Termination w.r.t. Q proof of AProVE_09_Inductive

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

0 QTRS

↳1 Overlay + Local Confluence (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 0 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 QReductionProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 QDPSizeChangeProof (⇔, 0 ms)

        ↳13 YES

    ↳14 QDP

        ↳15 UsableRulesProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 QReductionProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 QDPSizeChangeProof (⇔, 0 ms)

        ↳20 YES

    ↳21 QDP

        ↳22 UsableRulesProof (⇔, 0 ms)

        ↳23 QDP

        ↳24 QReductionProof (⇔, 0 ms)

        ↳25 QDP

        ↳26 QDPSizeChangeProof (⇔, 0 ms)

        ↳27 YES

    ↳28 QDP

        ↳29 UsableRulesProof (⇔, 0 ms)

        ↳30 QDP

        ↳31 QReductionProof (⇔, 0 ms)

        ↳32 QDP

        ↳33 QDPSizeChangeProof (⇔, 0 ms)

        ↳34 YES

    ↳35 QDP

        ↳36 UsableRulesProof (⇔, 0 ms)

        ↳37 QDP

        ↳38 QReductionProof (⇔, 0 ms)

        ↳39 QDP

        ↳40 QDPSizeChangeProof (⇔, 0 ms)

        ↳41 YES

    ↳42 QDP

        ↳43 UsableRulesProof (⇔, 0 ms)

        ↳44 QDP

        ↳45 QReductionProof (⇔, 0 ms)

        ↳46 QDP

        ↳47 Induction-Processor (⇒, 21 ms)

        ↳48 AND

            ↳49 QDP

                ↳50 UsableRulesProof (⇔, 0 ms)

                ↳51 QDP

                ↳52 QReductionProof (⇔, 0 ms)

                ↳53 QDP

                ↳54 Induction-Processor (⇒, 6 ms)

                ↳55 AND

                    ↳56 QDP

                        ↳57 PisEmptyProof (⇔, 0 ms)

                        ↳58 YES

                    ↳59 QTRS

                        ↳60 QTRSRRRProof (⇔, 0 ms)

                        ↳61 QTRS

                        ↳62 RisEmptyProof (⇔, 0 ms)

                        ↳63 YES

            ↳64 QTRS

                ↳65 QTRSRRRProof (⇔, 0 ms)

                ↳66 QTRS

                ↳67 QTRSRRRProof (⇔, 0 ms)

                ↳68 QTRS

                ↳69 RisEmptyProof (⇔, 0 ms)

                ↳70 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)) = 2·x₁
POL(cons(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(false_renamed) = 0
POL(filterhigh(x₁, x₂)) = x₂
POL(filterlow(x₁, x₂)) = x₂
POL(ge(x₁, x₂)) = 0
POL(if1(x₁, x₂, x₃, x₄)) = 2 + x₃ + 2·x₄
POL(if2(x₁, x₂, x₃, x₄)) = 2 + 3·x₁ + x₃ + 2·x₄
POL(last(x₁)) = 1 + x₁
POL(nil) = 0
POL(s(x₁)) = 3·x₁
POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if2(true_renamed, n3, x9, xs5) → filterhigh(n3, xs5)

The following formula is valid:
z0:sort[a37].(¬(z0 =nil)→filterhigh'(last(z0 ), z0 )=true)

The transformed set:
filterhigh'(n2, cons(x8, xs4)) → if2'(ge(x8, n2), n2, x8, xs4)
if2'(true_renamed, n3, x9, xs5) → true
if2'(false_renamed, n4, x10, xs6) → filterhigh'(n4, xs6)
filterhigh'(n5, nil) → false
last(cons(x'', nil)) → x''
last(cons(x1, cons(y, xs''))) → last(cons(y, xs''))
filterlow(n, cons(x2, xs1)) → if1(ge(n, x2), n, x2, xs1)
if1(true_renamed, n', x3, xs2) → filterlow(n', xs2)
ge(x4, 0) → true_renamed
ge(0, s(x5)) → false_renamed
ge(s(x6), s(y')) → ge(x6, y')
if1(false_renamed, n'', x7, xs3) → cons(x7, filterlow(n'', xs3))
filterlow(n1, nil) → nil
filterhigh(n2, cons(x8, xs4)) → if2(ge(x8, n2), n2, x8, xs4)
if2(true_renamed, n3, x9, xs5) → filterhigh(n3, xs5)
if2(false_renamed, n4, x10, xs6) → cons(x10, filterhigh(n4, xs6))
filterhigh(n5, 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(v26)) → false
equal_sort[a0](s(v27), 0) → false
equal_sort[a0](s(v27), s(v28)) → equal_sort[a0](v27, v28)
equal_sort[a37](cons(v29, v30), cons(v31, v32)) → and(equal_sort[a0](v29, v31), equal_sort[a37](v30, v32))
equal_sort[a37](cons(v29, v30), nil) → false
equal_sort[a37](nil, cons(v33, v34)) → false
equal_sort[a37](nil, nil) → true
equal_sort[a48](true_renamed, true_renamed) → true
equal_sort[a48](true_renamed, false_renamed) → false
equal_sort[a48](false_renamed, true_renamed) → false
equal_sort[a48](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

