YES Termination w.r.t. Q proof of AProVE_09_Inductive_qsort.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 QReductionProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 QDPSizeChangeProof (⇔, 0 ms)
        ↳34 YES
    ↳35 QDP
        ↳36 UsableRulesProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 QReductionProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 TransformationProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 TransformationProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 Induction-Processor (⇒, 11 ms)
        ↳45 AND
            ↳46 QDP
                ↳47 UsableRulesProof (⇔, 0 ms)
                ↳48 QDP
                ↳49 QReductionProof (⇔, 0 ms)
                ↳50 QDP
                ↳51 Induction-Processor (⇒, 1 ms)
                ↳52 AND
                    ↳53 QDP
                        ↳54 PisEmptyProof (⇔, 0 ms)
                        ↳55 YES
                    ↳56 QTRS
                        ↳57 QTRSRRRProof (⇔, 25 ms)
                        ↳58 QTRS
                        ↳59 QTRSRRRProof (⇔, 0 ms)
                        ↳60 QTRS
                        ↳61 RisEmptyProof (⇔, 0 ms)
                        ↳62 YES
            ↳63 QTRS
                ↳64 QTRSRRRProof (⇔, 30 ms)
                ↳65 QTRS
                ↳66 RisEmptyProof (⇔, 0 ms)
                ↳67 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(x, cons(x, xs))), cons(x, qsort(filterhigh(x, 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))

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(x, cons(x, xs))), cons(x, qsort(filterhigh(x, 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))

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)

(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(x, cons(x, xs))), cons(x, qsort(filterhigh(x, cons(x, xs)))))
QSORT(cons(x, xs)) → QSORT(filterlow(x, cons(x, xs)))
QSORT(cons(x, xs)) → FILTERLOW(x, cons(x, xs))
QSORT(cons(x, xs)) → QSORT(filterhigh(x, cons(x, xs)))
QSORT(cons(x, xs)) → FILTERHIGH(x, 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)

The TRS R consists of the following rules:

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

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)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) Obligation:

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

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(x, cons(x, xs))), cons(x, qsort(filterhigh(x, 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))

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)

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:

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)

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)

(11) 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.

(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:

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

(13) YES

(14) Obligation:

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

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(x, cons(x, xs))), cons(x, qsort(filterhigh(x, 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))

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)

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:

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)

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)

(18) 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.

(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:

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

(20) YES

(21) Obligation:

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

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(x, cons(x, xs))), cons(x, qsort(filterhigh(x, 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))

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)

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:

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)

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

(25) 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.

(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:

  • 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

(27) YES

(28) 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(x, cons(x, xs))), cons(x, qsort(filterhigh(x, 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))

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)

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:

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)

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)

(32) 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.

(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:

  • 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

(34) YES

(35) Obligation:

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

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

The TRS R consists of the following rules:

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

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)

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:

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

The TRS R consists of the following rules:

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)

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

(39) Obligation:

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

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

The TRS R consists of the following rules:

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))

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

(40) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QSORT(cons(x, xs)) → QSORT(filterhigh(x, cons(x, xs))) at position [0] we obtained the following new rules [LPAR04]:

QSORT(cons(x, xs)) → QSORT(if2(ge(x, x), x, x, xs)) → QSORT(cons(x, xs)) → QSORT(if2(ge(x, x), x, x, xs))

(41) Obligation:

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

QSORT(cons(x, xs)) → QSORT(filterlow(x, cons(x, xs)))
QSORT(cons(x, xs)) → QSORT(if2(ge(x, x), x, x, xs))

The TRS R consists of the following rules:

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))

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

(42) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule QSORT(cons(x, xs)) → QSORT(filterlow(x, cons(x, xs))) at position [0] we obtained the following new rules [LPAR04]:

QSORT(cons(x, xs)) → QSORT(if1(ge(x, x), x, x, xs)) → QSORT(cons(x, xs)) → QSORT(if1(ge(x, x), x, x, xs))

(43) Obligation:

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

QSORT(cons(x, xs)) → QSORT(if2(ge(x, x), x, x, xs))
QSORT(cons(x, xs)) → QSORT(if1(ge(x, x), x, x, xs))

The TRS R consists of the following rules:

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))

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

(44) Induction-Processor (SOUND transformation)


This DP could be deleted by the Induction-Processor:
QSORT(cons(x, xs)) → QSORT(if2(ge(x, x), x, x, xs))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(QSORT(x1)) = x1   
POL(cons(x1, x2)) = 1 + x2   
POL(false_renamed) = 1   
POL(filterhigh(x1, x2)) = x2   
POL(filterlow(x1, x2)) = x2   
POL(ge(x1, x2)) = 1   
POL(if1(x1, x2, x3, x4)) = 1 + x4   
POL(if2(x1, x2, x3, x4)) = x1 + x4   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(true_renamed) = 1   

