Termination w.r.t. Q proof of /tmp/tmpPLGoA1/qsortlast.trs

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

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔)

        ↳9 QDP

        ↳10 QReductionProof (⇔)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔)

        ↳16 QDP

        ↳17 QReductionProof (⇔)

        ↳18 QDP

        ↳19 QDPSizeChangeProof (⇔)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔)

        ↳23 QDP

        ↳24 QReductionProof (⇔)

        ↳25 QDP

        ↳26 QDPSizeChangeProof (⇔)

        ↳27 YES

    ↳28 QDP

        ↳29 UsableRulesProof (⇔)

        ↳30 QDP

        ↳31 QReductionProof (⇔)

        ↳32 QDP

        ↳33 QDPSizeChangeProof (⇔)

        ↳34 YES

    ↳35 QDP

        ↳36 UsableRulesProof (⇔)

        ↳37 QDP

        ↳38 QReductionProof (⇔)

        ↳39 QDP

        ↳40 QDPSizeChangeProof (⇔)

        ↳41 YES

    ↳42 QDP

        ↳43 UsableRulesProof (⇔)

        ↳44 QDP

        ↳45 QReductionProof (⇔)

        ↳46 QDP

        ↳47 Induction-Processor (⇐)

        ↳48 AND

            ↳49 QDP

                ↳50 UsableRulesProof (⇔)

                ↳51 QDP

                ↳52 QReductionProof (⇔)

                ↳53 QDP

                ↳54 Induction-Processor (⇐)

                ↳55 AND

                    ↳56 QDP

                        ↳57 PisEmptyProof (⇔)

                        ↳58 YES

                    ↳59 QTRS

                        ↳60 QTRSRRRProof (⇔)

                        ↳61 QTRS

                        ↳62 QTRSRRRProof (⇔)

                        ↳63 QTRS

                        ↳64 RisEmptyProof (⇔)

                        ↳65 YES

            ↳66 QTRS

                ↳67 QTRSRRRProof (⇔)

                ↳68 QTRS

                ↳69 QTRSRRRProof (⇔)

                ↳70 QTRS

                ↳71 RisEmptyProof (⇔)

                ↳72 YES

(0) Obligation:

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

qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(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))
last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(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(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(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))
last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))

The set Q consists of the following terms:

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

QSORT(cons(x, xs)) → APPEND(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(cons(x, xs)), cons(x, xs)))))
QSORT(cons(x, xs)) → QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))
QSORT(cons(x, xs)) → FILTERLOW(last(cons(x, xs)), cons(x, xs))
QSORT(cons(x, xs)) → LAST(cons(x, xs))
QSORT(cons(x, xs)) → QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))
QSORT(cons(x, xs)) → FILTERHIGH(last(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)
LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

The TRS R consists of the following rules:

qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(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))
last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))

The set Q consists of the following terms:

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 6 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

The TRS R consists of the following rules:

qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(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))
last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))

The set Q consists of the following terms:

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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:

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))

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

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

(11) Obligation:

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

LAST(cons(x, cons(y, xs))) → LAST(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:

LAST(cons(x, cons(y, xs))) → LAST(cons(y, xs))
The graph contains the following edges 1 > 1

(13) YES

(14) 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(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(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))
last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))

The set Q consists of the following terms:

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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:

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

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

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

(18) 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.

(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:

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

(20) YES

(21) Obligation:

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(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))
last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))

The set Q consists of the following terms:

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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:

GE(s(x), s(y)) → GE(x, y)

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

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

(25) 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.

(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:

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

(27) YES

(28) 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(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(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))
last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))

The set Q consists of the following terms:

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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:

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(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

(32) 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.

(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:

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

(34) YES

(35) 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(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(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))
last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))

The set Q consists of the following terms:

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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:

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(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

(39) 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.

(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:

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

(41) YES

(42) Obligation:

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

QSORT(cons(x, xs)) → QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))
QSORT(cons(x, xs)) → QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

qsort(nil) → nil
qsort(cons(x, xs)) → append(qsort(filterlow(last(cons(x, xs)), cons(x, xs))), cons(last(cons(x, xs)), qsort(filterhigh(last(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))
last(nil) → 0
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))

The set Q consists of the following terms:

qsort(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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:

QSORT(cons(x, xs)) → QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))
QSORT(cons(x, xs)) → QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(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(nil)
qsort(cons(x0, x1))
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)
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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(nil)
qsort(cons(x0, x1))
append(nil, ys)
append(cons(x0, x1), ys)

(46) Obligation:

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

QSORT(cons(x, xs)) → QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))
QSORT(cons(x, xs)) → QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(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))
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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

(47) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
QSORT(cons(x, xs)) → QSORT(filterhigh(last(cons(x, xs)), cons(x, xs)))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 1
POL(QSORT(x₁)) = x₁
POL(cons(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(false) = 1
POL(filterhigh(x₁, x₂)) = x₂
POL(filterlow(x₁, x₂)) = x₂
POL(ge(x₁, x₂)) = 1
POL(if1(x₁, x₂, x₃, x₄)) = 1 + 2·x₃ + x₄
POL(if2(x₁, x₂, x₃, x₄)) = x₁ + 2·x₃ + x₄
POL(last(x₁)) = x₁
POL(nil) = 2
POL(s(x₁)) = 0
POL(true) = 1

At least one of these decreasing rules is always used after the deleted DP:
if2(true, n834, x1174, xs734) → filterhigh(n834, xs734)

The following formula is valid:
z0:sort[a36].(¬(z0 =nil)→filterhigh'(last(z0 ), z0 )=true)

