/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 8 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) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) QReductionProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (27) YES
    (28) QDP
        (29) UsableRulesProof [EQUIVALENT, 0 ms]
        (30) QDP
        (31) QReductionProof [EQUIVALENT, 0 ms]
        (32) QDP
        (33) TransformationProof [EQUIVALENT, 0 ms]
        (34) QDP
        (35) DependencyGraphProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) TransformationProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) TransformationProof [EQUIVALENT, 0 ms]
        (40) QDP
        (41) TransformationProof [EQUIVALENT, 0 ms]
        (42) QDP
        (43) TransformationProof [EQUIVALENT, 0 ms]
        (44) QDP
        (45) TransformationProof [EQUIVALENT, 0 ms]
        (46) QDP
        (47) DependencyGraphProof [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) TransformationProof [EQUIVALENT, 0 ms]
        (56) QDP
        (57) TransformationProof [EQUIVALENT, 0 ms]
        (58) QDP
        (59) UsableRulesProof [EQUIVALENT, 0 ms]
        (60) QDP
        (61) QReductionProof [EQUIVALENT, 0 ms]
        (62) QDP
        (63) TransformationProof [EQUIVALENT, 0 ms]
        (64) QDP
        (65) UsableRulesProof [EQUIVALENT, 0 ms]
        (66) QDP
        (67) QReductionProof [EQUIVALENT, 0 ms]
        (68) QDP
        (69) TransformationProof [EQUIVALENT, 0 ms]
        (70) QDP
        (71) UsableRulesProof [EQUIVALENT, 0 ms]
        (72) QDP
        (73) QReductionProof [EQUIVALENT, 0 ms]
        (74) QDP
        (75) TransformationProof [EQUIVALENT, 0 ms]
        (76) QDP
        (77) UsableRulesProof [EQUIVALENT, 0 ms]
        (78) QDP
        (79) TransformationProof [EQUIVALENT, 0 ms]
        (80) QDP
        (81) QDPOrderProof [EQUIVALENT, 0 ms]
        (82) QDP
        (83) DependencyGraphProof [EQUIVALENT, 0 ms]
        (84) QDP
        (85) QDPOrderProof [EQUIVALENT, 17 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(cons(x, xs)) -> false
   isEmpty(nil) -> true
   isZero(0) -> true
   isZero(s(x)) -> false
   head(cons(x, xs)) -> x
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   addLists(xs, ys, zs) -> if(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   if(true, true, b, xs, ys, xs2, ys2, zs, zs2) -> zs
   if(true, false, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, true, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs2, ys2, zs)
   if(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs, ys, zs2)
   addList(xs, ys) -> addLists(xs, ys, 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(cons(x, xs)) -> false
   isEmpty(nil) -> true
   isZero(0) -> true
   isZero(s(x)) -> false
   head(cons(x, xs)) -> x
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   addLists(xs, ys, zs) -> if(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   if(true, true, b, xs, ys, xs2, ys2, zs, zs2) -> zs
   if(true, false, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, true, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs2, ys2, zs)
   if(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs, ys, zs2)
   addList(xs, ys) -> addLists(xs, ys, nil)

The set Q consists of the following terms:

   isEmpty(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)


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

(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)
   P(s(s(x))) -> P(s(x))
   INC(s(x)) -> INC(x)
   ADDLISTS(xs, ys, zs) -> IF(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   ADDLISTS(xs, ys, zs) -> ISEMPTY(xs)
   ADDLISTS(xs, ys, zs) -> ISEMPTY(ys)
   ADDLISTS(xs, ys, zs) -> ISZERO(head(xs))
   ADDLISTS(xs, ys, zs) -> HEAD(xs)
   ADDLISTS(xs, ys, zs) -> TAIL(xs)
   ADDLISTS(xs, ys, zs) -> TAIL(ys)
   ADDLISTS(xs, ys, zs) -> P(head(xs))
   ADDLISTS(xs, ys, zs) -> INC(head(ys))
   ADDLISTS(xs, ys, zs) -> HEAD(ys)
   ADDLISTS(xs, ys, zs) -> APPEND(zs, head(ys))
   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLIST(xs, ys) -> ADDLISTS(xs, ys, nil)

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   isZero(0) -> true
   isZero(s(x)) -> false
   head(cons(x, xs)) -> x
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   addLists(xs, ys, zs) -> if(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   if(true, true, b, xs, ys, xs2, ys2, zs, zs2) -> zs
   if(true, false, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, true, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs2, ys2, zs)
   if(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs, ys, zs2)
   addList(xs, ys) -> addLists(xs, ys, nil)

The set Q consists of the following terms:

   isEmpty(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 11 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INC(s(x)) -> INC(x)

