Termination w.r.t. Q proof of Secret_06_TRS

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 TransformationProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 DependencyGraphProof (⇔, 0 ms)

        ↳22 QDP

        ↳23 TransformationProof (⇔, 0 ms)

        ↳24 QDP

        ↳25 TransformationProof (⇔, 0 ms)

        ↳26 QDP

        ↳27 TransformationProof (⇔, 0 ms)

        ↳28 QDP

        ↳29 TransformationProof (⇔, 0 ms)

        ↳30 QDP

        ↳31 TransformationProof (⇔, 0 ms)

        ↳32 QDP

        ↳33 TransformationProof (⇔, 0 ms)

        ↳34 QDP

        ↳35 TransformationProof (⇔, 0 ms)

        ↳36 QDP

        ↳37 TransformationProof (⇔, 0 ms)

        ↳38 QDP

        ↳39 UsableRulesProof (⇔, 0 ms)

        ↳40 QDP

        ↳41 QReductionProof (⇔, 0 ms)

        ↳42 QDP

        ↳43 TransformationProof (⇔, 0 ms)

        ↳44 QDP

        ↳45 UsableRulesProof (⇔, 0 ms)

        ↳46 QDP

        ↳47 TransformationProof (⇔, 1 ms)

        ↳48 QDP

        ↳49 UsableRulesProof (⇔, 0 ms)

        ↳50 QDP

        ↳51 QReductionProof (⇔, 0 ms)

        ↳52 QDP

        ↳53 TransformationProof (⇔, 0 ms)

        ↳54 QDP

        ↳55 UsableRulesProof (⇔, 0 ms)

        ↳56 QDP

        ↳57 TransformationProof (⇔, 0 ms)

        ↳58 QDP

        ↳59 UsableRulesProof (⇔, 0 ms)

        ↳60 QDP

        ↳61 TransformationProof (⇔, 0 ms)

        ↳62 QDP

        ↳63 UsableRulesProof (⇔, 0 ms)

        ↳64 QDP

        ↳65 QReductionProof (⇔, 0 ms)

        ↳66 QDP

        ↳67 TransformationProof (⇔, 0 ms)

        ↳68 QDP

        ↳69 TransformationProof (⇔, 0 ms)

        ↳70 QDP

        ↳71 TransformationProof (⇔, 0 ms)

        ↳72 QDP

        ↳73 TransformationProof (⇔, 0 ms)

        ↳74 QDP

        ↳75 TransformationProof (⇔, 0 ms)

        ↳76 QDP

        ↳77 TransformationProof (⇔, 0 ms)

        ↳78 QDP

        ↳79 QDPQMonotonicMRRProof (⇔, 90 ms)

        ↳80 QDP

        ↳81 DependencyGraphProof (⇔, 0 ms)

        ↳82 TRUE

(0) Obligation:

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

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
listify(n, xs) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n))
if(true, b, n, m, xs, ys) → xs
if(false, false, n, m, xs, ys) → listify(m, xs)
if(false, true, n, m, xs, ys) → listify(n, ys)
toList(n) → listify(n, nil)

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:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
listify(n, xs) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n))
if(true, b, n, m, xs, ys) → xs
if(false, false, n, m, xs, ys) → listify(m, xs)
if(false, true, n, m, xs, ys) → listify(n, ys)
toList(n) → listify(n, nil)

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)
listify(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
toList(x0)

(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:

APPEND(cons(y, ys), x) → APPEND(ys, x)
LISTIFY(n, xs) → IF(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n))
LISTIFY(n, xs) → ISEMPTY(n)
LISTIFY(n, xs) → ISEMPTY(left(n))
LISTIFY(n, xs) → LEFT(n)
LISTIFY(n, xs) → RIGHT(n)
LISTIFY(n, xs) → LEFT(left(n))
LISTIFY(n, xs) → ELEM(left(n))
LISTIFY(n, xs) → RIGHT(left(n))
LISTIFY(n, xs) → ELEM(n)
LISTIFY(n, xs) → APPEND(xs, n)
IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
TOLIST(n) → LISTIFY(n, nil)

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
listify(n, xs) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n))
if(true, b, n, m, xs, ys) → xs
if(false, false, n, m, xs, ys) → listify(m, xs)
if(false, true, n, m, xs, ys) → listify(n, ys)
toList(n) → listify(n, nil)

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)
listify(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
toList(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 10 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
listify(n, xs) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n))
if(true, b, n, m, xs, ys) → xs
if(false, false, n, m, xs, ys) → listify(m, xs)
if(false, true, n, m, xs, ys) → listify(n, ys)
toList(n) → listify(n, nil)

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)
listify(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
toList(x0)

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(y, ys), x) → APPEND(ys, x)

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

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)
listify(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
toList(x0)

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].