At least one of these decreasing rules is always used after the deleted DP:
if2(true_renamed, n1, x5, xs2) → filterhigh(n1, xs2)


The following formula is valid:
x:sort[a0],xs:sort[a2].if2'(ge(x , x ), x , x , xs )=true


The transformed set:
if2'(true_renamed, n1, x5, xs2) → true
if2'(false_renamed, n2, x6, xs3) → filterhigh'(n2, xs3)
filterhigh'(n3, nil) → false
filterhigh'(n5, cons(x8, xs5)) → if2'(ge(x8, n5), n5, x8, xs5)
if1(true_renamed, n, x'', xs'') → filterlow(n, xs'')
ge(x1, 0) → true_renamed
ge(0, s(x2)) → false_renamed
ge(s(x3), s(y)) → ge(x3, y)
if1(false_renamed, n', x4, xs1) → cons(x4, filterlow(n', xs1))
filterlow(n'', nil) → nil
if2(true_renamed, n1, x5, xs2) → filterhigh(n1, xs2)
if2(false_renamed, n2, x6, xs3) → cons(x6, filterhigh(n2, xs3))
filterhigh(n3, nil) → nil
filterlow(n4, cons(x7, xs4)) → if1(ge(n4, x7), n4, x7, xs4)
filterhigh(n5, cons(x8, xs5)) → if2(ge(x8, n5), n5, x8, xs5)
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(v21)) → false
equal_sort[a0](s(v22), 0) → false
equal_sort[a0](s(v22), s(v23)) → equal_sort[a0](v22, v23)
equal_sort[a2](cons(v24, v25), cons(v26, v27)) → and(equal_sort[a0](v24, v26), equal_sort[a2](v25, v27))
equal_sort[a2](cons(v24, v25), nil) → false
equal_sort[a2](nil, cons(v28, v29)) → false
equal_sort[a2](nil, nil) → true
equal_sort[a33](true_renamed, true_renamed) → true
equal_sort[a33](true_renamed, false_renamed) → false
equal_sort[a33](false_renamed, true_renamed) → false
equal_sort[a33](false_renamed, false_renamed) → true
equal_sort[a57](witness_sort[a57], witness_sort[a57]) → 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, v21, v22, v23, v24, v25, v26, v27, v28, v29, n1, x5, xs2, n2, x6, x8, xs5, n3, n5, x1, x2, x3, y, n'', n4, x7, xs4, n, x'', n', x4]
   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(v21)) -> false
   equal_sort[a0](s(v22), 0) -> false
   equal_sort[a0](s(v22), s(v23)) -> equal_sort[a0](v22, v23)
   equal_sort[a2](cons(v24, v25), cons(v26, v27)) -> equal_sort[a0](v24, v26) and equal_sort[a2](v25, v27)
   equal_sort[a2](cons(v24, v25), nil) -> false
   equal_sort[a2](nil, cons(v28, v29)) -> false
   equal_sort[a2](nil, nil) -> true
   equal_sort[a33](true_renamed, true_renamed) -> true
   equal_sort[a33](true_renamed, false_renamed) -> false
   equal_sort[a33](false_renamed, true_renamed) -> false
   equal_sort[a33](false_renamed, false_renamed) -> true
   equal_sort[a57](witness_sort[a57], witness_sort[a57]) -> true
   if2'(true_renamed, n1, x5, xs2) -> true
   if2'(false_renamed, n2, x6, nil) -> false
   if2'(false_renamed, n2, x6, cons(x8, xs5)) -> if2'(ge(x8, n2), n2, x8, xs5)
   filterhigh'(n3, nil) -> false
   equal_sort[a33](ge(x8, n5), true_renamed) -> true | filterhigh'(n5, cons(x8, xs5)) -> true
   equal_sort[a33](ge(x8, n5), true_renamed) -> false | filterhigh'(n5, cons(x8, xs5)) -> filterhigh'(n5, xs5)
   ge(x1, 0) -> true_renamed
   ge(0, s(x2)) -> false_renamed
   ge(s(x3), s(y)) -> ge(x3, y)
   filterlow(n'', nil) -> nil
   equal_sort[a33](ge(n4, x7), true_renamed) -> true | filterlow(n4, cons(x7, xs4)) -> filterlow(n4, xs4)
   equal_sort[a33](ge(n4, x7), true_renamed) -> false | filterlow(n4, cons(x7, xs4)) -> cons(x7, filterlow(n4, xs4))
   filterhigh(n3, nil) -> nil
   equal_sort[a33](ge(x8, n5), true_renamed) -> true | filterhigh(n5, cons(x8, xs5)) -> filterhigh(n5, xs5)
   equal_sort[a33](ge(x8, n5), true_renamed) -> false | filterhigh(n5, cons(x8, xs5)) -> cons(x8, filterhigh(n5, xs5))
   if1(true_renamed, n, x'', nil) -> nil
   if1(true_renamed, n, x'', cons(x7, xs4)) -> if1(ge(n, x7), n, x7, xs4)
   if1(false_renamed, n', x4, nil) -> cons(x4, nil)
   if1(false_renamed, n', x4, cons(x7, xs4)) -> cons(x4, if1(ge(n', x7), n', x7, xs4))
   if2(true_renamed, n1, x5, nil) -> nil
   if2(true_renamed, n1, x5, cons(x8, xs5)) -> if2(ge(x8, n1), n1, x8, xs5)
   if2(false_renamed, n2, x6, nil) -> cons(x6, nil)
   if2(false_renamed, n2, x6, cons(x8, xs5)) -> cons(x6, if2(ge(x8, n2), n2, x8, xs5))