The proof given by the theorem prover:
The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 3a20a6ef7432c3f292db1a8838479c42bf5e3b22 root 20240618 unpublished


Partial correctness of the following Program

   [x, v26, v27, v28, v29, v30, v31, v32, v33, v34, n3, x9, xs5, n4, x10, x8, xs4, n5, n2, x'', x1, y, xs'', x4, x5, x6, y', n1, n, x2, xs1, n', x3, n'', x7]
   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(v26)) -> false
   equal_sort[a0](s(v27), 0) -> false
   equal_sort[a0](s(v27), s(v28)) -> equal_sort[a0](v27, v28)
   equal_sort[a37](cons(v29, v30), cons(v31, v32)) -> equal_sort[a0](v29, v31) and equal_sort[a37](v30, v32)
   equal_sort[a37](cons(v29, v30), nil) -> false
   equal_sort[a37](nil, cons(v33, v34)) -> false
   equal_sort[a37](nil, nil) -> true
   equal_sort[a48](true_renamed, true_renamed) -> true
   equal_sort[a48](true_renamed, false_renamed) -> false
   equal_sort[a48](false_renamed, true_renamed) -> false
   equal_sort[a48](false_renamed, false_renamed) -> true
   equal_sort[a63](witness_sort[a63], witness_sort[a63]) -> true
   if2'(true_renamed, n3, x9, xs5) -> true
   if2'(false_renamed, n4, x10, cons(x8, xs4)) -> if2'(ge(x8, n4), n4, x8, xs4)
   if2'(false_renamed, n4, x10, nil) -> false
   filterhigh'(n5, nil) -> false
   equal_sort[a48](ge(x8, n2), true_renamed) -> true | filterhigh'(n2, cons(x8, xs4)) -> true
   equal_sort[a48](ge(x8, n2), true_renamed) -> false | filterhigh'(n2, cons(x8, xs4)) -> filterhigh'(n2, xs4)
   last(cons(x'', nil)) -> x''
   last(cons(x1, cons(y, xs''))) -> last(cons(y, xs''))
   last(nil) -> 0
   ge(x4, 0) -> true_renamed
   ge(0, s(x5)) -> false_renamed
   ge(s(x6), s(y')) -> ge(x6, y')
   filterlow(n1, nil) -> nil
   equal_sort[a48](ge(n, x2), true_renamed) -> true | filterlow(n, cons(x2, xs1)) -> filterlow(n, xs1)
   equal_sort[a48](ge(n, x2), true_renamed) -> false | filterlow(n, cons(x2, xs1)) -> cons(x2, filterlow(n, xs1))
   filterhigh(n5, nil) -> nil
   equal_sort[a48](ge(x8, n2), true_renamed) -> true | filterhigh(n2, cons(x8, xs4)) -> filterhigh(n2, xs4)
   equal_sort[a48](ge(x8, n2), true_renamed) -> false | filterhigh(n2, cons(x8, xs4)) -> cons(x8, filterhigh(n2, xs4))
   if1(true_renamed, n', x3, cons(x2, xs1)) -> if1(ge(n', x2), n', x2, xs1)
   if1(true_renamed, n', x3, nil) -> nil
   if1(false_renamed, n'', x7, cons(x2, xs1)) -> cons(x7, if1(ge(n'', x2), n'', x2, xs1))
   if1(false_renamed, n'', x7, nil) -> cons(x7, nil)
   if2(true_renamed, n3, x9, cons(x8, xs4)) -> if2(ge(x8, n3), n3, x8, xs4)
   if2(true_renamed, n3, x9, nil) -> nil
   if2(false_renamed, n4, x10, cons(x8, xs4)) -> cons(x10, if2(ge(x8, n4), n4, x8, xs4))
   if2(false_renamed, n4, x10, nil) -> cons(x10, nil)

using the following formula:
z0:sort[a37].(~(z0=nil)->filterhigh'(last(z0), z0)=true)