The transformed set:
filterhigh'(n74, cons(x105, xs65)) → if2'(ge(x105, n74), n74, x105, xs65)
if2'(true, n83, x117, xs73) → true
if2'(false, n92, x129, xs81) → filterhigh'(n92, xs81)
filterhigh'(n101, nil) → false
last(cons(x, nil)) → x
last(cons(x10, cons(y'', xs5))) → last(cons(y'', xs5))
filterlow(n14, cons(x22, xs13)) → if1(ge(n14, x22), n14, x22, xs13)
if1(true, n23, x34, xs21) → filterlow(n23, xs21)
ge(x46, 0) → true
ge(0, s(x58)) → false
ge(s(x70), s(y11)) → ge(x70, y11)
if1(false, n56, x82, xs50) → cons(x82, filterlow(n56, xs50))
filterlow(n65, nil) → nil
filterhigh(n74, cons(x105, xs65)) → if2(ge(x105, n74), n74, x105, xs65)
if2(true, n83, x117, xs73) → filterhigh(n83, xs73)
if2(false, n92, x129, xs81) → cons(x129, filterhigh(n92, xs81))
filterhigh(n101, nil) → nil
last(nil) → 0
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](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, n83, x117, xs73, n92, x129, x105, xs65, n101, n74, x10, y'', xs5, x46, x58, x70, y11, n65, n14, x22, xs13, n23, x34, n56, x82]
   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[a36](cons(x0, x1), cons(x2, x3)) -> equal_sort[a36](x0, x2) and equal_sort[a36](x1, x3)
   equal_sort[a36](cons(x0, x1), nil) -> false
   equal_sort[a36](nil, cons(x0, x1)) -> false
   equal_sort[a36](nil, nil) -> true
   equal_sort[a62](witness_sort[a62], witness_sort[a62]) -> true
   if2'(true, n83, x117, xs73) -> true
   if2'(false, n92, x129, cons(x105, xs65)) -> if2'(ge(x105, n92), n92, x105, xs65)
   if2'(false, n92, x129, nil) -> false
   filterhigh'(n101, nil) -> false
   equal_bool(ge(x105, n74), true) -> true | filterhigh'(n74, cons(x105, xs65)) -> true
   equal_bool(ge(x105, n74), true) -> false | filterhigh'(n74, cons(x105, xs65)) -> filterhigh'(n74, xs65)
   last(cons(x, nil)) -> x
   last(cons(x10, cons(y'', xs5))) -> last(cons(y'', xs5))
   last(nil) -> 0
   ge(x46, 0) -> true
   ge(0, s(x58)) -> false
   ge(s(x70), s(y11)) -> ge(x70, y11)
   filterlow(n65, nil) -> nil
   equal_bool(ge(n14, x22), true) -> true | filterlow(n14, cons(x22, xs13)) -> filterlow(n14, xs13)
   equal_bool(ge(n14, x22), true) -> false | filterlow(n14, cons(x22, xs13)) -> cons(x22, filterlow(n14, xs13))
   filterhigh(n101, nil) -> nil
   equal_bool(ge(x105, n74), true) -> true | filterhigh(n74, cons(x105, xs65)) -> filterhigh(n74, xs65)
   equal_bool(ge(x105, n74), true) -> false | filterhigh(n74, cons(x105, xs65)) -> cons(x105, filterhigh(n74, xs65))
   if1(true, n23, x34, cons(x22, xs13)) -> if1(ge(n23, x22), n23, x22, xs13)
   if1(true, n23, x34, nil) -> nil
   if1(false, n56, x82, cons(x22, xs13)) -> cons(x82, if1(ge(n56, x22), n56, x22, xs13))
   if1(false, n56, x82, nil) -> cons(x82, nil)
   if2(true, n83, x117, cons(x105, xs65)) -> if2(ge(x105, n83), n83, x105, xs65)
   if2(true, n83, x117, nil) -> nil
   if2(false, n92, x129, cons(x105, xs65)) -> cons(x129, if2(ge(x105, n92), n92, x105, xs65))
   if2(false, n92, x129, nil) -> cons(x129, 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, n83, x117, xs73, n92, x129, x105, xs65, n101, n74, x10, y'', xs5, x46, x58, x70, y11, n65, n14, x22, xs13, n23, x34, n56, x82]
   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[a36](cons(x0, x1), cons(x2, x3)) -> equal_sort[a36](x0, x2) and equal_sort[a36](x1, x3)
   equal_sort[a36](cons(x0, x1), nil) -> false
   equal_sort[a36](nil, cons(x0, x1)) -> false
   equal_sort[a36](nil, nil) -> true
   equal_sort[a62](witness_sort[a62], witness_sort[a62]) -> true
   if2'(true, n83, x117, xs73) -> true
   if2'(false, n92, x129, cons(x105, xs65)) -> if2'(ge(x105, n92), n92, x105, xs65)
   if2'(false, n92, x129, nil) -> false
   filterhigh'(n101, nil) -> false
   equal_bool(ge(x105, n74), true) -> true | filterhigh'(n74, cons(x105, xs65)) -> true
   equal_bool(ge(x105, n74), true) -> false | filterhigh'(n74, cons(x105, xs65)) -> filterhigh'(n74, xs65)
   last(cons(x, nil)) -> x
   last(cons(x10, cons(y'', xs5))) -> last(cons(y'', xs5))
   last(nil) -> 0
   ge(x46, 0) -> true
   ge(0, s(x58)) -> false
   ge(s(x70), s(y11)) -> ge(x70, y11)
   filterlow(n65, nil) -> nil
   equal_bool(ge(n14, x22), true) -> true | filterlow(n14, cons(x22, xs13)) -> filterlow(n14, xs13)
   equal_bool(ge(n14, x22), true) -> false | filterlow(n14, cons(x22, xs13)) -> cons(x22, filterlow(n14, xs13))
   filterhigh(n101, nil) -> nil
   equal_bool(ge(x105, n74), true) -> true | filterhigh(n74, cons(x105, xs65)) -> filterhigh(n74, xs65)
   equal_bool(ge(x105, n74), true) -> false | filterhigh(n74, cons(x105, xs65)) -> cons(x105, filterhigh(n74, xs65))
   if1(true, n23, x34, cons(x22, xs13)) -> if1(ge(n23, x22), n23, x22, xs13)
   if1(true, n23, x34, nil) -> nil
   if1(false, n56, x82, cons(x22, xs13)) -> cons(x82, if1(ge(n56, x22), n56, x22, xs13))
   if1(false, n56, x82, nil) -> cons(x82, nil)
   if2(true, n83, x117, cons(x105, xs65)) -> if2(ge(x105, n83), n83, x105, xs65)
   if2(true, n83, x117, nil) -> nil
   if2(false, n92, x129, cons(x105, xs65)) -> cons(x129, if2(ge(x105, n92), n92, x105, xs65))
   if2(false, n92, x129, nil) -> cons(x129, nil)