using the following formula:
x:sort[a0],xs:sort[a2].if2'(ge(x, x), x, x, xs)=true

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


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

(0)
Obligation:
Formula:
x:sort[a0],xs:sort[a2].if2'(ge(x, x), x, x, xs)=true

There are no hypotheses.




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

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



1. Base Case:
Formula:
xs:sort[a2].if2'(ge(0, 0), 0, 0, xs)=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0],xs:sort[a2].if2'(ge(s(n), s(n)), s(n), s(n), xs)=true

Hypotheses:
n:sort[a0],!xs:sort[a2].if2'(ge(n, n), n, n, xs)=true






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
xs:sort[a2].if2'(ge(0, 0), 0, 0, xs)=true

There are no hypotheses.




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

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

(5)
YES

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

(6)
Obligation:
Formula:
n:sort[a0],xs:sort[a2].if2'(ge(s(n), s(n)), s(n), s(n), xs)=true

Hypotheses:
n:sort[a0],!xs:sort[a2].if2'(ge(n, n), n, n, xs)=true




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

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

(8)
Obligation:
Formula:
n:sort[a0],xs:sort[a2].if2'(ge(n, n), s(n), s(n), xs)=true

Hypotheses:
n:sort[a0],!xs:sort[a2].if2'(ge(n, n), n, n, xs)=true




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

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

Formula:
n:sort[a0],x_1:sort[a0],x_2:sort[a2].if2'(ge(n, n), s(n), s(n), cons(x_1, x_2))=true

Hypotheses:
n:sort[a0],!xs:sort[a2].if2'(ge(n, n), n, n, xs)=true





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

Hypotheses:
n:sort[a0],!xs:sort[a2].if2'(ge(n, n), n, n, xs)=true






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

(10)
Complex Obligation (AND)

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

(11)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0],x_2:sort[a2].if2'(ge(n, n), s(n), s(n), cons(x_1, x_2))=true

Hypotheses:
n:sort[a0],!xs:sort[a2].if2'(ge(n, n), n, n, xs)=true




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

(12) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0],x_1:sort[a0],x_2:sort[a2].if2'(ge(n, n), n', n', cons(x_1, x_2))=true

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


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

(13)
Obligation:
Formula:
n:sort[a0],n':sort[a0],x_1:sort[a0],x_2:sort[a2].if2'(ge(n, n), n', n', cons(x_1, x_2))=true

Hypotheses:
n:sort[a0],!xs:sort[a2].if2'(ge(n, n), n, n, xs)=true




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

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



1. Base Case:
Formula:
n':sort[a0],x_1:sort[a0],x_2:sort[a2].if2'(ge(0, 0), n', n', cons(x_1, x_2))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0],x_2:sort[a2].if2'(ge(s(n''), s(n'')), n', n', cons(x_1, x_2))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_2:sort[a2].if2'(ge(n'', n''), n', n', cons(x_1, x_2))=true






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

(15)
Complex Obligation (AND)

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

(16)
Obligation:
Formula:
n':sort[a0],x_1:sort[a0],x_2:sort[a2].if2'(ge(0, 0), n', n', cons(x_1, x_2))=true

There are no hypotheses.




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

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

(18)
YES

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

(19)
Obligation:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0],x_2:sort[a2].if2'(ge(s(n''), s(n'')), n', n', cons(x_1, x_2))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_2:sort[a2].if2'(ge(n'', n''), n', n', cons(x_1, x_2))=true




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

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

n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_2:sort[a2].if2'(ge(n'', n''), n', n', cons(x_1, x_2))=true

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

(21)
YES

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

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

Hypotheses:
n:sort[a0],!xs:sort[a2].if2'(ge(n, n), n, n, xs)=true




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

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

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


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

(24)
Obligation:
Formula:
n:sort[a0],n':sort[a0].if2'(ge(n, n), n', n', nil)=true

Hypotheses:
n:sort[a0],!xs:sort[a2].if2'(ge(n, n), n, n, xs)=true




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

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



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

There are no hypotheses.