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


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

(0)
Obligation:
Formula:
z0:sort[a37].(~(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:
x1:sort[a0],y:sort[a0],xs'':sort[a37].(~(cons(x1, cons(y, xs''))=nil)->filterhigh'(last(cons(x1, cons(y, xs''))), cons(x1, cons(y, xs'')))=true)

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=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 shown by simple symbolic evaluation.
----------------------------------------

(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_sort[a48](ge(s(n), s(n)), true_renamed)=true





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

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false






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

(13)
Complex Obligation (AND)

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

(14)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=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_sort[a48](ge(s(n), s(n)), true_renamed)=false




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

(18) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(19)
Obligation:
Formula:
False

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=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_sort[a48](ge(s(n), s(n)), true_renamed)=false)->False)

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false




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

(21)
Obligation:
Formula:
n:sort[a0].((filterhigh'(n, cons(n, nil))=true/\equal_sort[a48](ge(s(n), s(n)), true_renamed)=false)->False)

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false




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

(22) Symbolic evaluation under hypothesis (SOUND)
Could be reduced by symbolic evaluation under hypothesis to:
n:sort[a0].~(equal_sort[a48](ge(n, n), true_renamed)=false)

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

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

(23)
Obligation:
Formula:
n:sort[a0].~(equal_sort[a48](ge(n, n), true_renamed)=false)

Hypotheses:
n:sort[a0].filterhigh'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a48](ge(s(n), s(n)), true_renamed)=false




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

(24) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false))

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




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

(25)
Obligation:
Formula:
n:sort[a0].(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=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_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

There are no hypotheses.




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

(27)
Obligation:
Formula:
n:sort[a0].(filterhigh'(n, cons(n, nil))=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=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_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=true





Formula:
n:sort[a0].(filterhigh'(n, nil)=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=false






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

(29)
Complex Obligation (AND)

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

(30)
Obligation:
Formula:
n:sort[a0].(true=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=true




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

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

n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=true

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

(32)
YES

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

(33)
Obligation:
Formula:
n:sort[a0].(filterhigh'(n, nil)=true->(equal_sort[a48](ge(n, n), true_renamed)=false->~(equal_sort[a48](ge(n, n), true_renamed)=false)))

Hypotheses:
n:sort[a0].equal_sort[a48](ge(n, n), true_renamed)=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:
x1:sort[a0],y:sort[a0],xs'':sort[a37].(~(cons(x1, cons(y, xs''))=nil)->filterhigh'(last(cons(x1, cons(y, xs''))), cons(x1, cons(y, xs'')))=true)

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true




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

(40) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(41)
Obligation:
Formula:
y:sort[a0],xs'':sort[a37],x1:sort[a0].filterhigh'(last(cons(y, xs'')), cons(x1, cons(y, xs'')))=true

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=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],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true
x1:sort[a0],y:sort[a0],xs'':sort[a37].equal_sort[a48](ge(x1, last(cons(y, xs''))), true_renamed)=true





Formula:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true
x1:sort[a0],y:sort[a0],xs'':sort[a37].equal_sort[a48](ge(x1, last(cons(y, xs''))), true_renamed)=false






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

(43)
Complex Obligation (AND)

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

(44)
Obligation:
Formula:
true=true

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true
x1:sort[a0],y:sort[a0],xs'':sort[a37].equal_sort[a48](ge(x1, last(cons(y, xs''))), true_renamed)=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],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true

Hypotheses:
y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=true
x1:sort[a0],y:sort[a0],xs'':sort[a37].equal_sort[a48](ge(x1, last(cons(y, xs''))), true_renamed)=false




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

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

y:sort[a0],xs'':sort[a37].filterhigh'(last(cons(y, xs'')), cons(y, xs''))=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) = 3
POL(QSORT(x₁)) = 2·x₁
POL(cons(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(false_renamed) = 0
POL(filterlow(x₁, x₂)) = x₂
POL(ge(x₁, x₂)) = 0
POL(if1(x₁, x₂, x₃, x₄)) = 1 + x₁ + x₃ + 2·x₄
POL(last(x₁)) = 2 + x₁
POL(nil) = 1
POL(s(x₁)) = 3·x₁
POL(true_renamed) = 0

At least one of these decreasing rules is always used after the deleted DP:
if1(true_renamed, n1, x6, xs2) → filterlow(n1, xs2)