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   isZero(0) -> true
   isZero(s(x)) -> false
   head(cons(x, xs)) -> x
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   addLists(xs, ys, zs) -> if(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   if(true, true, b, xs, ys, xs2, ys2, zs, zs2) -> zs
   if(true, false, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, true, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs2, ys2, zs)
   if(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs, ys, zs2)
   addList(xs, ys) -> addLists(xs, ys, nil)

The set Q consists of the following terms:

   isEmpty(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)

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:

   INC(s(x)) -> INC(x)

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

   isEmpty(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)

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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)


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

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   INC(s(x)) -> INC(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:
*INC(s(x)) -> INC(x)
The graph contains the following edges 1 > 1


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

(13)
YES

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

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P(s(s(x))) -> P(s(x))

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   isZero(0) -> true
   isZero(s(x)) -> false
   head(cons(x, xs)) -> x
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   addLists(xs, ys, zs) -> if(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   if(true, true, b, xs, ys, xs2, ys2, zs, zs2) -> zs
   if(true, false, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, true, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs2, ys2, zs)
   if(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs, ys, zs2)
   addList(xs, ys) -> addLists(xs, ys, nil)

The set Q consists of the following terms:

   isEmpty(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)

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:

   P(s(s(x))) -> P(s(x))

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

   isEmpty(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)

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].

   isEmpty(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)


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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   P(s(s(x))) -> P(s(x))

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

(19) 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:
*P(s(s(x))) -> P(s(x))
The graph contains the following edges 1 > 1


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

(20)
YES

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

(21)
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(cons(x, xs)) -> false
   isEmpty(nil) -> true
   isZero(0) -> true
   isZero(s(x)) -> false
   head(cons(x, xs)) -> x
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   addLists(xs, ys, zs) -> if(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   if(true, true, b, xs, ys, xs2, ys2, zs, zs2) -> zs
   if(true, false, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, true, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs2, ys2, zs)
   if(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs, ys, zs2)
   addList(xs, ys) -> addLists(xs, ys, nil)

The set Q consists of the following terms:

   isEmpty(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)

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

(22) 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.
----------------------------------------

(23)
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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)

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

(24) 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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)


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

(25)
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.
----------------------------------------

(26) 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


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

(27)
YES

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

(28)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ADDLISTS(xs, ys, zs) -> IF(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   isZero(0) -> true
   isZero(s(x)) -> false
   head(cons(x, xs)) -> x
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   addLists(xs, ys, zs) -> if(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   if(true, true, b, xs, ys, xs2, ys2, zs, zs2) -> zs
   if(true, false, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, true, b, xs, ys, xs2, ys2, zs, zs2) -> differentLengthError
   if(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs2, ys2, zs)
   if(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> addLists(xs, ys, zs2)
   addList(xs, ys) -> addLists(xs, ys, nil)

The set Q consists of the following terms:

   isEmpty(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)

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

(29) 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.
----------------------------------------

(30)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ADDLISTS(xs, ys, zs) -> IF(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)
   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)

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

(31) 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].

   addLists(x0, x1, x2)
   if(true, true, x0, x1, x2, x3, x4, x5, x6)
   if(true, false, x0, x1, x2, x3, x4, x5, x6)
   if(false, true, x0, x1, x2, x3, x4, x5, x6)
   if(false, false, false, x0, x1, x2, x3, x4, x5)
   if(false, false, true, x0, x1, x2, x3, x4, x5)
   addList(x0, x1)


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

(32)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ADDLISTS(xs, ys, zs) -> IF(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys)))
   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(33) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ADDLISTS(xs, ys, zs) -> IF(isEmpty(xs), isEmpty(ys), isZero(head(xs)), tail(xs), tail(ys), cons(p(head(xs)), tail(xs)), cons(inc(head(ys)), tail(ys)), zs, append(zs, head(ys))) at position [0] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(head(cons(x0, x1))), tail(cons(x0, x1)), tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))),ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(head(cons(x0, x1))), tail(cons(x0, x1)), tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))))
   (ADDLISTS(nil, y1, y2) -> IF(true, isEmpty(y1), isZero(head(nil)), tail(nil), tail(y1), cons(p(head(nil)), tail(nil)), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))),ADDLISTS(nil, y1, y2) -> IF(true, isEmpty(y1), isZero(head(nil)), tail(nil), tail(y1), cons(p(head(nil)), tail(nil)), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))))


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

(34)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(head(cons(x0, x1))), tail(cons(x0, x1)), tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1)))
   ADDLISTS(nil, y1, y2) -> IF(true, isEmpty(y1), isZero(head(nil)), tail(nil), tail(y1), cons(p(head(nil)), tail(nil)), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1)))