using the following formula:
z0:sort[a36].(~(z0=nil)->filterhigh'(last(z0), z0)=true)

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT]
        (5) Formula
        (6) Induction by data structure [EQUIVALENT]
        (7) AND
            (8) Formula
                (9) Symbolic evaluation [EQUIVALENT]
                (10) YES
            (11) Formula
                (12) Conditional Evaluation [EQUIVALENT]
                (13) AND
                    (14) Formula
                        (15) Symbolic evaluation [EQUIVALENT]
                        (16) YES
                    (17) Formula
                        (18) Symbolic evaluation [EQUIVALENT]
                        (19) Formula
                        (20) Hypothesis Lifting [EQUIVALENT]
                        (21) Formula
                        (22) Symbolic evaluation under hypothesis [SOUND]
                        (23) Formula
                        (24) Hypothesis Lifting [EQUIVALENT]
                        (25) Formula
                        (26) Hypothesis Lifting [EQUIVALENT]
                        (27) Formula
                        (28) Conditional Evaluation [EQUIVALENT]
                        (29) AND
                            (30) Formula
                                (31) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (32) YES
                            (33) Formula
                                (34) Symbolic evaluation [EQUIVALENT]
                                (35) YES
    (36) Formula
        (37) Symbolic evaluation [EQUIVALENT]
        (38) YES
    (39) Formula
        (40) Symbolic evaluation [EQUIVALENT]
        (41) Formula
        (42) Conditional Evaluation [EQUIVALENT]
        (43) AND
            (44) Formula
                (45) Symbolic evaluation [EQUIVALENT]
                (46) YES
            (47) Formula
                (48) Symbolic evaluation under hypothesis [EQUIVALENT]
                (49) YES


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

(0)
Obligation:
Formula:
z0:sort[a36].(~(z0=nil)->filterhigh'(last(z0), z0)=true)

There are no hypotheses.




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

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm last(z0) generates the following cases:

1. Base Case:
Formula:
x:sort[a0].(~(cons(x, nil)=nil)->filterhigh'(last(cons(x, nil)), cons(x, nil))=true)

There are no hypotheses.





2. Base Case:
Formula:
(~(nil=nil)->filterhigh'(last(nil), nil)=true)

There are no hypotheses.





1. Step Case:
Formula:
x10:sort[a0],y'':sort[a0],xs5:sort[a36].(~(cons(x10, cons(y'', xs5))=nil)->filterhigh'(last(cons(x10, cons(y'', xs5))), cons(x10, cons(y'', xs5)))=true)

Hypotheses:
y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
x:sort[a0].(~(cons(x, nil)=nil)->filterhigh'(last(cons(x, nil)), cons(x, nil))=true)

There are no hypotheses.




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

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x:sort[a0].filterhigh'(x, cons(x, nil))=true
----------------------------------------

(5)
Obligation:
Formula:
x:sort[a0].filterhigh'(x, cons(x, nil))=true

There are no hypotheses.




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

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



1. Base Case:
Formula:
filterhigh'(0, cons(0, nil))=true

There are no hypotheses.





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

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true






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

(7)
Complex Obligation (AND)

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

(8)
Obligation:
Formula:
filterhigh'(0, cons(0, nil))=true

There are no hypotheses.




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

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

(10)
YES

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

(11)
Obligation:
Formula:
n:sort[a0].filterhigh'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true




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

(12) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=true





Formula:
n:sort[a0].filterhigh'(s(n), nil)=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false






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

(13)
Complex Obligation (AND)

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

(14)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=true




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

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

(16)
YES

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

(17)
Obligation:
Formula:
n:sort[a0].filterhigh'(s(n), nil)=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

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

(19)
Obligation:
Formula:
False

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

(20) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].((filterhigh'(n, cons(n, nil))=true/\equal_bool(ge(s(n), s(n)), true)=false)->False)

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

(21)
Obligation:
Formula:
n:sort[a0].((filterhigh'(n, cons(n, nil))=true/\equal_bool(ge(s(n), s(n)), true)=false)->False)

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

(22) Symbolic evaluation under hypothesis (SOUND)
Could be reduced by symbolic evaluation under hypothesis to:
n:sort[a0].~(equal_bool(ge(n, n), true)=false)

By using the following hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true

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

(23)
Obligation:
Formula:
n:sort[a0].~(equal_bool(ge(n, n), true)=false)

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

(24) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false))

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true




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

(25)
Obligation:
Formula:
n:sort[a0].(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false))

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true




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

(26) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(filterhigh'(n, cons(n, nil))=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

There are no hypotheses.




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

(27)
Obligation:
Formula:
n:sort[a0].(filterhigh'(n, cons(n, nil))=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

There are no hypotheses.




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

(28) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
n:sort[a0].(true=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(ge(n, n), true)=true





Formula:
n:sort[a0].(filterhigh'(n, nil)=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(ge(n, n), true)=false






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

(29)
Complex Obligation (AND)

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

(30)
Obligation:
Formula:
n:sort[a0].(true=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(ge(n, n), true)=true




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

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

n:sort[a0].equal_bool(ge(n, n), true)=true

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

(32)
YES

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

(33)
Obligation:
Formula:
n:sort[a0].(filterhigh'(n, nil)=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(ge(n, n), true)=false




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

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

(35)
YES

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

(36)
Obligation:
Formula:
(~(nil=nil)->filterhigh'(last(nil), nil)=true)

There are no hypotheses.