The following formula is valid:
z0:sort[a25].(¬(z0 =nil)→filterlow'(last(z0 ), z0 )=true)

The transformed set:
filterlow'(n, cons(x1, xs'')) → if1'(ge(n, x1), n, x1, xs'')
filterlow'(n', nil) → false
if1'(false_renamed, n'', x5, xs1) → filterlow'(n'', xs1)
if1'(true_renamed, n1, x6, xs2) → true
last(cons(x', nil)) → x'
last(cons(x'', cons(y, xs'))) → last(cons(y, xs'))
filterlow(n, cons(x1, xs'')) → if1(ge(n, x1), n, x1, xs'')
filterlow(n', nil) → nil
ge(x2, 0) → true_renamed
ge(0, s(x3)) → false_renamed
ge(s(x4), s(y')) → ge(x4, y')
if1(false_renamed, n'', x5, xs1) → cons(x5, filterlow(n'', xs1))
if1(true_renamed, n1, x6, xs2) → filterlow(n1, xs2)
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(v18)) → false
equal_sort[a0](s(v19), 0) → false
equal_sort[a0](s(v19), s(v20)) → equal_sort[a0](v19, v20)
equal_sort[a25](cons(v21, v22), cons(v23, v24)) → and(equal_sort[a0](v21, v23), equal_sort[a25](v22, v24))
equal_sort[a25](cons(v21, v22), nil) → false
equal_sort[a25](nil, cons(v25, v26)) → false
equal_sort[a25](nil, nil) → true
equal_sort[a36](true_renamed, true_renamed) → true
equal_sort[a36](true_renamed, false_renamed) → false
equal_sort[a36](false_renamed, true_renamed) → false
equal_sort[a36](false_renamed, false_renamed) → true
equal_sort[a43](witness_sort[a43], witness_sort[a43]) → true

The proof given by the theorem prover:
The following output was given by the internal theorem prover:

proof of internal
# AProVE Commit ID: 3a20a6ef7432c3f292db1a8838479c42bf5e3b22 root 20240618 unpublished


Partial correctness of the following Program

   [x, v18, v19, v20, v21, v22, v23, v24, v25, v26, n', n, x1, xs'', n1, x6, xs2, n'', x5, x', x'', y, xs', x2, x3, x4, y']
   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(v18)) -> false
   equal_sort[a0](s(v19), 0) -> false
   equal_sort[a0](s(v19), s(v20)) -> equal_sort[a0](v19, v20)
   equal_sort[a25](cons(v21, v22), cons(v23, v24)) -> equal_sort[a0](v21, v23) and equal_sort[a25](v22, v24)
   equal_sort[a25](cons(v21, v22), nil) -> false
   equal_sort[a25](nil, cons(v25, v26)) -> false
   equal_sort[a25](nil, nil) -> true
   equal_sort[a36](true_renamed, true_renamed) -> true
   equal_sort[a36](true_renamed, false_renamed) -> false
   equal_sort[a36](false_renamed, true_renamed) -> false
   equal_sort[a36](false_renamed, false_renamed) -> true
   equal_sort[a43](witness_sort[a43], witness_sort[a43]) -> true
   filterlow'(n', nil) -> false
   equal_sort[a36](ge(n, x1), false_renamed) -> true | filterlow'(n, cons(x1, xs'')) -> filterlow'(n, xs'')
   equal_sort[a36](ge(n, x1), false_renamed) -> false | filterlow'(n, cons(x1, xs'')) -> true
   if1'(true_renamed, n1, x6, xs2) -> true
   if1'(false_renamed, n'', x5, cons(x1, xs'')) -> if1'(ge(n'', x1), n'', x1, xs'')
   if1'(false_renamed, n'', x5, nil) -> false
   last(cons(x', nil)) -> x'
   last(cons(x'', cons(y, xs'))) -> last(cons(y, xs'))
   last(nil) -> 0
   filterlow(n', nil) -> nil
   equal_sort[a36](ge(n, x1), false_renamed) -> true | filterlow(n, cons(x1, xs'')) -> cons(x1, filterlow(n, xs''))
   equal_sort[a36](ge(n, x1), false_renamed) -> false | filterlow(n, cons(x1, xs'')) -> filterlow(n, xs'')
   ge(x2, 0) -> true_renamed
   ge(0, s(x3)) -> false_renamed
   ge(s(x4), s(y')) -> ge(x4, y')
   if1(false_renamed, n'', x5, cons(x1, xs'')) -> cons(x5, if1(ge(n'', x1), n'', x1, xs''))
   if1(false_renamed, n'', x5, nil) -> cons(x5, nil)
   if1(true_renamed, n1, x6, cons(x1, xs'')) -> if1(ge(n1, x1), n1, x1, xs'')
   if1(true_renamed, n1, x6, nil) -> nil

using the following formula:
z0:sort[a25].(~(z0=nil)->filterlow'(last(z0), z0)=true)