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

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






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

(26)
Complex Obligation (AND)

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

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

There are no hypotheses.




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

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

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




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

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

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

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

(32)
YES

(45) Complex Obligation (AND)

(46) Obligation:

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

QSORT(cons(x, xs)) → QSORT(if1(ge(x, x), x, x, xs))

The TRS R consists of the following rules:

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))

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

(47) 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.

(48) Obligation:

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

QSORT(cons(x, xs)) → QSORT(if1(ge(x, x), x, x, xs))

The TRS R consists of the following rules:

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

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))

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

(49) 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)

(50) Obligation:

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

QSORT(cons(x, xs)) → QSORT(if1(ge(x, x), x, x, xs))

The TRS R consists of the following rules:

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

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))

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

(51) Induction-Processor (SOUND transformation)


This DP could be deleted by the Induction-Processor:
QSORT(cons(x, xs)) → QSORT(if1(ge(x, x), x, x, xs))


This order was computed:
Polynomial interpretation [POLO]:

POL(0) = 1   
POL(QSORT(x1)) = x1   
POL(cons(x1, x2)) = 1 + x2   
POL(false_renamed) = 1   
POL(filterlow(x1, x2)) = x2   
POL(ge(x1, x2)) = 1   
POL(if1(x1, x2, x3, x4)) = x1 + x4   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(true_renamed) = 1   

At least one of these decreasing rules is always used after the deleted DP:
if1(true_renamed, n, x1, xs') → filterlow(n, xs')


The following formula is valid:
x:sort[a0],xs:sort[a5].if1'(ge(x , x ), x , x , xs )=true


The transformed set:
if1'(true_renamed, n, x1, xs') → true
filterlow'(n', cons(x2, xs'')) → if1'(ge(n', x2), n', x2, xs'')
if1'(false_renamed, n'', x3, xs1) → filterlow'(n'', xs1)
filterlow'(n1, nil) → false
ge(x', 0) → true_renamed
ge(s(x''), s(y)) → ge(x'', y)
if1(true_renamed, n, x1, xs') → filterlow(n, xs')
filterlow(n', cons(x2, xs'')) → if1(ge(n', x2), n', x2, xs'')
if1(false_renamed, n'', x3, xs1) → cons(x3, filterlow(n'', xs1))
filterlow(n1, nil) → nil
ge(0, s(x4)) → false_renamed
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(v13)) → false
equal_sort[a0](s(v14), 0) → false
equal_sort[a0](s(v14), s(v15)) → equal_sort[a0](v14, v15)
equal_sort[a5](cons(v16, v17), cons(v18, v19)) → and(equal_sort[a0](v16, v18), equal_sort[a5](v17, v19))
equal_sort[a5](cons(v16, v17), nil) → false
equal_sort[a5](nil, cons(v20, v21)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a18](true_renamed, true_renamed) → true
equal_sort[a18](true_renamed, false_renamed) → false
equal_sort[a18](false_renamed, true_renamed) → false
equal_sort[a18](false_renamed, false_renamed) → true
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → 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, v13, v14, v15, v16, v17, v18, v19, v20, v21, n, x1, xs', n'', x3, x2, xs'', n1, n', x', x'', y, x4]
   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(v13)) -> false
   equal_sort[a0](s(v14), 0) -> false
   equal_sort[a0](s(v14), s(v15)) -> equal_sort[a0](v14, v15)
   equal_sort[a5](cons(v16, v17), cons(v18, v19)) -> equal_sort[a0](v16, v18) and equal_sort[a5](v17, v19)
   equal_sort[a5](cons(v16, v17), nil) -> false
   equal_sort[a5](nil, cons(v20, v21)) -> false
   equal_sort[a5](nil, nil) -> true
   equal_sort[a18](true_renamed, true_renamed) -> true
   equal_sort[a18](true_renamed, false_renamed) -> false
   equal_sort[a18](false_renamed, true_renamed) -> false
   equal_sort[a18](false_renamed, false_renamed) -> true
   equal_sort[a37](witness_sort[a37], witness_sort[a37]) -> true
   if1'(true_renamed, n, x1, xs') -> true
   if1'(false_renamed, n'', x3, cons(x2, xs'')) -> if1'(ge(n'', x2), n'', x2, xs'')
   if1'(false_renamed, n'', x3, nil) -> false
   filterlow'(n1, nil) -> false
   equal_sort[a18](ge(n', x2), true_renamed) -> true | filterlow'(n', cons(x2, xs'')) -> true
   equal_sort[a18](ge(n', x2), true_renamed) -> false | filterlow'(n', cons(x2, xs'')) -> filterlow'(n', xs'')
   ge(x', 0) -> true_renamed
   ge(s(x''), s(y)) -> ge(x'', y)
   ge(0, s(x4)) -> false_renamed
   filterlow(n1, nil) -> nil
   equal_sort[a18](ge(n', x2), true_renamed) -> true | filterlow(n', cons(x2, xs'')) -> filterlow(n', xs'')
   equal_sort[a18](ge(n', x2), true_renamed) -> false | filterlow(n', cons(x2, xs'')) -> cons(x2, filterlow(n', xs''))
   if1(true_renamed, n, x1, cons(x2, xs'')) -> if1(ge(n, x2), n, x2, xs'')
   if1(true_renamed, n, x1, nil) -> nil
   if1(false_renamed, n'', x3, cons(x2, xs'')) -> cons(x3, if1(ge(n'', x2), n'', x2, xs''))
   if1(false_renamed, n'', x3, nil) -> cons(x3, nil)