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

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

(38)
YES

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

(39)
Obligation:
Formula:
x10:sort[a0],y'':sort[a0],xs5:sort[a36].(~(cons(x10, cons(y'', xs5))=nil)->filterhigh'(last(cons(x10, cons(y'', xs5))), cons(x10, cons(y'', xs5)))=true)

Hypotheses:
y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true




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

(40) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
y'':sort[a0],xs5:sort[a36],x10:sort[a0].filterhigh'(last(cons(y'', xs5)), cons(x10, cons(y'', xs5)))=true
----------------------------------------

(41)
Obligation:
Formula:
y'':sort[a0],xs5:sort[a36],x10:sort[a0].filterhigh'(last(cons(y'', xs5)), cons(x10, cons(y'', xs5)))=true

Hypotheses:
y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true




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

(42) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true
x10:sort[a0],y'':sort[a0],xs5:sort[a36].equal_bool(ge(x10, last(cons(y'', xs5))), true)=true





Formula:
y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true

Hypotheses:
y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true
x10:sort[a0],y'':sort[a0],xs5:sort[a36].equal_bool(ge(x10, last(cons(y'', xs5))), true)=false






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

(43)
Complex Obligation (AND)

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

(44)
Obligation:
Formula:
true=true

Hypotheses:
y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true
x10:sort[a0],y'':sort[a0],xs5:sort[a36].equal_bool(ge(x10, last(cons(y'', xs5))), true)=true




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

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

(46)
YES

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

(47)
Obligation:
Formula:
y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true

Hypotheses:
y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true
x10:sort[a0],y'':sort[a0],xs5:sort[a36].equal_bool(ge(x10, last(cons(y'', xs5))), true)=false




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

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

y'':sort[a0],xs5:sort[a36].filterhigh'(last(cons(y'', xs5)), cons(y'', xs5))=true

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

(49)
YES

(48) Complex Obligation (AND)

(49) Obligation:

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

QSORT(cons(x, xs)) → QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(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))
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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:

QSORT(cons(x, xs)) → QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, 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
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))

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))
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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].

filterhigh(x0, nil)
filterhigh(x0, cons(x1, x2))
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)

(53) Obligation:

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

QSORT(cons(x, xs)) → QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

The TRS R consists of the following rules:

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, 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
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))

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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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

(54) Induction-Processor (SOUND transformation)

This DP could be deleted by the Induction-Processor:
QSORT(cons(x, xs)) → QSORT(filterlow(last(cons(x, xs)), cons(x, xs)))

This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 0
POL(QSORT(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(false) = 0
POL(filterlow(x₁, x₂)) = x₂
POL(ge(x₁, x₂)) = 3·x₁
POL(if1(x₁, x₂, x₃, x₄)) = 2 + x₃ + 2·x₄
POL(last(x₁)) = 2·x₁
POL(nil) = 1
POL(s(x₁)) = 3·x₁
POL(true) = 0

At least one of these decreasing rules is always used after the deleted DP:
if1(true, n5', x15', xs6') → filterlow(n5', xs6')

The following formula is valid:
z0:sort[a24].(¬(z0 =nil)→filterlow'(last(z0 ), z0 )=true)

The transformed set:
if1'(true, n5, x15, xs6) → true
filterlow'(n10, cons(x24, xs11)) → if1'(ge(n10, x24), n10, x24, xs11)
filterlow'(n15, nil) → false
if1'(false, n32, x68, xs32) → filterlow'(n32, xs32)
last(cons(x, nil)) → x
last(cons(x6, cons(y'', xs1))) → last(cons(y'', xs1))
if1(true, n5, x15, xs6) → filterlow(n5, xs6)
filterlow(n10, cons(x24, xs11)) → if1(ge(n10, x24), n10, x24, xs11)
filterlow(n15, nil) → nil
ge(x41, 0) → true
ge(0, s(x50)) → false
ge(s(x59), s(y13)) → ge(x59, y13)
if1(false, n32, x68, xs32) → cons(x68, filterlow(n32, xs32))
last(nil) → 0
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[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](nil, nil) → true
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