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


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

(0)
Obligation:
Formula:
z0:sort[a25].(~(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:
x'':sort[a0],y:sort[a0],xs':sort[a25].(~(cons(x'', cons(y, xs'))=nil)->filterlow'(last(cons(x'', cons(y, xs'))), cons(x'', cons(y, xs')))=true)

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=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 shown by simple symbolic evaluation.
----------------------------------------

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

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true





Formula:
true=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=false






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

(13)
Complex Obligation (AND)

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

(14)
Obligation:
Formula:
n:sort[a0].filterlow'(s(n), nil)=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




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

(15) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(16)
Obligation:
Formula:
False

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




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

(17) 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_sort[a36](ge(s(n), s(n)), false_renamed)=true)->False)

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




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

(18)
Obligation:
Formula:
n:sort[a0].((filterlow'(n, cons(n, nil))=true/\equal_sort[a36](ge(s(n), s(n)), false_renamed)=true)->False)

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




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

(19) Symbolic evaluation under hypothesis (SOUND)
Could be reduced by symbolic evaluation under hypothesis to:
n:sort[a0].~(equal_sort[a36](ge(n, n), false_renamed)=true)

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

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

(20)
Obligation:
Formula:
n:sort[a0].~(equal_sort[a36](ge(n, n), false_renamed)=true)

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=true




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

(21) Hypothesis Lifting (EQUIVALENT)
Formula could be generalised by hypothesis lifting to the following new obligation:
Formula:
n:sort[a0].(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true))

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




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

(22)
Obligation:
Formula:
n:sort[a0].(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true))

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




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

(23) 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_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

There are no hypotheses.




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

(24)
Obligation:
Formula:
n:sort[a0].(filterlow'(n, cons(n, nil))=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

There are no hypotheses.




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

(25) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
n:sort[a0].(filterlow'(n, nil)=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

Hypotheses:
n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=true





Formula:
n:sort[a0].(true=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

Hypotheses:
n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=false






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

(26)
Complex Obligation (AND)

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

(27)
Obligation:
Formula:
n:sort[a0].(filterlow'(n, nil)=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

Hypotheses:
n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=true




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

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

(29)
YES

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

(30)
Obligation:
Formula:
n:sort[a0].(true=true->(equal_sort[a36](ge(n, n), false_renamed)=true->~(equal_sort[a36](ge(n, n), false_renamed)=true)))

Hypotheses:
n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=false




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

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

n:sort[a0].equal_sort[a36](ge(n, n), false_renamed)=false

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

(32)
YES

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

(33)
Obligation:
Formula:
true=true

Hypotheses:
n:sort[a0].filterlow'(n, cons(n, nil))=true
n:sort[a0].equal_sort[a36](ge(s(n), s(n)), false_renamed)=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:
x'':sort[a0],y:sort[a0],xs':sort[a25].(~(cons(x'', cons(y, xs'))=nil)->filterlow'(last(cons(x'', cons(y, xs'))), cons(x'', cons(y, xs')))=true)

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true




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

(40) Symbolic evaluation (EQUIVALENT)
Could be shown by simple symbolic evaluation.
----------------------------------------

(41)
Obligation:
Formula:
y:sort[a0],xs':sort[a25],x'':sort[a0].filterlow'(last(cons(y, xs')), cons(x'', cons(y, xs')))=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true




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

(42) Conditional Evaluation (EQUIVALENT)
The formula could be reduced to the following new obligations by conditional evaluation:
Formula:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](ge(last(cons(y, xs')), x''), false_renamed)=true





Formula:
true=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](ge(last(cons(y, xs')), x''), false_renamed)=false






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

(43)
Complex Obligation (AND)

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

(44)
Obligation:
Formula:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](ge(last(cons(y, xs')), x''), false_renamed)=true




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

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

y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true

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

(46)
YES

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

(47)
Obligation:
Formula:
true=true

Hypotheses:
y:sort[a0],xs':sort[a25].filterlow'(last(cons(y, xs')), cons(y, xs'))=true
y:sort[a0],xs':sort[a25],x'':sort[a0].equal_sort[a36](ge(last(cons(y, xs')), x''), false_renamed)=false