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)
listify(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
toList(x0)

(11) Obligation:

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

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

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(y, ys), x) → APPEND(ys, x)
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:

LISTIFY(n, xs) → IF(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n))
IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
listify(n, xs) → if(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n))
if(true, b, n, m, xs, ys) → xs
if(false, false, n, m, xs, ys) → listify(m, xs)
if(false, true, n, m, xs, ys) → listify(n, ys)
toList(n) → listify(n, nil)

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)
listify(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
toList(x0)

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:

LISTIFY(n, xs) → IF(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n))
IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)
listify(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
toList(x0)

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].

listify(x0, x1)
if(true, x0, x1, x2, x3, x4)
if(false, false, x0, x1, x2, x3)
if(false, true, x0, x1, x2, x3)
toList(x0)

(18) Obligation:

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

LISTIFY(n, xs) → IF(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n))
IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(19) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LISTIFY(n, xs) → IF(isEmpty(n), isEmpty(left(n)), right(n), node(left(left(n)), elem(left(n)), node(right(left(n)), elem(n), right(n))), xs, append(xs, n)) at position [0] we obtained the following new rules [LPAR04]:

LISTIFY(empty, y1) → IF(true, isEmpty(left(empty)), right(empty), node(left(left(empty)), elem(left(empty)), node(right(left(empty)), elem(empty), right(empty))), y1, append(y1, empty)) → LISTIFY(empty, y1) → IF(true, isEmpty(left(empty)), right(empty), node(left(left(empty)), elem(left(empty)), node(right(left(empty)), elem(empty), right(empty))), y1, append(y1, empty))
LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(left(node(x0, x1, x2))), right(node(x0, x1, x2)), node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) → LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(left(node(x0, x1, x2))), right(node(x0, x1, x2)), node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

(20) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(empty, y1) → IF(true, isEmpty(left(empty)), right(empty), node(left(left(empty)), elem(left(empty)), node(right(left(empty)), elem(empty), right(empty))), y1, append(y1, empty))
LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(left(node(x0, x1, x2))), right(node(x0, x1, x2)), node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(left(node(x0, x1, x2))), right(node(x0, x1, x2)), node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))
IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(23) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(left(node(x0, x1, x2))), right(node(x0, x1, x2)), node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) at position [1,0] we obtained the following new rules [LPAR04]:

LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), right(node(x0, x1, x2)), node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) → LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), right(node(x0, x1, x2)), node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

(24) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), right(node(x0, x1, x2)), node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(25) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), right(node(x0, x1, x2)), node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) at position [2] we obtained the following new rules [LPAR04]:

LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) → LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

(26) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(27) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(left(node(x0, x1, x2))), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) at position [3,0,0] we obtained the following new rules [LPAR04]:

LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) → LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

(28) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(29) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(left(node(x0, x1, x2))), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) at position [3,1,0] we obtained the following new rules [LPAR04]:

LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) → LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

(30) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(31) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(left(node(x0, x1, x2))), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) at position [3,2,0,0] we obtained the following new rules [LPAR04]:

LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) → LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

(32) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(33) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), elem(node(x0, x1, x2)), right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) at position [3,2,1] we obtained the following new rules [LPAR04]:

LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), x1, right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) → LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), x1, right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

(34) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), x1, right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2)))

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(35) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), x1, right(node(x0, x1, x2)))), y1, append(y1, node(x0, x1, x2))) at position [3,2,2] we obtained the following new rules [LPAR04]:

LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), x1, x2)), y1, append(y1, node(x0, x1, x2))) → LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), x1, x2)), y1, append(y1, node(x0, x1, x2)))

(36) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), x1, x2)), y1, append(y1, node(x0, x1, x2)))

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(37) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LISTIFY(node(x0, x1, x2), y1) → IF(false, isEmpty(x0), x2, node(left(x0), elem(x0), node(right(x0), x1, x2)), y1, append(y1, node(x0, x1, x2))) at position [1] we obtained the following new rules [LPAR04]:

LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(left(empty), elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2))) → LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(left(empty), elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(left(node(x0, x1, x2)), elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2))) → LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(left(node(x0, x1, x2)), elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

(38) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(left(empty), elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(left(node(x0, x1, x2)), elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

isEmpty(empty) → true
isEmpty(node(l, x, r)) → false
left(empty) → empty
left(node(l, x, r)) → l
right(empty) → empty
right(node(l, x, r)) → r
elem(node(l, x, r)) → x
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(39) 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.