using the following formula:
x:sort[a0],xs:sort[a5].if1'(ge(x, x), x, x, xs)=true

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


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

(0)
Obligation:
Formula:
x:sort[a0],xs:sort[a5].if1'(ge(x, x), x, x, xs)=true

There are no hypotheses.




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

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



1. Base Case:
Formula:
xs:sort[a5].if1'(ge(0, 0), 0, 0, xs)=true

There are no hypotheses.





1. Step Case:
Formula:
n:sort[a0],xs:sort[a5].if1'(ge(s(n), s(n)), s(n), s(n), xs)=true

Hypotheses:
n:sort[a0],!xs:sort[a5].if1'(ge(n, n), n, n, xs)=true






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

(2)
Complex Obligation (AND)

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

(3)
Obligation:
Formula:
xs:sort[a5].if1'(ge(0, 0), 0, 0, xs)=true

There are no hypotheses.




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

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

(5)
YES

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

(6)
Obligation:
Formula:
n:sort[a0],xs:sort[a5].if1'(ge(s(n), s(n)), s(n), s(n), xs)=true

Hypotheses:
n:sort[a0],!xs:sort[a5].if1'(ge(n, n), n, n, xs)=true




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

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

(8)
Obligation:
Formula:
n:sort[a0],xs:sort[a5].if1'(ge(n, n), s(n), s(n), xs)=true

Hypotheses:
n:sort[a0],!xs:sort[a5].if1'(ge(n, n), n, n, xs)=true




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

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

Formula:
n:sort[a0],x_1:sort[a0],x_2:sort[a5].if1'(ge(n, n), s(n), s(n), cons(x_1, x_2))=true

Hypotheses:
n:sort[a0],!xs:sort[a5].if1'(ge(n, n), n, n, xs)=true





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

Hypotheses:
n:sort[a0],!xs:sort[a5].if1'(ge(n, n), n, n, xs)=true






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

(10)
Complex Obligation (AND)

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

(11)
Obligation:
Formula:
n:sort[a0],x_1:sort[a0],x_2:sort[a5].if1'(ge(n, n), s(n), s(n), cons(x_1, x_2))=true

Hypotheses:
n:sort[a0],!xs:sort[a5].if1'(ge(n, n), n, n, xs)=true




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

(12) Inverse Substitution (SOUND)
The formula could be generalised by inverse substitution to:
n:sort[a0],n':sort[a0],x_1:sort[a0],x_2:sort[a5].if1'(ge(n, n), n', n', cons(x_1, x_2))=true

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


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

(13)
Obligation:
Formula:
n:sort[a0],n':sort[a0],x_1:sort[a0],x_2:sort[a5].if1'(ge(n, n), n', n', cons(x_1, x_2))=true

Hypotheses:
n:sort[a0],!xs:sort[a5].if1'(ge(n, n), n, n, xs)=true




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

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



1. Base Case:
Formula:
n':sort[a0],x_1:sort[a0],x_2:sort[a5].if1'(ge(0, 0), n', n', cons(x_1, x_2))=true

There are no hypotheses.





1. Step Case:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0],x_2:sort[a5].if1'(ge(s(n''), s(n'')), n', n', cons(x_1, x_2))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_2:sort[a5].if1'(ge(n'', n''), n', n', cons(x_1, x_2))=true






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

(15)
Complex Obligation (AND)

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

(16)
Obligation:
Formula:
n':sort[a0],x_1:sort[a0],x_2:sort[a5].if1'(ge(0, 0), n', n', cons(x_1, x_2))=true

There are no hypotheses.