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

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

filterlow'(n, cons(x1, xs'')) → if1'(ge(n, x1), n, x1, xs'')
filterlow'(n', nil) → false
if1'(false_renamed, n'', x5, xs1) → filterlow'(n'', xs1)
if1'(true_renamed, n1, x6, xs2) → true
last(cons(x', nil)) → x'
last(cons(x'', cons(y, xs'))) → last(cons(y, xs'))
filterlow(n, cons(x1, xs'')) → if1(ge(n, x1), n, x1, xs'')
filterlow(n', nil) → nil
ge(x2, 0) → true_renamed
ge(0, s(x3)) → false_renamed
ge(s(x4), s(y')) → ge(x4, y')
if1(false_renamed, n'', x5, xs1) → cons(x5, filterlow(n'', xs1))
if1(true_renamed, n1, x6, xs2) → filterlow(n1, xs2)
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(v18)) → false
equal_sort[a0](s(v19), 0) → false
equal_sort[a0](s(v19), s(v20)) → equal_sort[a0](v19, v20)
equal_sort[a25](cons(v21, v22), cons(v23, v24)) → and(equal_sort[a0](v21, v23), equal_sort[a25](v22, v24))
equal_sort[a25](cons(v21, v22), nil) → false
equal_sort[a25](nil, cons(v25, v26)) → false
equal_sort[a25](nil, nil) → true
equal_sort[a36](true_renamed, true_renamed) → true
equal_sort[a36](true_renamed, false_renamed) → false
equal_sort[a36](false_renamed, true_renamed) → false
equal_sort[a36](false_renamed, false_renamed) → true
equal_sort[a43](witness_sort[a43], witness_sort[a43]) → true

Q is empty.

(60) QTRSRRRProof (EQUIVALENT transformation)

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

[filterlow'₂, if1'₄] > [cons₂, ge₂, last₁] > true_renamed
[filterlow'₂, if1'₄] > [cons₂, ge₂, last₁] > 0
[filterlow'₂, if1'₄] > [cons₂, ge₂, last₁] > and₂
[filterlow'₂, if1'₄] > [true, equal_sort[a0]₂] > false
s₁ > false_renamed > false
s₁ > false_renamed > [filterlow₂, if1₄] > [cons₂, ge₂, last₁] > true_renamed
s₁ > false_renamed > [filterlow₂, if1₄] > [cons₂, ge₂, last₁] > 0
s₁ > false_renamed > [filterlow₂, if1₄] > [cons₂, ge₂, last₁] > and₂
s₁ > false_renamed > [filterlow₂, if1₄] > nil > 0
equal_bool₂ > [true, equal_sort[a0]₂] > false
not₁ > [true, equal_sort[a0]₂] > false
isa_false₁ > [true, equal_sort[a0]₂] > false
equal_sort[a25]₂ > [true, equal_sort[a0]₂] > false
equal_sort[a25]₂ > and₂
equal_sort[a36]₂ > [true, equal_sort[a0]₂] > false
witness_sort[a43] > [true, equal_sort[a0]₂] > false

Status:

filterlow'₂: [1,2]
cons₂: multiset
if1'₄: [2,4,1,3]
ge₂: [2,1]
nil: multiset
false: multiset
false_renamed: multiset
true_renamed: multiset
true: multiset
last₁: multiset
filterlow₂: [1,2]
if1₄: [2,4,1,3]
0: multiset
s₁: [1]
equal_bool₂: multiset
and₂: [2,1]
or₂: multiset
not₁: multiset
isa_true₁: [1]
isa_false₁: [1]
equal_sort[a0]₂: [1,2]
equal_sort[a25]₂: multiset
equal_sort[a36]₂: multiset
equal_sort[a43]₂: multiset
witness_sort[a43]: multiset

