/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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

(0) QTRS
(1) Overlay + Local Confluence [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) TransformationProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) DependencyGraphProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) TransformationProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) TransformationProof [EQUIVALENT, 0 ms]
        (26) QDP
        (27) TransformationProof [EQUIVALENT, 0 ms]
        (28) QDP
        (29) TransformationProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) TransformationProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) TransformationProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) TransformationProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) TransformationProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) UsableRulesProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) QReductionProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) TransformationProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) UsableRulesProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) TransformationProof [EQUIVALENT, 0 ms]
        (48) QDP
        (49) UsableRulesProof [EQUIVALENT, 0 ms]
        (50) QDP
        (51) QReductionProof [EQUIVALENT, 0 ms]
        (52) QDP
        (53) TransformationProof [EQUIVALENT, 0 ms]
        (54) QDP
        (55) UsableRulesProof [EQUIVALENT, 0 ms]
        (56) QDP
        (57) TransformationProof [EQUIVALENT, 0 ms]
        (58) QDP
        (59) UsableRulesProof [EQUIVALENT, 0 ms]
        (60) QDP
        (61) TransformationProof [EQUIVALENT, 0 ms]
        (62) QDP
        (63) UsableRulesProof [EQUIVALENT, 0 ms]
        (64) QDP
        (65) QReductionProof [EQUIVALENT, 0 ms]
        (66) QDP
        (67) TransformationProof [EQUIVALENT, 0 ms]
        (68) QDP
        (69) TransformationProof [EQUIVALENT, 0 ms]
        (70) QDP
        (71) TransformationProof [EQUIVALENT, 0 ms]
        (72) QDP
        (73) TransformationProof [EQUIVALENT, 0 ms]
        (74) QDP
        (75) TransformationProof [EQUIVALENT, 0 ms]
        (76) QDP
        (77) TransformationProof [EQUIVALENT, 0 ms]
        (78) QDP
        (79) QDPQMonotonicMRRProof [EQUIVALENT, 184 ms]
        (80) QDP
        (81) DependencyGraphProof [EQUIVALENT, 0 ms]
        (82) QDP
        (83) QReductionProof [EQUIVALENT, 0 ms]
        (84) QDP
        (85) QDPQMonotonicMRRProof [EQUIVALENT, 6 ms]
        (86) QDP
        (87) PisEmptyProof [EQUIVALENT, 0 ms]
        (88) YES


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

(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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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:

   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(x_1, x_2, x_3, x_4, x_5, x_6)) = x_2 + 2*x_4
   POL(LISTIFY(x_1, x_2)) = 2 + 2*x_1
   POL(append(x_1, x_2)) = 0
   POL(cons(x_1, x_2)) = 0
   POL(elem(x_1)) = 0
   POL(empty) = 1
   POL(false) = 2
   POL(nil) = 0
   POL(node(x_1, x_2, x_3)) = x_1 + x_3
   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(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(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)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.
----------------------------------------

(82)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   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, 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.
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(83) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   elem(node(x0, x1, x2))


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(84)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   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, 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:

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

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


Used ordering: Polynomial interpretation [POLO]:

   POL(IF(x_1, x_2, x_3, x_4, x_5, x_6)) = 2*x_1 + 2*x_4 + 2*x_5
   POL(LISTIFY(x_1, x_2)) = 2 + 2*x_1 + 2*x_2
   POL(append(x_1, x_2)) = 0
   POL(cons(x_1, x_2)) = 0
   POL(false) = 2
   POL(nil) = 0
   POL(node(x_1, x_2, x_3)) = 2 + 2*x_1
   POL(y) = 0


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(86)
Obligation:
Q DP problem:
P is empty.
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:

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

We have to consider all minimal (P,Q,R)-chains.
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(87) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(88)
YES