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

   [x, x0, x1, x2, x3, n5, x15, xs6, n32, x68, x24, xs11, n15, n10, x6, y'', xs1, x41, x50, x59, y13]
   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[a24](cons(x0, x1), cons(x2, x3)) -> equal_sort[a24](x0, x2) and equal_sort[a24](x1, x3)
   equal_sort[a24](cons(x0, x1), nil) -> false
   equal_sort[a24](nil, cons(x0, x1)) -> false
   equal_sort[a24](nil, nil) -> true
   equal_sort[a42](witness_sort[a42], witness_sort[a42]) -> true
   if1'(true, n5, x15, xs6) -> true
   if1'(false, n32, x68, cons(x24, xs11)) -> if1'(ge(n32, x24), n32, x24, xs11)
   if1'(false, n32, x68, nil) -> false
   filterlow'(n15, nil) -> false
   equal_bool(ge(n10, x24), true) -> true | filterlow'(n10, cons(x24, xs11)) -> true
   equal_bool(ge(n10, x24), true) -> false | filterlow'(n10, cons(x24, xs11)) -> filterlow'(n10, xs11)
   last(cons(x, nil)) -> x
   last(cons(x6, cons(y'', xs1))) -> last(cons(y'', xs1))
   last(nil) -> 0
   filterlow(n15, nil) -> nil
   equal_bool(ge(n10, x24), true) -> true | filterlow(n10, cons(x24, xs11)) -> filterlow(n10, xs11)
   equal_bool(ge(n10, x24), true) -> false | filterlow(n10, cons(x24, xs11)) -> cons(x24, filterlow(n10, xs11))
   ge(x41, 0) -> true
   ge(0, s(x50)) -> false
   ge(s(x59), s(y13)) -> ge(x59, y13)
   if1(true, n5, x15, cons(x24, xs11)) -> if1(ge(n5, x24), n5, x24, xs11)
   if1(true, n5, x15, nil) -> nil
   if1(false, n32, x68, cons(x24, xs11)) -> cons(x68, if1(ge(n32, x24), n32, x24, xs11))
   if1(false, n32, x68, nil) -> cons(x68, 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, n5, x15, xs6, n32, x68, x24, xs11, n15, n10, x6, y'', xs1, x41, x50, x59, y13]
   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[a24](cons(x0, x1), cons(x2, x3)) -> equal_sort[a24](x0, x2) and equal_sort[a24](x1, x3)
   equal_sort[a24](cons(x0, x1), nil) -> false
   equal_sort[a24](nil, cons(x0, x1)) -> false
   equal_sort[a24](nil, nil) -> true
   equal_sort[a42](witness_sort[a42], witness_sort[a42]) -> true
   if1'(true, n5, x15, xs6) -> true
   if1'(false, n32, x68, cons(x24, xs11)) -> if1'(ge(n32, x24), n32, x24, xs11)
   if1'(false, n32, x68, nil) -> false
   filterlow'(n15, nil) -> false
   equal_bool(ge(n10, x24), true) -> true | filterlow'(n10, cons(x24, xs11)) -> true
   equal_bool(ge(n10, x24), true) -> false | filterlow'(n10, cons(x24, xs11)) -> filterlow'(n10, xs11)
   last(cons(x, nil)) -> x
   last(cons(x6, cons(y'', xs1))) -> last(cons(y'', xs1))
   last(nil) -> 0
   filterlow(n15, nil) -> nil
   equal_bool(ge(n10, x24), true) -> true | filterlow(n10, cons(x24, xs11)) -> filterlow(n10, xs11)
   equal_bool(ge(n10, x24), true) -> false | filterlow(n10, cons(x24, xs11)) -> cons(x24, filterlow(n10, xs11))
   ge(x41, 0) -> true
   ge(0, s(x50)) -> false
   ge(s(x59), s(y13)) -> ge(x59, y13)
   if1(true, n5, x15, cons(x24, xs11)) -> if1(ge(n5, x24), n5, x24, xs11)
   if1(true, n5, x15, nil) -> nil
   if1(false, n32, x68, cons(x24, xs11)) -> cons(x68, if1(ge(n32, x24), n32, x24, xs11))
   if1(false, n32, x68, nil) -> cons(x68, nil)

using the following formula:
z0:sort[a24].(~(z0=nil)->filterlow'(last(z0), z0)=true)

could be successfully shown:
(0) Formula
(1) Induction by algorithm [EQUIVALENT]
(2) AND
    (3) Formula
        (4) Symbolic evaluation [EQUIVALENT]
        (5) Formula
        (6) Induction by data structure [EQUIVALENT]
        (7) AND
            (8) Formula
                (9) Symbolic evaluation [EQUIVALENT]
                (10) YES
            (11) Formula
                (12) Conditional Evaluation [EQUIVALENT]
                (13) AND
                    (14) Formula
                        (15) Symbolic evaluation [EQUIVALENT]
                        (16) YES
                    (17) Formula
                        (18) Symbolic evaluation [EQUIVALENT]
                        (19) Formula
                        (20) Hypothesis Lifting [EQUIVALENT]
                        (21) Formula
                        (22) Symbolic evaluation under hypothesis [SOUND]
                        (23) Formula
                        (24) Hypothesis Lifting [EQUIVALENT]
                        (25) Formula
                        (26) Hypothesis Lifting [EQUIVALENT]
                        (27) Formula
                        (28) Conditional Evaluation [EQUIVALENT]
                        (29) AND
                            (30) Formula
                                (31) Symbolic evaluation under hypothesis [EQUIVALENT]
                                (32) YES
                            (33) Formula
                                (34) Symbolic evaluation [EQUIVALENT]
                                (35) YES
    (36) Formula
        (37) Symbolic evaluation [EQUIVALENT]
        (38) YES
    (39) Formula
        (40) Symbolic evaluation [EQUIVALENT]
        (41) Formula
        (42) Conditional Evaluation [EQUIVALENT]
        (43) AND
            (44) Formula
                (45) Symbolic evaluation [EQUIVALENT]
                (46) YES
            (47) Formula
                (48) Symbolic evaluation under hypothesis [EQUIVALENT]
                (49) YES


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

(0)
Obligation:
Formula:
z0:sort[a24].(~(z0=nil)->filterlow'(last(z0), z0)=true)

There are no hypotheses.




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

(1) Induction by algorithm (EQUIVALENT)
Induction by algorithm last(z0) generates the following cases:

1. Base Case:
Formula:
x:sort[a0].(~(cons(x, nil)=nil)->filterlow'(last(cons(x, nil)), cons(x, nil))=true)

There are no hypotheses.





2. Base Case:
Formula:
(~(nil=nil)->filterlow'(last(nil), nil)=true)

There are no hypotheses.





1. Step Case:
Formula:
x6:sort[a0],y'':sort[a0],xs1:sort[a24].(~(cons(x6, cons(y'', xs1))=nil)->filterlow'(last(cons(x6, cons(y'', xs1))), cons(x6, cons(y'', xs1)))=true)

Hypotheses:
y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
x:sort[a0].(~(cons(x, nil)=nil)->filterlow'(last(cons(x, nil)), cons(x, nil))=true)

There are no hypotheses.




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

(4) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
x:sort[a0].filterlow'(x, cons(x, nil))=true
----------------------------------------

(5)
Obligation:
Formula:
x:sort[a0].filterlow'(x, cons(x, nil))=true

There are no hypotheses.




