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > and₂
[if2₄, filterhigh₂] > cons₂
[if2₄, filterhigh₂] > ge₂
equal_bool2 > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if2'₄, filterhigh'₂] > ge₂
equal_bool2 > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > cons₂
equal_bool2 > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > ge₂
equal_bool2 > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > and₂
or₂ > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if2'₄, filterhigh'₂] > ge₂
or₂ > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > cons₂
or₂ > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > ge₂
or₂ > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > and₂
witness_sort[a67] > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if2'₄, filterhigh'₂] > ge₂
witness_sort[a67] > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > cons₂
witness_sort[a67] > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > [if1₄, filterlow₂] > ge₂
witness_sort[a67] > [true, false, equal_sort[a0]₂, equal_sort[a5]₂] > and₂

Status:

if2'₄: [4,2,3,1]
true: multiset
filterhigh'₂: [2,1]
cons₂: [2,1]
ge₂: multiset
false: multiset
nil: multiset
0: multiset
s₁: [1]
if1₄: [2,4,1,3]
filterlow₂: [1,2]
if2₄: [2,4,3,1]
filterhigh₂: [1,2]
equal_bool2: [2,1]
and₂: [2,1]
or₂: [2,1]
not₁: [1]
equal_sort[a0]₂: [1,2]
equal_sort[a5]₂: [1,2]
equal_sort[a67]₂: [2,1]
witness_sort[a67]: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

if2'(true, n82, x105, xs61) → true
filterhigh'(n91, cons(x116, xs68)) → if2'(ge(x116, n91), n91, x116, xs68)
if2'(false, n100, x127, xs75) → filterhigh'(n100, xs75)
filterhigh'(n109, nil) → false
ge(x, 0) → true
ge(s(x9), s(y')) → ge(x9, y')
if1(true, n14, x20, xs10) → filterlow(n14, xs10)
filterlow(n23, cons(x31, xs17)) → if1(ge(n23, x31), n23, x31, xs17)
if1(false, n32, x42, xs24) → cons(x42, filterlow(n32, xs24))
filterlow(n41, nil) → nil
ge(0, s(x63)) → false
half(s(0)) → 0
half(s(s(x94))) → s(half(x94))
if2(true, n82, x105, xs61) → filterhigh(n82, xs61)
filterhigh(n91, cons(x116, xs68)) → if2(ge(x116, n91), n91, x116, xs68)
if2(false, n100, x127, xs75) → cons(x127, filterhigh(n100, xs75))
filterhigh(n109, nil) → nil
equal_bool(true, false) → false
equal_bool(false, true) → false
equal_bool(true, true) → true
equal_bool(false, false) → true
and(true, x) → x
and(false, x) → false
or(true, x) → true
or(false, x) → x
not(false) → true
not(true) → false
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(x0)) → false
equal_sort[a0](s(x0), 0) → false
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)
equal_sort[a5](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a5](x0, x2), equal_sort[a5](x1, x3))
equal_sort[a5](cons(x0, x1), nil) → false
equal_sort[a5](nil, cons(x0, x1)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a67](witness_sort[a67], witness_sort[a67]) → true

(150) Obligation:

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

half(0) → 0
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true

Q is empty.

(151) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(false) = 2   
POL(half(x₁)) = 2 + 2·x₁   
POL(isa_false(x₁)) = 2 + 2·x₁   
POL(isa_true(x₁)) = 1 + 2·x₁   
POL(true) = 1   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

half(0) → 0
isa_true(true) → true
isa_true(false) → false
isa_false(true) → false
isa_false(false) → true

(152) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(153) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(154) YES