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

filterlow'(n, cons(x1, xs'')) → if1'(ge(n, x1), n, x1, xs'')
filterlow'(n', nil) → false
if1'(false_renamed, n'', x5, xs1) → filterlow'(n'', xs1)
if1'(true_renamed, n1, x6, xs2) → true
last(cons(x', nil)) → x'
last(cons(x'', cons(y, xs'))) → last(cons(y, xs'))
filterlow(n, cons(x1, xs'')) → if1(ge(n, x1), n, x1, xs'')
filterlow(n', nil) → nil
ge(x2, 0) → true_renamed
ge(0, s(x3)) → false_renamed
ge(s(x4), s(y')) → ge(x4, y')
if1(false_renamed, n'', x5, xs1) → cons(x5, filterlow(n'', xs1))
if1(true_renamed, n1, x6, xs2) → filterlow(n1, xs2)
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(v18)) → false
equal_sort[a0](s(v19), 0) → false
equal_sort[a0](s(v19), s(v20)) → equal_sort[a0](v19, v20)
equal_sort[a25](cons(v21, v22), cons(v23, v24)) → and(equal_sort[a0](v21, v23), equal_sort[a25](v22, v24))
equal_sort[a25](cons(v21, v22), nil) → false
equal_sort[a25](nil, cons(v25, v26)) → false
equal_sort[a25](nil, nil) → true
equal_sort[a36](true_renamed, true_renamed) → true
equal_sort[a36](true_renamed, false_renamed) → false
equal_sort[a36](false_renamed, true_renamed) → false
equal_sort[a36](false_renamed, false_renamed) → true
equal_sort[a43](witness_sort[a43], witness_sort[a43]) → true

(61) Obligation:

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

(62) RisEmptyProof (EQUIVALENT transformation)

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

(63) YES

(64) Obligation:

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

filterhigh'(n2, cons(x8, xs4)) → if2'(ge(x8, n2), n2, x8, xs4)
if2'(true_renamed, n3, x9, xs5) → true
if2'(false_renamed, n4, x10, xs6) → filterhigh'(n4, xs6)
filterhigh'(n5, nil) → false
last(cons(x'', nil)) → x''
last(cons(x1, cons(y, xs''))) → last(cons(y, xs''))
filterlow(n, cons(x2, xs1)) → if1(ge(n, x2), n, x2, xs1)
if1(true_renamed, n', x3, xs2) → filterlow(n', xs2)
ge(x4, 0) → true_renamed
ge(0, s(x5)) → false_renamed
ge(s(x6), s(y')) → ge(x6, y')
if1(false_renamed, n'', x7, xs3) → cons(x7, filterlow(n'', xs3))
filterlow(n1, nil) → nil
filterhigh(n2, cons(x8, xs4)) → if2(ge(x8, n2), n2, x8, xs4)
if2(true_renamed, n3, x9, xs5) → filterhigh(n3, xs5)
if2(false_renamed, n4, x10, xs6) → cons(x10, filterhigh(n4, xs6))
filterhigh(n5, 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(v26)) → false
equal_sort[a0](s(v27), 0) → false
equal_sort[a0](s(v27), s(v28)) → equal_sort[a0](v27, v28)
equal_sort[a37](cons(v29, v30), cons(v31, v32)) → and(equal_sort[a0](v29, v31), equal_sort[a37](v30, v32))
equal_sort[a37](cons(v29, v30), nil) → false
equal_sort[a37](nil, cons(v33, v34)) → false
equal_sort[a37](nil, nil) → true
equal_sort[a48](true_renamed, true_renamed) → true
equal_sort[a48](true_renamed, false_renamed) → false
equal_sort[a48](false_renamed, true_renamed) → false
equal_sort[a48](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

Q is empty.

(65) 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_renamed = true_renamed
true = true
false_renamed = false_renamed
nil = nil
false = false
last(x1) = last(x1)
filterlow(x1, x2) = filterlow(x1, x2)
if1(x1, x2, x3, x4) = if1(x1, x2, x3, x4)
0 = 0
s(x1) = s(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) = x1
isa_false(x1) = isa_false(x1)
equal_sort[a0](x1, x2) = equal_sort[a0](x1, x2)
equal_sort[a37](x1, x2) = equal_sort[a37](x1, x2)
equal_sort[a48](x1, x2) = equal_sort[a48](x1, x2)
equal_sort[a63](x1, x2) = equal_sort[a63](x1, x2)
witness_sort[a63] = witness_sort[a63]