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

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

(18)
YES

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

(19)
Obligation:
Formula:
n'':sort[a0],n':sort[a0],x_1:sort[a0],x_2:sort[a5].if1'(ge(s(n''), s(n'')), n', n', cons(x_1, x_2))=true

Hypotheses:
n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_2:sort[a5].if1'(ge(n'', n''), n', n', cons(x_1, x_2))=true




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

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

n'':sort[a0],!n':sort[a0],!x_1:sort[a0],!x_2:sort[a5].if1'(ge(n'', n''), n', n', cons(x_1, x_2))=true

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

(21)
YES

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

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

Hypotheses:
n:sort[a0],!xs:sort[a5].if1'(ge(n, n), n, n, xs)=true




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

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

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


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

(24)
Obligation:
Formula:
n:sort[a0],n':sort[a0].if1'(ge(n, n), n', n', nil)=true

Hypotheses:
n:sort[a0],!xs:sort[a5].if1'(ge(n, n), n, n, xs)=true




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

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



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

There are no hypotheses.





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

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






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

(26)
Complex Obligation (AND)

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

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

There are no hypotheses.




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

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

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




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

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

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

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

(32)
YES

(52) Complex Obligation (AND)

(53) Obligation:

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

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

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))

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

(54) PisEmptyProof (EQUIVALENT transformation)

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

(55) YES

(56) Obligation:

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

if1'(true_renamed, n, x1, xs') → true
filterlow'(n', cons(x2, xs'')) → if1'(ge(n', x2), n', x2, xs'')
if1'(false_renamed, n'', x3, xs1) → filterlow'(n'', xs1)
filterlow'(n1, nil) → false
ge(x', 0) → true_renamed
ge(s(x''), s(y)) → ge(x'', y)
if1(true_renamed, n, x1, xs') → filterlow(n, xs')
filterlow(n', cons(x2, xs'')) → if1(ge(n', x2), n', x2, xs'')
if1(false_renamed, n'', x3, xs1) → cons(x3, filterlow(n'', xs1))
filterlow(n1, nil) → nil
ge(0, s(x4)) → false_renamed
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(v13)) → false
equal_sort[a0](s(v14), 0) → false
equal_sort[a0](s(v14), s(v15)) → equal_sort[a0](v14, v15)
equal_sort[a5](cons(v16, v17), cons(v18, v19)) → and(equal_sort[a0](v16, v18), equal_sort[a5](v17, v19))
equal_sort[a5](cons(v16, v17), nil) → false
equal_sort[a5](nil, cons(v20, v21)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a18](true_renamed, true_renamed) → true
equal_sort[a18](true_renamed, false_renamed) → false
equal_sort[a18](false_renamed, true_renamed) → false
equal_sort[a18](false_renamed, false_renamed) → true
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true

Q is empty.

(57) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
if1'(x1, x2, x3, x4)  =  if1'(x1, x2, x3, x4)
true_renamed  =  true_renamed
true  =  true
filterlow'(x1, x2)  =  filterlow'(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
ge(x1, x2)  =  ge(x1, x2)
false_renamed  =  false_renamed
nil  =  nil
false  =  false
0  =  0
s(x1)  =  s(x1)
if1(x1, x2, x3, x4)  =  if1(x1, x2, x3, x4)
filterlow(x1, x2)  =  filterlow(x1, x2)
equal_bool(x1, x2)  =  equal_bool(x1, x2)
and(x1, x2)  =  and(x1, x2)
or(x1, x2)  =  or(x1, x2)
not(x1)  =  not(x1)
isa_true(x1)  =  x1
isa_false(x1)  =  isa_false(x1)
equal_sort[a0](x1, x2)  =  equal_sort[a0](x1, x2)
equal_sort[a5](x1, x2)  =  equal_sort[a5](x1, x2)
equal_sort[a18](x1, x2)  =  equal_sort[a18](x1, x2)
equal_sort[a37](x1, x2)  =  equal_sort[a37](x1, x2)
witness_sort[a37]  =  witness_sort[a37]