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

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



1. Base Case:
Formula:
filterlow'(0, cons(0, nil))=true

There are no hypotheses.





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

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true






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

(7)
Complex Obligation (AND)

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

(8)
Obligation:
Formula:
filterlow'(0, cons(0, nil))=true

There are no hypotheses.




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

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

(10)
YES

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

(11)
Obligation:
Formula:
n:sort[a0].filterlow'(s(n), cons(s(n), nil))=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true




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

(12) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=true





Formula:
n:sort[a0].filterlow'(s(n), nil)=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false






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

(13)
Complex Obligation (AND)

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

(14)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=true




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

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

(16)
YES

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

(17)
Obligation:
Formula:
n:sort[a0].filterlow'(s(n), nil)=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

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

(19)
Obligation:
Formula:
False

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

(20) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].((filterlow'(n, cons(n, nil))=true/\equal_bool(ge(s(n), s(n)), true)=false)->False)

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

(21)
Obligation:
Formula:
n:sort[a0].((filterlow'(n, cons(n, nil))=true/\equal_bool(ge(s(n), s(n)), true)=false)->False)

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

(22) Symbolic evaluation under hypothesis (SOUND)
Could be reduced by symbolic evaluation under hypothesis to:
n:sort[a0].~(equal_bool(ge(n, n), true)=false)

By using the following hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true

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

(23)
Obligation:
Formula:
n:sort[a0].~(equal_bool(ge(n, n), true)=false)

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_bool(ge(s(n), s(n)), true)=false




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

(24) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false))

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true




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

(25)
Obligation:
Formula:
n:sort[a0].(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false))

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true




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

(26) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(filterlow'(n, cons(n, nil))=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

There are no hypotheses.




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

(27)
Obligation:
Formula:
n:sort[a0].(filterlow'(n, cons(n, nil))=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

There are no hypotheses.




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

(28) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
n:sort[a0].(true=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(ge(n, n), true)=true





Formula:
n:sort[a0].(filterlow'(n, nil)=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(ge(n, n), true)=false






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

(29)
Complex Obligation (AND)

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

(30)
Obligation:
Formula:
n:sort[a0].(true=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(ge(n, n), true)=true




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

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

n:sort[a0].equal_bool(ge(n, n), true)=true

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

(32)
YES

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

(33)
Obligation:
Formula:
n:sort[a0].(filterlow'(n, nil)=true->(equal_bool(ge(n, n), true)=false->~(equal_bool(ge(n, n), true)=false)))

Hypotheses:
n:sort[a0].equal_bool(ge(n, n), true)=false




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

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

(35)
YES

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

(36)
Obligation:
Formula:
(~(nil=nil)->filterlow'(last(nil), nil)=true)

There are no hypotheses.




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

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

(38)
YES

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

(39)
Obligation:
Formula:
x6:sort[a0],y'':sort[a0],xs1:sort[a24].(~(cons(x6, cons(y'', xs1))=nil)->filterlow'(last(cons(x6, cons(y'', xs1))), cons(x6, cons(y'', xs1)))=true)

Hypotheses:
y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true




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

(40) Symbolic evaluation (EQUIVALENT)
Could be reduced to the following new obligation by simple symbolic evaluation:
y'':sort[a0],xs1:sort[a24],x6:sort[a0].filterlow'(last(cons(y'', xs1)), cons(x6, cons(y'', xs1)))=true
----------------------------------------

(41)
Obligation:
Formula:
y'':sort[a0],xs1:sort[a24],x6:sort[a0].filterlow'(last(cons(y'', xs1)), cons(x6, cons(y'', xs1)))=true

Hypotheses:
y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true




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

(42) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
true=true

Hypotheses:
y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true
y'':sort[a0],xs1:sort[a24],x6:sort[a0].equal_bool(ge(last(cons(y'', xs1)), x6), true)=true





Formula:
y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true

Hypotheses:
y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true
y'':sort[a0],xs1:sort[a24],x6:sort[a0].equal_bool(ge(last(cons(y'', xs1)), x6), true)=false






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

(43)
Complex Obligation (AND)

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

(44)
Obligation:
Formula:
true=true

Hypotheses:
y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true
y'':sort[a0],xs1:sort[a24],x6:sort[a0].equal_bool(ge(last(cons(y'', xs1)), x6), true)=true




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

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

(46)
YES

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

(47)
Obligation:
Formula:
y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true

Hypotheses:
y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true
y'':sort[a0],xs1:sort[a24],x6:sort[a0].equal_bool(ge(last(cons(y'', xs1)), x6), true)=false




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

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

y'':sort[a0],xs1:sort[a24].filterlow'(last(cons(y'', xs1)), cons(y'', xs1))=true

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

(49)
YES

(55) Complex Obligation (AND)

(56) Obligation:

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

last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, 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
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))

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)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
last(nil)
last(cons(x0, nil))
last(cons(x0, cons(x1, x2)))

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

(57) PisEmptyProof (EQUIVALENT transformation)

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

(58) YES

(59) Obligation:

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

if1'(true, n5, x15, xs6) → true
filterlow'(n10, cons(x24, xs11)) → if1'(ge(n10, x24), n10, x24, xs11)
filterlow'(n15, nil) → false
if1'(false, n32, x68, xs32) → filterlow'(n32, xs32)
last(cons(x, nil)) → x
last(cons(x6, cons(y'', xs1))) → last(cons(y'', xs1))
if1(true, n5, x15, xs6) → filterlow(n5, xs6)
filterlow(n10, cons(x24, xs11)) → if1(ge(n10, x24), n10, x24, xs11)
filterlow(n15, nil) → nil
ge(x41, 0) → true
ge(0, s(x50)) → false
ge(s(x59), s(y13)) → ge(x59, y13)
if1(false, n32, x68, xs32) → cons(x68, filterlow(n32, xs32))
last(nil) → 0
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[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](nil, nil) → true
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

Q is empty.