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

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

(36)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(head(cons(x0, x1))), tail(cons(x0, x1)), tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1)))
   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(37) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(head(cons(x0, x1))), tail(cons(x0, x1)), tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))) at position [2,0] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), tail(cons(x0, x1)), tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))),ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), tail(cons(x0, x1)), tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))))


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

(38)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), tail(cons(x0, x1)), tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1)))

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(39) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), tail(cons(x0, x1)), tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))) at position [3] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))),ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))))


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

(40)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1)))

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(41) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(head(cons(x0, x1))), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))) at position [5,0,0] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(x0), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))),ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(x0), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))))


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

(42)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(x0), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1)))

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(43) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(x0), tail(cons(x0, x1))), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))) at position [5,1] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(x0), x1), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))),ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(x0), x1), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))))


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

(44)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(x0), x1), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1)))

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(45) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ADDLISTS(cons(x0, x1), y1, y2) -> IF(false, isEmpty(y1), isZero(x0), x1, tail(y1), cons(p(x0), x1), cons(inc(head(y1)), tail(y1)), y2, append(y2, head(y1))) at position [1] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, tail(cons(x0, x1)), cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1)))),ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, tail(cons(x0, x1)), cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1)))))
   (ADDLISTS(cons(y0, y1), nil, y3) -> IF(false, true, isZero(y0), y1, tail(nil), cons(p(y0), y1), cons(inc(head(nil)), tail(nil)), y3, append(y3, head(nil))),ADDLISTS(cons(y0, y1), nil, y3) -> IF(false, true, isZero(y0), y1, tail(nil), cons(p(y0), y1), cons(inc(head(nil)), tail(nil)), y3, append(y3, head(nil))))


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

(46)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, tail(cons(x0, x1)), cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1))))
   ADDLISTS(cons(y0, y1), nil, y3) -> IF(false, true, isZero(y0), y1, tail(nil), cons(p(y0), y1), cons(inc(head(nil)), tail(nil)), y3, append(y3, head(nil)))

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

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

(48)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, tail(cons(x0, x1)), cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1))))
   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)

The TRS R consists of the following rules:

   isEmpty(cons(x, xs)) -> false
   isEmpty(nil) -> true
   head(cons(x, xs)) -> x
   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   tail(nil) -> nil
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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:

   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, tail(cons(x0, x1)), cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1))))
   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   head(cons(x, xs)) -> x
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   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(cons(x0, x1))
   isEmpty(nil)
   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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].

   isEmpty(cons(x0, x1))
   isEmpty(nil)


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

(52)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, tail(cons(x0, x1)), cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1))))
   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   head(cons(x, xs)) -> x
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(53) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, tail(cons(x0, x1)), cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1)))) at position [4] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1)))),ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1)))))


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

(54)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   head(cons(x, xs)) -> x
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(55) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(head(cons(x0, x1))), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1)))) at position [6,0,0] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1)))),ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1)))))


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

(56)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   head(cons(x, xs)) -> x
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(57) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), tail(cons(x0, x1))), y3, append(y3, head(cons(x0, x1)))) at position [6,1] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, head(cons(x0, x1)))),ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, head(cons(x0, x1)))))


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

(58)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, head(cons(x0, x1))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   tail(cons(x, xs)) -> xs
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   head(cons(x, xs)) -> x
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, head(cons(x0, x1))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   head(cons(x, xs)) -> 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:

   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   tail(cons(x0, x1))
   tail(nil)
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(61) 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].

   tail(cons(x0, x1))
   tail(nil)


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

(62)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, head(cons(x0, x1))))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   head(cons(x, xs)) -> 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:

   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(63) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, head(cons(x0, x1)))) at position [8,1] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, x0)),ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, x0)))


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

(64)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, x0))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   head(cons(x, xs)) -> 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:

   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(65) 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.
----------------------------------------

(66)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, x0))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   head(cons(x0, x1))
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(67) 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].

   head(cons(x0, x1))


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

(68)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, x0))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(69) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule ADDLISTS(cons(y0, y1), cons(x0, x1), y3) -> IF(false, false, isZero(y0), y1, x1, cons(p(y0), y1), cons(inc(x0), x1), y3, append(y3, x0)) at position [2] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(p(0), y1), cons(inc(y2), y3), y4, append(y4, y2)),ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(p(0), y1), cons(inc(y2), y3), y4, append(y4, y2)))
   (ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2)),ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2)))


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

(70)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(p(0), y1), cons(inc(y2), y3), y4, append(y4, y2))
   ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2))

The TRS R consists of the following rules:

   isZero(0) -> true
   isZero(s(x)) -> false
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   p(0) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(71) 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.
----------------------------------------

(72)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(p(0), y1), cons(inc(y2), y3), y4, append(y4, y2))
   ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2))