Recursive path order with status [RPO].
Quasi-Precedence:
[if1'4, filterlow'2] > [truerenamed, ge2, false, if14, filterlow2, not1] > [true, witnesssort[a37]] > and2
[if1'4, filterlow'2] > [truerenamed, ge2, false, if14, filterlow2, not1] > cons2 > and2
[if1'4, filterlow'2] > [truerenamed, ge2, false, if14, filterlow2, not1] > nil > and2
0 > [true, witnesssort[a37]] > and2
s1 > falserenamed > [truerenamed, ge2, false, if14, filterlow2, not1] > [true, witnesssort[a37]] > and2
s1 > falserenamed > [truerenamed, ge2, false, if14, filterlow2, not1] > cons2 > and2
s1 > falserenamed > [truerenamed, ge2, false, if14, filterlow2, not1] > nil > and2
s1 > equalsort[a0]2 > [truerenamed, ge2, false, if14, filterlow2, not1] > [true, witnesssort[a37]] > and2
s1 > equalsort[a0]2 > [truerenamed, ge2, false, if14, filterlow2, not1] > cons2 > and2
s1 > equalsort[a0]2 > [truerenamed, ge2, false, if14, filterlow2, not1] > nil > and2
equalbool2 > and2
or2 > [true, witnesssort[a37]] > and2
isafalse1 > [truerenamed, ge2, false, if14, filterlow2, not1] > [true, witnesssort[a37]] > and2
isafalse1 > [truerenamed, ge2, false, if14, filterlow2, not1] > cons2 > and2
isafalse1 > [truerenamed, ge2, false, if14, filterlow2, not1] > nil > and2
equalsort[a5]2 > equalsort[a0]2 > [truerenamed, ge2, false, if14, filterlow2, not1] > [true, witnesssort[a37]] > and2
equalsort[a5]2 > equalsort[a0]2 > [truerenamed, ge2, false, if14, filterlow2, not1] > cons2 > and2
equalsort[a5]2 > equalsort[a0]2 > [truerenamed, ge2, false, if14, filterlow2, not1] > nil > and2
equalsort[a18]2 > [true, witnesssort[a37]] > and2
equalsort[a37]2 > [true, witnesssort[a37]] > and2

Status:
if1'4: [4,2,3,1]
truerenamed: multiset
true: multiset
filterlow'2: [2,1]
cons2: multiset
ge2: [2,1]
falserenamed: multiset
nil: multiset
false: multiset
0: multiset
s1: [1]
if14: [4,2,1,3]
filterlow2: [2,1]
equalbool2: multiset
and2: multiset
or2: multiset
not1: multiset
isafalse1: multiset
equalsort[a0]2: [1,2]
equalsort[a5]2: multiset
equalsort[a18]2: multiset
equalsort[a37]2: multiset
witnesssort[a37]: multiset

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

if1'(true_renamed, n, x1, xs') → true
filterlow'(n', cons(x2, xs'')) → if1'(ge(n', x2), n', x2, xs'')
if1'(false_renamed, n'', x3, xs1) → filterlow'(n'', xs1)
filterlow'(n1, nil) → false
ge(x', 0) → true_renamed
ge(s(x''), s(y)) → ge(x'', y)
if1(true_renamed, n, x1, xs') → filterlow(n, xs')
filterlow(n', cons(x2, xs'')) → if1(ge(n', x2), n', x2, xs'')
if1(false_renamed, n'', x3, xs1) → cons(x3, filterlow(n'', xs1))
filterlow(n1, nil) → nil
ge(0, s(x4)) → false_renamed
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(v13)) → false
equal_sort[a0](s(v14), 0) → false
equal_sort[a0](s(v14), s(v15)) → equal_sort[a0](v14, v15)
equal_sort[a5](cons(v16, v17), cons(v18, v19)) → and(equal_sort[a0](v16, v18), equal_sort[a5](v17, v19))
equal_sort[a5](cons(v16, v17), nil) → false
equal_sort[a5](nil, cons(v20, v21)) → false
equal_sort[a5](nil, nil) → true
equal_sort[a18](true_renamed, true_renamed) → true
equal_sort[a18](true_renamed, false_renamed) → false
equal_sort[a18](false_renamed, true_renamed) → false
equal_sort[a18](false_renamed, false_renamed) → true
equal_sort[a37](witness_sort[a37], witness_sort[a37]) → true


(58) Obligation:

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

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

Q is empty.

(59) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:
isatrue1 > 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


(60) Obligation:

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

(61) RisEmptyProof (EQUIVALENT transformation)

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

(62) YES

(63) Obligation:

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

if2'(true_renamed, n1, x5, xs2) → true
if2'(false_renamed, n2, x6, xs3) → filterhigh'(n2, xs3)
filterhigh'(n3, nil) → false
filterhigh'(n5, cons(x8, xs5)) → if2'(ge(x8, n5), n5, x8, xs5)
if1(true_renamed, n, x'', xs'') → filterlow(n, xs'')
ge(x1, 0) → true_renamed
ge(0, s(x2)) → false_renamed
ge(s(x3), s(y)) → ge(x3, y)
if1(false_renamed, n', x4, xs1) → cons(x4, filterlow(n', xs1))
filterlow(n'', nil) → nil
if2(true_renamed, n1, x5, xs2) → filterhigh(n1, xs2)
if2(false_renamed, n2, x6, xs3) → cons(x6, filterhigh(n2, xs3))
filterhigh(n3, nil) → nil
filterlow(n4, cons(x7, xs4)) → if1(ge(n4, x7), n4, x7, xs4)
filterhigh(n5, cons(x8, xs5)) → if2(ge(x8, n5), n5, x8, xs5)
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(v21)) → false
equal_sort[a0](s(v22), 0) → false
equal_sort[a0](s(v22), s(v23)) → equal_sort[a0](v22, v23)
equal_sort[a2](cons(v24, v25), cons(v26, v27)) → and(equal_sort[a0](v24, v26), equal_sort[a2](v25, v27))
equal_sort[a2](cons(v24, v25), nil) → false
equal_sort[a2](nil, cons(v28, v29)) → false
equal_sort[a2](nil, nil) → true
equal_sort[a33](true_renamed, true_renamed) → true
equal_sort[a33](true_renamed, false_renamed) → false
equal_sort[a33](false_renamed, true_renamed) → false
equal_sort[a33](false_renamed, false_renamed) → true
equal_sort[a57](witness_sort[a57], witness_sort[a57]) → true