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

[filterhigh'₂, if2'₄] > [true, witness_sort[a63]] > [nil, false, s₁] > [ge₂, false_renamed]
last₁ > cons₂ > [nil, false, s₁] > [ge₂, false_renamed]
last₁ > [true_renamed, 0] > [nil, false, s₁] > [ge₂, false_renamed]
[filterlow₂, if1₄] > cons₂ > [nil, false, s₁] > [ge₂, false_renamed]
[filterhigh₂, if2₄] > cons₂ > [nil, false, s₁] > [ge₂, false_renamed]
equal_bool₂ > [true, witness_sort[a63]] > [nil, false, s₁] > [ge₂, false_renamed]
or₂ > [true, witness_sort[a63]] > [nil, false, s₁] > [ge₂, false_renamed]
not₁ > [true, witness_sort[a63]] > [nil, false, s₁] > [ge₂, false_renamed]
isa_false₁ > [true, witness_sort[a63]] > [nil, false, s₁] > [ge₂, false_renamed]
equal_sort[a37]₂ > and₂ > [nil, false, s₁] > [ge₂, false_renamed]
equal_sort[a37]₂ > equal_sort[a0]₂ > [true, witness_sort[a63]] > [nil, false, s₁] > [ge₂, false_renamed]
equal_sort[a48]₂ > [true, witness_sort[a63]] > [nil, false, s₁] > [ge₂, false_renamed]
equal_sort[a63]₂ > [ge₂, false_renamed]

Status:

filterhigh'₂: [1,2]
cons₂: multiset
if2'₄: [2,4,1,3]
ge₂: [1,2]
true_renamed: multiset
true: multiset
false_renamed: multiset
nil: multiset
false: multiset
last₁: multiset
filterlow₂: [1,2]
if1₄: [2,4,3,1]
0: multiset
s₁: multiset
filterhigh₂: [2,1]
if2₄: [4,2,1,3]
equal_bool₂: multiset
and₂: multiset
or₂: multiset
not₁: [1]
isa_false₁: [1]
equal_sort[a0]₂: [2,1]
equal_sort[a37]₂: multiset
equal_sort[a48]₂: multiset
equal_sort[a63]₂: multiset
witness_sort[a63]: multiset

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

filterhigh'(n2, cons(x8, xs4)) → if2'(ge(x8, n2), n2, x8, xs4)
if2'(true_renamed, n3, x9, xs5) → true
if2'(false_renamed, n4, x10, xs6) → filterhigh'(n4, xs6)
filterhigh'(n5, nil) → false
last(cons(x'', nil)) → x''
last(cons(x1, cons(y, xs''))) → last(cons(y, xs''))
filterlow(n, cons(x2, xs1)) → if1(ge(n, x2), n, x2, xs1)
if1(true_renamed, n', x3, xs2) → filterlow(n', xs2)
ge(x4, 0) → true_renamed
ge(0, s(x5)) → false_renamed
ge(s(x6), s(y')) → ge(x6, y')
if1(false_renamed, n'', x7, xs3) → cons(x7, filterlow(n'', xs3))
filterlow(n1, nil) → nil
filterhigh(n2, cons(x8, xs4)) → if2(ge(x8, n2), n2, x8, xs4)
if2(true_renamed, n3, x9, xs5) → filterhigh(n3, xs5)
if2(false_renamed, n4, x10, xs6) → cons(x10, filterhigh(n4, xs6))
filterhigh(n5, 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_false(true) → false
isa_false(false) → true
equal_sort[a0](0, 0) → true
equal_sort[a0](0, s(v26)) → false
equal_sort[a0](s(v27), 0) → false
equal_sort[a0](s(v27), s(v28)) → equal_sort[a0](v27, v28)
equal_sort[a37](cons(v29, v30), cons(v31, v32)) → and(equal_sort[a0](v29, v31), equal_sort[a37](v30, v32))
equal_sort[a37](cons(v29, v30), nil) → false
equal_sort[a37](nil, cons(v33, v34)) → false
equal_sort[a37](nil, nil) → true
equal_sort[a48](true_renamed, true_renamed) → true
equal_sort[a48](true_renamed, false_renamed) → false
equal_sort[a48](false_renamed, true_renamed) → false
equal_sort[a48](false_renamed, false_renamed) → true
equal_sort[a63](witness_sort[a63], witness_sort[a63]) → true

(66) Obligation:

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

isa_true(true) → true
isa_true(false) → false

Q is empty.

(67) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:

isa_true₁ > false > true

and weight map:

true=1
false=1
isa_true_1=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

isa_true(true) → true
isa_true(false) → false

(68) Obligation:

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

(69) RisEmptyProof (EQUIVALENT transformation)

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

(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) RisEmptyProof (EQUIVALENT transformation)

(63) YES

(64) Obligation:

(65) QTRSRRRProof (EQUIVALENT transformation)

(66) Obligation:

(67) QTRSRRRProof (EQUIVALENT transformation)

(68) Obligation:

(69) RisEmptyProof (EQUIVALENT transformation)

(70) YES