The TRS R consists of the following rules:

   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(0) -> 0

The set Q consists of the following terms:

   isZero(0)
   isZero(s(x0))
   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(73) 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].

   isZero(0)
   isZero(s(x0))


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

(74)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(p(0), y1), cons(inc(y2), y3), y4, append(y4, y2))
   ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2))

The TRS R consists of the following rules:

   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(0) -> 0

The set Q consists of the following terms:

   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(75) TransformationProof (EQUIVALENT)
By rewriting [LPAR04] the rule ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(p(0), y1), cons(inc(y2), y3), y4, append(y4, y2)) at position [5,0] we obtained the following new rules [LPAR04]:

   (ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(0, y1), cons(inc(y2), y3), y4, append(y4, y2)),ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(0, y1), cons(inc(y2), y3), y4, append(y4, y2)))


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

(76)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2))
   ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(0, y1), cons(inc(y2), y3), y4, append(y4, y2))

The TRS R consists of the following rules:

   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0
   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(0) -> 0

The set Q consists of the following terms:

   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(77) 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.
----------------------------------------

(78)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2)
   ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2))
   ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(0, y1), cons(inc(y2), y3), y4, append(y4, y2))

The TRS R consists of the following rules:

   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0

The set Q consists of the following terms:

   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

(79) TransformationProof (EQUIVALENT)
By instantiating [LPAR04] the rule IF(false, false, true, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs, ys, zs2) we obtained the following new rules [LPAR04]:

   (IF(false, false, true, z0, z2, cons(0, z0), cons(y_0, z2), z3, y_1) -> ADDLISTS(z0, z2, y_1),IF(false, false, true, z0, z2, cons(0, z0), cons(y_0, z2), z3, y_1) -> ADDLISTS(z0, z2, y_1))


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

(80)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2))
   ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(0, y1), cons(inc(y2), y3), y4, append(y4, y2))
   IF(false, false, true, z0, z2, cons(0, z0), cons(y_0, z2), z3, y_1) -> ADDLISTS(z0, z2, y_1)

The TRS R consists of the following rules:

   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0

The set Q consists of the following terms:

   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

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


The following pairs can be oriented strictly and are deleted.

   IF(false, false, true, z0, z2, cons(0, z0), cons(y_0, z2), z3, y_1) -> ADDLISTS(z0, z2, y_1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Combined order from the following AFS and order.
IF(x1, x2, x3, x4, x5, x6, x7, x8, x9)  =  x7

ADDLISTS(x1, x2, x3)  =  x2

cons(x1, x2)  =  cons(x2)


Knuth-Bendix order [KBO] with precedence:trivial

and weight map:

   dummyConstant=1
   cons_1=1

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


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

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

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)
   ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2))
   ADDLISTS(cons(0, y1), cons(y2, y3), y4) -> IF(false, false, true, y1, y3, cons(0, y1), cons(inc(y2), y3), y4, append(y4, y2))

The TRS R consists of the following rules:

   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0

The set Q consists of the following terms:

   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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

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

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

   ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2))
   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)

The TRS R consists of the following rules:

   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0

The set Q consists of the following terms:

   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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.

   ADDLISTS(cons(s(x0), y1), cons(y2, y3), y4) -> IF(false, false, false, y1, y3, cons(p(s(x0)), y1), cons(inc(y2), y3), y4, append(y4, y2))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( IF_9(x_1, ..., x_9) ) = 2x_1 + 2x_2 + 2x_3 + 2x_6 + 2x_8
POL( cons_2(x_1, x_2) ) = max{0, 2x_1 + 2x_2 - 1}
POL( p_1(x_1) ) = max{0, x_1 - 2}
POL( s_1(x_1) ) = 2x_1 + 1
POL( 0 ) = 0
POL( inc_1(x_1) ) = 0
POL( append_2(x_1, x_2) ) = 0
POL( nil ) = 0
POL( ADDLISTS_3(x_1, ..., x_3) ) = 2x_1 + 2x_3
POL( false ) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0


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

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

   IF(false, false, false, xs, ys, xs2, ys2, zs, zs2) -> ADDLISTS(xs2, ys2, zs)

The TRS R consists of the following rules:

   inc(s(x)) -> s(inc(x))
   inc(0) -> s(0)
   append(nil, x) -> cons(x, nil)
   append(cons(y, ys), x) -> cons(y, append(ys, x))
   p(s(s(x))) -> s(p(s(x)))
   p(s(0)) -> 0

The set Q consists of the following terms:

   append(nil, x0)
   append(cons(x0, x1), x2)
   p(s(s(x0)))
   p(s(0))
   p(0)
   inc(s(x0))
   inc(0)

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.
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(88)
TRUE