(40) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(left(empty), elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(left(node(x0, x1, x2)), elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

left(node(l, x, r)) → l
elem(node(l, x, r)) → x
right(node(l, x, r)) → r
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
left(empty) → empty
right(empty) → empty

The set Q consists of the following terms:

isEmpty(empty)
isEmpty(node(x0, x1, x2))
left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(41) 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].

isEmpty(empty)
isEmpty(node(x0, x1, x2))

(42) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(left(empty), elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(left(node(x0, x1, x2)), elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

left(node(l, x, r)) → l
elem(node(l, x, r)) → x
right(node(l, x, r)) → r
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
left(empty) → empty
right(empty) → empty

The set Q consists of the following terms:

left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(43) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(left(empty), elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2))) at position [3,0] we obtained the following new rules [LPAR04]:

LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2))) → LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))

(44) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(left(node(x0, x1, x2)), elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))

The TRS R consists of the following rules:

left(node(l, x, r)) → l
elem(node(l, x, r)) → x
right(node(l, x, r)) → r
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
left(empty) → empty
right(empty) → empty

The set Q consists of the following terms:

left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(45) 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.

(46) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(left(node(x0, x1, x2)), elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))

The TRS R consists of the following rules:

right(empty) → empty
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
left(node(l, x, r)) → l
elem(node(l, x, r)) → x
right(node(l, x, r)) → r

The set Q consists of the following terms:

left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(47) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(left(node(x0, x1, x2)), elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2))) at position [3,0] we obtained the following new rules [LPAR04]:

LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2))) → LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

(48) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

right(empty) → empty
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
left(node(l, x, r)) → l
elem(node(l, x, r)) → x
right(node(l, x, r)) → r

The set Q consists of the following terms:

left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(49) 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.

(50) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

elem(node(l, x, r)) → x
right(node(l, x, r)) → r
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
right(empty) → empty

The set Q consists of the following terms:

left(empty)
left(node(x0, x1, x2))
right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(51) 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].

left(empty)
left(node(x0, x1, x2))

(52) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

elem(node(l, x, r)) → x
right(node(l, x, r)) → r
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
right(empty) → empty

The set Q consists of the following terms:

right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(53) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(right(empty), y1, y2)), y3, append(y3, node(empty, y1, y2))) at position [3,2,0] we obtained the following new rules [LPAR04]:

LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2))) → LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))

(54) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))

The TRS R consists of the following rules:

elem(node(l, x, r)) → x
right(node(l, x, r)) → r
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
right(empty) → empty

The set Q consists of the following terms:

right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(55) 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.

(56) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
elem(node(l, x, r)) → x
right(node(l, x, r)) → r

The set Q consists of the following terms:

right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(57) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, elem(node(x0, x1, x2)), node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2))) at position [3,1] we obtained the following new rules [LPAR04]:

LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2))) → LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

(58) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))
elem(node(l, x, r)) → x
right(node(l, x, r)) → r

The set Q consists of the following terms:

right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(59) 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.

(60) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

right(node(l, x, r)) → r
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(61) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(right(node(x0, x1, x2)), y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2))) at position [3,2,0] we obtained the following new rules [LPAR04]:

LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(x2, y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2))) → LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(x2, y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

(62) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(x2, y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

right(node(l, x, r)) → r
append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(63) 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.

(64) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(x2, y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

right(empty)
right(node(x0, x1, x2))
elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(65) 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].

right(empty)
right(node(x0, x1, x2))

(66) Obligation:

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

IF(false, false, n, m, xs, ys) → LISTIFY(m, xs)
IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(x2, y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(67) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, false, n, m, xs, ys) → LISTIFY(m, xs) we obtained the following new rules [LPAR04]:

IF(false, false, z4, node(z0, z1, node(z2, z3, z4)), z5, y_0) → LISTIFY(node(z0, z1, node(z2, z3, z4)), z5) → IF(false, false, z4, node(z0, z1, node(z2, z3, z4)), z5, y_0) → LISTIFY(node(z0, z1, node(z2, z3, z4)), z5)

(68) Obligation:

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

IF(false, true, n, m, xs, ys) → LISTIFY(n, ys)
LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(x2, y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))
IF(false, false, z4, node(z0, z1, node(z2, z3, z4)), z5, y_0) → LISTIFY(node(z0, z1, node(z2, z3, z4)), z5)