Q is empty.

(64) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:
[if2'4, filterhigh'2] > [ge2, if24, filterhigh2] > [cons2, and2]
nil > [truerenamed, true, false, equalsort[a57]2] > [if14, filterlow2] > [ge2, if24, filterhigh2] > [cons2, and2]
0 > [truerenamed, true, false, equalsort[a57]2] > [if14, filterlow2] > [ge2, if24, filterhigh2] > [cons2, and2]
0 > falserenamed > [ge2, if24, filterhigh2] > [cons2, and2]
s1 > [ge2, if24, filterhigh2] > [cons2, and2]
equalbool2 > [truerenamed, true, false, equalsort[a57]2] > [if14, filterlow2] > [ge2, if24, filterhigh2] > [cons2, and2]
or2 > [cons2, and2]
not1 > [cons2, and2]
isatrue1 > [cons2, and2]
isafalse1 > [cons2, and2]
[equalsort[a0]2, equalsort[a2]2] > [truerenamed, true, false, equalsort[a57]2] > [if14, filterlow2] > [ge2, if24, filterhigh2] > [cons2, and2]
equalsort[a33]2 > [truerenamed, true, false, equalsort[a57]2] > [if14, filterlow2] > [ge2, if24, filterhigh2] > [cons2, and2]
witnesssort[a57] > [cons2, and2]

Status:
if2'4: [2,4,1,3]
truerenamed: multiset
true: multiset
falserenamed: multiset
filterhigh'2: [1,2]
nil: multiset
false: multiset
cons2: multiset
ge2: [1,2]
if14: [2,4,3,1]
filterlow2: [1,2]
0: multiset
s1: [1]
if24: [4,2,1,3]
filterhigh2: [2,1]
equalbool2: [1,2]
and2: multiset
or2: multiset
not1: [1]
isatrue1: [1]
isafalse1: [1]
equalsort[a0]2: [2,1]
equalsort[a2]2: [2,1]
equalsort[a33]2: [1,2]
equalsort[a57]2: multiset
witnesssort[a57]: multiset

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

if2'(true_renamed, n1, x5, xs2) → true
if2'(false_renamed, n2, x6, xs3) → filterhigh'(n2, xs3)
filterhigh'(n3, nil) → false
filterhigh'(n5, cons(x8, xs5)) → if2'(ge(x8, n5), n5, x8, xs5)
if1(true_renamed, n, x'', xs'') → filterlow(n, xs'')
ge(x1, 0) → true_renamed
ge(0, s(x2)) → false_renamed
ge(s(x3), s(y)) → ge(x3, y)
if1(false_renamed, n', x4, xs1) → cons(x4, filterlow(n', xs1))
filterlow(n'', nil) → nil
if2(true_renamed, n1, x5, xs2) → filterhigh(n1, xs2)
if2(false_renamed, n2, x6, xs3) → cons(x6, filterhigh(n2, xs3))
filterhigh(n3, nil) → nil
filterlow(n4, cons(x7, xs4)) → if1(ge(n4, x7), n4, x7, xs4)
filterhigh(n5, cons(x8, xs5)) → if2(ge(x8, n5), n5, x8, xs5)
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(v21)) → false
equal_sort[a0](s(v22), 0) → false
equal_sort[a0](s(v22), s(v23)) → equal_sort[a0](v22, v23)
equal_sort[a2](cons(v24, v25), cons(v26, v27)) → and(equal_sort[a0](v24, v26), equal_sort[a2](v25, v27))
equal_sort[a2](cons(v24, v25), nil) → false
equal_sort[a2](nil, cons(v28, v29)) → false
equal_sort[a2](nil, nil) → true
equal_sort[a33](true_renamed, true_renamed) → true
equal_sort[a33](true_renamed, false_renamed) → false
equal_sort[a33](false_renamed, true_renamed) → false
equal_sort[a33](false_renamed, false_renamed) → true
equal_sort[a57](witness_sort[a57], witness_sort[a57]) → true


(65) Obligation:

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

(66) RisEmptyProof (EQUIVALENT transformation)

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

(67) YES