(60) 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)
nil = nil
false = false
last(x1) = x1
if1(x1, x2, x3, x4) = if1(x1, x2, x3, x4)
filterlow(x1, x2) = filterlow(x1, x2)
0 = 0
s(x1) = x1
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[a24](x1, x2) = equal_sort[a24](x1, x2)
equal_sort[a42](x1, x2) = equal_sort[a42](x1, x2)
witness_sort[a42] = witness_sort[a42]

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

nil > [true, false, 0] > [if1₄, filterlow₂] > cons₂ > [if1'₄, filterlow'₂, ge₂]
equal_bool₂ > [if1'₄, filterlow'₂, ge₂]
or₂ > [if1'₄, filterlow'₂, ge₂]
not₁ > [if1'₄, filterlow'₂, ge₂]
isa_true₁ > [if1'₄, filterlow'₂, ge₂]
isa_false₁ > [if1'₄, filterlow'₂, ge₂]
equal_sort[a0]₂ > [if1'₄, filterlow'₂, ge₂]
equal_sort[a24]₂ > and₂ > [if1'₄, filterlow'₂, ge₂]
equal_sort[a42]₂ > [true, false, 0] > [if1₄, filterlow₂] > cons₂ > [if1'₄, filterlow'₂, ge₂]
witness_sort[a42] > [if1'₄, filterlow'₂, ge₂]

Status:

if1'₄: [4,2,3,1]
true: multiset
filterlow'₂: [2,1]
cons₂: multiset
ge₂: [2,1]
nil: multiset
false: multiset
if1₄: [4,2,1,3]
filterlow₂: [2,1]
0: multiset
equal_bool₂: [2,1]
and₂: [1,2]
or₂: [1,2]
not₁: [1]
isa_true₁: [1]
isa_false₁: multiset
equal_sort[a0]₂: [2,1]
equal_sort[a24]₂: multiset
equal_sort[a42]₂: [2,1]
witness_sort[a42]: multiset

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

if1'(true, n5, x15, xs6) → true
filterlow'(n10, cons(x24, xs11)) → if1'(ge(n10, x24), n10, x24, xs11)
filterlow'(n15, nil) → false
if1'(false, n32, x68, xs32) → filterlow'(n32, xs32)
last(cons(x, nil)) → x
last(cons(x6, cons(y'', xs1))) → last(cons(y'', xs1))
if1(true, n5, x15, xs6) → filterlow(n5, xs6)
filterlow(n10, cons(x24, xs11)) → if1(ge(n10, x24), n10, x24, xs11)
filterlow(n15, nil) → nil
ge(x41, 0) → true
ge(0, s(x50)) → false
if1(false, n32, x68, xs32) → cons(x68, filterlow(n32, xs32))
last(nil) → 0
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[a24](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a24](x0, x2), equal_sort[a24](x1, x3))
equal_sort[a24](cons(x0, x1), nil) → false
equal_sort[a24](nil, cons(x0, x1)) → false
equal_sort[a24](nil, nil) → true
equal_sort[a42](witness_sort[a42], witness_sort[a42]) → true

(61) Obligation:

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

ge(s(x59), s(y13)) → ge(x59, y13)
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

Q is empty.

(62) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(equal_sort[a0](x₁, x₂)) = x₁ + x₂
POL(ge(x₁, x₂)) = 2·x₁ + x₂
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(x59), s(y13)) → ge(x59, y13)
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

(63) Obligation:

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

(64) RisEmptyProof (EQUIVALENT transformation)

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

(65) YES

(66) Obligation:

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

filterhigh'(n74, cons(x105, xs65)) → if2'(ge(x105, n74), n74, x105, xs65)
if2'(true, n83, x117, xs73) → true
if2'(false, n92, x129, xs81) → filterhigh'(n92, xs81)
filterhigh'(n101, nil) → false
last(cons(x, nil)) → x
last(cons(x10, cons(y'', xs5))) → last(cons(y'', xs5))
filterlow(n14, cons(x22, xs13)) → if1(ge(n14, x22), n14, x22, xs13)
if1(true, n23, x34, xs21) → filterlow(n23, xs21)
ge(x46, 0) → true
ge(0, s(x58)) → false
ge(s(x70), s(y11)) → ge(x70, y11)
if1(false, n56, x82, xs50) → cons(x82, filterlow(n56, xs50))
filterlow(n65, nil) → nil
filterhigh(n74, cons(x105, xs65)) → if2(ge(x105, n74), n74, x105, xs65)
if2(true, n83, x117, xs73) → filterhigh(n83, xs73)
if2(false, n92, x129, xs81) → cons(x129, filterhigh(n92, xs81))
filterhigh(n101, nil) → nil
last(nil) → 0
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

Q is empty.

(67) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
filterhigh'(x1, x2) = filterhigh'(x1, x2)
cons(x1, x2) = cons(x1, x2)
if2'(x1, x2, x3, x4) = if2'(x1, x2, x3, x4)
ge(x1, x2) = ge(x1, x2)
true = true
false = false
nil = nil
last(x1) = last(x1)
filterlow(x1, x2) = filterlow(x1, x2)
if1(x1, x2, x3, x4) = if1(x1, x2, x3, x4)
0 = 0
s(x1) = x1
filterhigh(x1, x2) = filterhigh(x1, x2)
if2(x1, x2, x3, x4) = if2(x1, x2, x3, x4)
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[a36](x1, x2) = equal_sort[a36](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:

[filterhigh'₂, if2'₄] > [ge₂, true, equal_sort[a36]₂] > and₂ > false
[last₁, 0] > cons₂ > false
[filterlow₂, if1₄] > cons₂ > false
[filterlow₂, if1₄] > [ge₂, true, equal_sort[a36]₂] > and₂ > false
[filterlow₂, if1₄] > nil > false
[filterhigh₂, if2₄] > cons₂ > false
[filterhigh₂, if2₄] > [ge₂, true, equal_sort[a36]₂] > and₂ > false
[filterhigh₂, if2₄] > nil > false
equal_bool₂ > [ge₂, true, equal_sort[a36]₂] > and₂ > false
or₂ > [ge₂, true, equal_sort[a36]₂] > and₂ > false
not₁ > [ge₂, true, equal_sort[a36]₂] > and₂ > false
isa_true₁ > [ge₂, true, equal_sort[a36]₂] > and₂ > false
isa_false₁ > [ge₂, true, equal_sort[a36]₂] > and₂ > false
equal_sort[a0]₂ > [ge₂, true, equal_sort[a36]₂] > and₂ > false
witness_sort[a62] > [ge₂, true, equal_sort[a36]₂] > and₂ > false

Status:

filterhigh'₂: [2,1]
cons₂: multiset
if2'₄: [4,2,3,1]
ge₂: [1,2]
true: multiset
false: multiset
nil: multiset
last₁: multiset
filterlow₂: [1,2]
if1₄: [2,4,3,1]
0: multiset
filterhigh₂: [2,1]
if2₄: [4,2,1,3]
equal_bool₂: multiset
and₂: [1,2]
or₂: [2,1]
not₁: [1]
isa_true₁: multiset
isa_false₁: [1]
equal_sort[a0]₂: [1,2]
equal_sort[a36]₂: [1,2]
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:

filterhigh'(n74, cons(x105, xs65)) → if2'(ge(x105, n74), n74, x105, xs65)
if2'(true, n83, x117, xs73) → true
if2'(false, n92, x129, xs81) → filterhigh'(n92, xs81)
filterhigh'(n101, nil) → false
last(cons(x, nil)) → x
last(cons(x10, cons(y'', xs5))) → last(cons(y'', xs5))
filterlow(n14, cons(x22, xs13)) → if1(ge(n14, x22), n14, x22, xs13)
if1(true, n23, x34, xs21) → filterlow(n23, xs21)
ge(x46, 0) → true
ge(0, s(x58)) → false
if1(false, n56, x82, xs50) → cons(x82, filterlow(n56, xs50))
filterlow(n65, nil) → nil
filterhigh(n74, cons(x105, xs65)) → if2(ge(x105, n74), n74, x105, xs65)
if2(true, n83, x117, xs73) → filterhigh(n83, xs73)
if2(false, n92, x129, xs81) → cons(x129, filterhigh(n92, xs81))
filterhigh(n101, nil) → nil
last(nil) → 0
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[a36](cons(x0, x1), cons(x2, x3)) → and(equal_sort[a36](x0, x2), equal_sort[a36](x1, x3))
equal_sort[a36](cons(x0, x1), nil) → false
equal_sort[a36](nil, cons(x0, x1)) → false
equal_sort[a36](nil, nil) → true
equal_sort[a62](witness_sort[a62], witness_sort[a62]) → true

(68) Obligation:

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

ge(s(x70), s(y11)) → ge(x70, y11)
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

Q is empty.

(69) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(equal_sort[a0](x₁, x₂)) = x₁ + x₂
POL(ge(x₁, x₂)) = 2·x₁ + x₂
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(x70), s(y11)) → ge(x70, y11)
equal_sort[a0](s(x0), s(x1)) → equal_sort[a0](x0, x1)

(0) Obligation:

(1) Overlay + Local Confluence (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) UsableRulesProof (EQUIVALENT transformation)

(9) Obligation:

(10) QReductionProof (EQUIVALENT transformation)

(11) Obligation:

(12) QDPSizeChangeProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) UsableRulesProof (EQUIVALENT transformation)

(16) Obligation:

(17) QReductionProof (EQUIVALENT transformation)

(18) Obligation:

(19) QDPSizeChangeProof (EQUIVALENT transformation)

(20) YES

(21) Obligation:

(22) UsableRulesProof (EQUIVALENT transformation)

(23) Obligation:

(24) QReductionProof (EQUIVALENT transformation)

(25) Obligation:

(26) QDPSizeChangeProof (EQUIVALENT transformation)

(27) YES

(28) Obligation:

(29) UsableRulesProof (EQUIVALENT transformation)

(30) Obligation:

(31) QReductionProof (EQUIVALENT transformation)

(32) Obligation:

(33) QDPSizeChangeProof (EQUIVALENT transformation)

(34) YES

(35) Obligation:

(36) UsableRulesProof (EQUIVALENT transformation)

(37) Obligation:

(38) QReductionProof (EQUIVALENT transformation)

(39) Obligation:

(40) QDPSizeChangeProof (EQUIVALENT transformation)

(41) YES

(42) Obligation:

(43) UsableRulesProof (EQUIVALENT transformation)

(44) Obligation:

(45) QReductionProof (EQUIVALENT transformation)

(46) Obligation:

(47) Induction-Processor (SOUND transformation)

(48) Complex Obligation (AND)

(49) Obligation:

(50) UsableRulesProof (EQUIVALENT transformation)

(51) Obligation:

(52) QReductionProof (EQUIVALENT transformation)

(53) Obligation:

(54) Induction-Processor (SOUND transformation)

(55) Complex Obligation (AND)

(56) Obligation:

(57) PisEmptyProof (EQUIVALENT transformation)

(58) YES

(59) Obligation:

(60) QTRSRRRProof (EQUIVALENT transformation)

(61) Obligation:

(62) QTRSRRRProof (EQUIVALENT transformation)

(63) Obligation:

(64) RisEmptyProof (EQUIVALENT transformation)

(65) YES

(66) Obligation:

(67) QTRSRRRProof (EQUIVALENT transformation)

(68) Obligation:

(69) QTRSRRRProof (EQUIVALENT transformation)

(70) Obligation:

(71) RisEmptyProof (EQUIVALENT transformation)

(72) YES