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


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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) QDPOrderProof [EQUIVALENT, 313 ms]
        (80) QDP
        (81) DependencyGraphProof [EQUIVALENT, 0 ms]
        (82) QDP
        (83) QReductionProof [EQUIVALENT, 0 ms]
        (84) QDP
        (85) QDPOrderProof [EQUIVALENT, 0 ms]
        (86) QDP
        (87) DependencyGraphProof [EQUIVALENT, 0 ms]
        (88) 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)
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) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   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)))
   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 remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( IF_6(x_1, ..., x_6) ) = max{0, 2x_2 + x_4 - 2}
POL( append_2(x_1, x_2) ) = max{0, -2}
POL( nil ) = 2
POL( cons_2(x_1, x_2) ) = x_1 + 2x_2 + 2
POL( y ) = 1
POL( false ) = 1
POL( node_3(x_1, ..., x_3) ) = 2x_1 + x_3
POL( empty ) = 1
POL( LISTIFY_2(x_1, x_2) ) = x_1
POL( true ) = 0
POL( elem_1(x_1) ) = max{0, -2}

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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

(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(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 4 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.
----------------------------------------

(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))


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

(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.
----------------------------------------

(85) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   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 remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( IF_6(x_1, ..., x_6) ) = max{0, 2x_1 + 2x_2 + x_4 - 2}
POL( append_2(x_1, x_2) ) = 2
POL( nil ) = 0
POL( cons_2(x_1, x_2) ) = 2x_1 + x_2 + 2
POL( y ) = 2
POL( LISTIFY_2(x_1, x_2) ) = max{0, x_1 - 2}
POL( node_3(x_1, ..., x_3) ) = x_1 + 1
POL( false ) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none


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

(86)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   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.
----------------------------------------

(87) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(88)
TRUE