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(69) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(false, true, n, m, xs, ys) → LISTIFY(n, ys) we obtained the following new rules [LPAR04]:

IF(false, true, z1, node(empty, elem(empty), node(empty, z0, z1)), z2, y_0) → LISTIFY(z1, y_0) → IF(false, true, z1, node(empty, elem(empty), node(empty, z0, z1)), z2, y_0) → LISTIFY(z1, y_0)

(70) Obligation:

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

LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(x2, y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))
IF(false, false, z4, node(z0, z1, node(z2, z3, z4)), z5, y_0) → LISTIFY(node(z0, z1, node(z2, z3, z4)), z5)
IF(false, true, z1, node(empty, elem(empty), node(empty, z0, z1)), z2, y_0) → LISTIFY(z1, y_0)

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(71) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(false, false, z4, node(z0, z1, node(z2, z3, z4)), z5, y_0) → LISTIFY(node(z0, z1, node(z2, z3, z4)), z5) we obtained the following new rules [LPAR04]:

IF(false, false, x0, node(empty, x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(empty, x2, node(x3, x4, x0)), x5) → IF(false, false, x0, node(empty, x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(empty, x2, node(x3, x4, x0)), x5)
IF(false, false, x0, node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5) → IF(false, false, x0, node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5)

(72) Obligation:

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

LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(x2, y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2)))
IF(false, true, z1, node(empty, elem(empty), node(empty, z0, z1)), z2, y_0) → LISTIFY(z1, y_0)
IF(false, false, x0, node(empty, x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(empty, x2, node(x3, x4, x0)), x5)
IF(false, false, x0, node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5)

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(73) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule LISTIFY(node(node(x0, x1, x2), y1, y2), y3) → IF(false, false, y2, node(x0, x1, node(x2, y1, y2)), y3, append(y3, node(node(x0, x1, x2), y1, y2))) we obtained the following new rules [LPAR04]:

LISTIFY(node(node(empty, x1, x2), x3, x4), x5) → IF(false, false, x4, node(empty, x1, node(x2, x3, x4)), x5, append(x5, node(node(empty, x1, x2), x3, x4))) → LISTIFY(node(node(empty, x1, x2), x3, x4), x5) → IF(false, false, x4, node(empty, x1, node(x2, x3, x4)), x5, append(x5, node(node(empty, x1, x2), x3, x4)))
LISTIFY(node(node(node(y_1, y_2, y_3), x1, x2), x3, x4), x5) → IF(false, false, x4, node(node(y_1, y_2, y_3), x1, node(x2, x3, x4)), x5, append(x5, node(node(node(y_1, y_2, y_3), x1, x2), x3, x4))) → LISTIFY(node(node(node(y_1, y_2, y_3), x1, x2), x3, x4), x5) → IF(false, false, x4, node(node(y_1, y_2, y_3), x1, node(x2, x3, x4)), x5, append(x5, node(node(node(y_1, y_2, y_3), x1, x2), x3, x4)))

(74) Obligation:

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

LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
IF(false, true, z1, node(empty, elem(empty), node(empty, z0, z1)), z2, y_0) → LISTIFY(z1, y_0)
IF(false, false, x0, node(empty, x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(empty, x2, node(x3, x4, x0)), x5)
IF(false, false, x0, node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5)
LISTIFY(node(node(empty, x1, x2), x3, x4), x5) → IF(false, false, x4, node(empty, x1, node(x2, x3, x4)), x5, append(x5, node(node(empty, x1, x2), x3, x4)))
LISTIFY(node(node(node(y_1, y_2, y_3), x1, x2), x3, x4), x5) → IF(false, false, x4, node(node(y_1, y_2, y_3), x1, node(x2, x3, x4)), x5, append(x5, node(node(node(y_1, y_2, y_3), x1, x2), x3, x4)))

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(75) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(false, true, z1, node(empty, elem(empty), node(empty, z0, z1)), z2, y_0) → LISTIFY(z1, y_0) we obtained the following new rules [LPAR04]:

IF(false, true, node(empty, y_0, y_1), node(empty, elem(empty), node(empty, x1, node(empty, y_0, y_1))), x2, x3) → LISTIFY(node(empty, y_0, y_1), x3) → IF(false, true, node(empty, y_0, y_1), node(empty, elem(empty), node(empty, x1, node(empty, y_0, y_1))), x2, x3) → LISTIFY(node(empty, y_0, y_1), x3)
IF(false, true, node(node(empty, y_0, y_1), y_2, y_3), node(empty, elem(empty), node(empty, x1, node(node(empty, y_0, y_1), y_2, y_3))), x2, x3) → LISTIFY(node(node(empty, y_0, y_1), y_2, y_3), x3) → IF(false, true, node(node(empty, y_0, y_1), y_2, y_3), node(empty, elem(empty), node(empty, x1, node(node(empty, y_0, y_1), y_2, y_3))), x2, x3) → LISTIFY(node(node(empty, y_0, y_1), y_2, y_3), x3)
IF(false, true, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), node(empty, elem(empty), node(empty, x1, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))), x2, x3) → LISTIFY(node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), x3) → IF(false, true, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), node(empty, elem(empty), node(empty, x1, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))), x2, x3) → LISTIFY(node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), x3)

(76) Obligation:

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

LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2)))
IF(false, false, x0, node(empty, x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(empty, x2, node(x3, x4, x0)), x5)
IF(false, false, x0, node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5)
LISTIFY(node(node(empty, x1, x2), x3, x4), x5) → IF(false, false, x4, node(empty, x1, node(x2, x3, x4)), x5, append(x5, node(node(empty, x1, x2), x3, x4)))
LISTIFY(node(node(node(y_1, y_2, y_3), x1, x2), x3, x4), x5) → IF(false, false, x4, node(node(y_1, y_2, y_3), x1, node(x2, x3, x4)), x5, append(x5, node(node(node(y_1, y_2, y_3), x1, x2), x3, x4)))
IF(false, true, node(empty, y_0, y_1), node(empty, elem(empty), node(empty, x1, node(empty, y_0, y_1))), x2, x3) → LISTIFY(node(empty, y_0, y_1), x3)
IF(false, true, node(node(empty, y_0, y_1), y_2, y_3), node(empty, elem(empty), node(empty, x1, node(node(empty, y_0, y_1), y_2, y_3))), x2, x3) → LISTIFY(node(node(empty, y_0, y_1), y_2, y_3), x3)
IF(false, true, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), node(empty, elem(empty), node(empty, x1, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))), x2, x3) → LISTIFY(node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), x3)

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(77) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule LISTIFY(node(empty, y1, y2), y3) → IF(false, true, y2, node(empty, elem(empty), node(empty, y1, y2)), y3, append(y3, node(empty, y1, y2))) we obtained the following new rules [LPAR04]:

LISTIFY(node(empty, x0, node(empty, y_0, y_1)), x2) → IF(false, true, node(empty, y_0, y_1), node(empty, elem(empty), node(empty, x0, node(empty, y_0, y_1))), x2, append(x2, node(empty, x0, node(empty, y_0, y_1)))) → LISTIFY(node(empty, x0, node(empty, y_0, y_1)), x2) → IF(false, true, node(empty, y_0, y_1), node(empty, elem(empty), node(empty, x0, node(empty, y_0, y_1))), x2, append(x2, node(empty, x0, node(empty, y_0, y_1))))
LISTIFY(node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3)), x2) → IF(false, true, node(node(empty, y_0, y_1), y_2, y_3), node(empty, elem(empty), node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3))), x2, append(x2, node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3)))) → LISTIFY(node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3)), x2) → IF(false, true, node(node(empty, y_0, y_1), y_2, y_3), node(empty, elem(empty), node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3))), x2, append(x2, node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3))))
LISTIFY(node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6)), x2) → IF(false, true, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), node(empty, elem(empty), node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))), x2, append(x2, node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6)))) → LISTIFY(node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6)), x2) → IF(false, true, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), node(empty, elem(empty), node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))), x2, append(x2, node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))))

(78) Obligation:

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

IF(false, false, x0, node(empty, x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(empty, x2, node(x3, x4, x0)), x5)
IF(false, false, x0, node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5)
LISTIFY(node(node(empty, x1, x2), x3, x4), x5) → IF(false, false, x4, node(empty, x1, node(x2, x3, x4)), x5, append(x5, node(node(empty, x1, x2), x3, x4)))
LISTIFY(node(node(node(y_1, y_2, y_3), x1, x2), x3, x4), x5) → IF(false, false, x4, node(node(y_1, y_2, y_3), x1, node(x2, x3, x4)), x5, append(x5, node(node(node(y_1, y_2, y_3), x1, x2), x3, x4)))
IF(false, true, node(empty, y_0, y_1), node(empty, elem(empty), node(empty, x1, node(empty, y_0, y_1))), x2, x3) → LISTIFY(node(empty, y_0, y_1), x3)
IF(false, true, node(node(empty, y_0, y_1), y_2, y_3), node(empty, elem(empty), node(empty, x1, node(node(empty, y_0, y_1), y_2, y_3))), x2, x3) → LISTIFY(node(node(empty, y_0, y_1), y_2, y_3), x3)
IF(false, true, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), node(empty, elem(empty), node(empty, x1, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))), x2, x3) → LISTIFY(node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), x3)
LISTIFY(node(empty, x0, node(empty, y_0, y_1)), x2) → IF(false, true, node(empty, y_0, y_1), node(empty, elem(empty), node(empty, x0, node(empty, y_0, y_1))), x2, append(x2, node(empty, x0, node(empty, y_0, y_1))))
LISTIFY(node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3)), x2) → IF(false, true, node(node(empty, y_0, y_1), y_2, y_3), node(empty, elem(empty), node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3))), x2, append(x2, node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3))))
LISTIFY(node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6)), x2) → IF(false, true, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), node(empty, elem(empty), node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))), x2, append(x2, node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))))

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(79) QDPQMonotonicMRRProof (EQUIVALENT transformation)

By using the Q-monotonic rule removal processor with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented such that it always occurs at a strongly monotonic position in a (P,Q,R)-chain.
Strictly oriented dependency pairs:

LISTIFY(node(node(empty, x1, x2), x3, x4), x5) → IF(false, false, x4, node(empty, x1, node(x2, x3, x4)), x5, append(x5, node(node(empty, x1, x2), x3, x4)))
LISTIFY(node(node(node(y_1, y_2, y_3), x1, x2), x3, x4), x5) → IF(false, false, x4, node(node(y_1, y_2, y_3), x1, node(x2, x3, x4)), x5, append(x5, node(node(node(y_1, y_2, y_3), x1, x2), x3, x4)))
IF(false, true, node(empty, y_0, y_1), node(empty, elem(empty), node(empty, x1, node(empty, y_0, y_1))), x2, x3) → LISTIFY(node(empty, y_0, y_1), x3)
IF(false, true, node(node(empty, y_0, y_1), y_2, y_3), node(empty, elem(empty), node(empty, x1, node(node(empty, y_0, y_1), y_2, y_3))), x2, x3) → LISTIFY(node(node(empty, y_0, y_1), y_2, y_3), x3)
IF(false, true, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), node(empty, elem(empty), node(empty, x1, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))), x2, x3) → LISTIFY(node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), x3)

Used ordering: Polynomial interpretation [POLO]:

POL(IF(x₁, x₂, x₃, x₄, x₅, x₆)) = 2·x₂ + 2·x₄ + x₆
POL(LISTIFY(x₁, x₂)) = 2 + 2·x₁
POL(append(x₁, x₂)) = 0
POL(cons(x₁, x₂)) = 0
POL(elem(x₁)) = 1 + x₁
POL(empty) = 0
POL(false) = 1
POL(nil) = 0
POL(node(x₁, x₂, x₃)) = 1 + 2·x₁ + x₃
POL(true) = 0
POL(y) = 0

(80) Obligation:

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

IF(false, false, x0, node(empty, x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(empty, x2, node(x3, x4, x0)), x5)
IF(false, false, x0, node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5, x6) → LISTIFY(node(node(y_0, y_1, y_2), x2, node(x3, x4, x0)), x5)
LISTIFY(node(empty, x0, node(empty, y_0, y_1)), x2) → IF(false, true, node(empty, y_0, y_1), node(empty, elem(empty), node(empty, x0, node(empty, y_0, y_1))), x2, append(x2, node(empty, x0, node(empty, y_0, y_1))))
LISTIFY(node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3)), x2) → IF(false, true, node(node(empty, y_0, y_1), y_2, y_3), node(empty, elem(empty), node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3))), x2, append(x2, node(empty, x0, node(node(empty, y_0, y_1), y_2, y_3))))
LISTIFY(node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6)), x2) → IF(false, true, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6), node(empty, elem(empty), node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))), x2, append(x2, node(empty, x0, node(node(node(y_0, y_1, y_2), y_3, y_4), y_5, y_6))))

The TRS R consists of the following rules:

append(nil, x) → cons(x, nil)
append(cons(y, ys), x) → cons(y, append(ys, x))

The set Q consists of the following terms:

elem(node(x0, x1, x2))
append(nil, x0)
append(cons(y, x0), x1)

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

(81) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 5 less nodes.