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


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 1 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 1 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, 1 ms]
        (32) QDP
        (33) NonInfProof [EQUIVALENT, 52 ms]
        (34) QDP
        (35) DependencyGraphProof [EQUIVALENT, 0 ms]
        (36) TRUE


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

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   rev(x) -> if(x, eq(0, length(x)), nil, 0, length(x))
   if(x, true, z, c, l) -> z
   if(x, false, z, c, l) -> help(s(c), l, x, z)
   help(c, l, cons(x, y), z) -> if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))

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:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   rev(x) -> if(x, eq(0, length(x)), nil, 0, length(x))
   if(x, true, z, c, l) -> z
   if(x, false, z, c, l) -> help(s(c), l, x, z)
   help(c, l, cons(x, y), z) -> if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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:

   GE(s(x), s(y)) -> GE(x, y)
   REV(x) -> IF(x, eq(0, length(x)), nil, 0, length(x))
   REV(x) -> LENGTH(x)
   IF(x, false, z, c, l) -> HELP(s(c), l, x, z)
   HELP(c, l, cons(x, y), z) -> IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   HELP(c, l, cons(x, y), z) -> APPEND(y, cons(x, nil))
   HELP(c, l, cons(x, y), z) -> GE(c, l)
   APPEND(cons(x, y), z) -> APPEND(y, z)
   LENGTH(cons(x, y)) -> LENGTH(y)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   rev(x) -> if(x, eq(0, length(x)), nil, 0, length(x))
   if(x, true, z, c, l) -> z
   if(x, false, z, c, l) -> help(s(c), l, x, z)
   help(c, l, cons(x, y), z) -> if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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 4 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   LENGTH(cons(x, y)) -> LENGTH(y)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   rev(x) -> if(x, eq(0, length(x)), nil, 0, length(x))
   if(x, true, z, c, l) -> z
   if(x, false, z, c, l) -> help(s(c), l, x, z)
   help(c, l, cons(x, y), z) -> if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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:

   LENGTH(cons(x, y)) -> LENGTH(y)

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

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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].

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(x0, x1))


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

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

   LENGTH(cons(x, y)) -> LENGTH(y)

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:
*LENGTH(cons(x, y)) -> LENGTH(y)
The graph contains the following edges 1 > 1


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

(13)
YES

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

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

   APPEND(cons(x, y), z) -> APPEND(y, z)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   rev(x) -> if(x, eq(0, length(x)), nil, 0, length(x))
   if(x, true, z, c, l) -> z
   if(x, false, z, c, l) -> help(s(c), l, x, z)
   help(c, l, cons(x, y), z) -> if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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:

   APPEND(cons(x, y), z) -> APPEND(y, z)

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

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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].

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(x0, x1))


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

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

   APPEND(cons(x, y), z) -> APPEND(y, z)

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:
*APPEND(cons(x, y), z) -> APPEND(y, z)
The graph contains the following edges 1 > 1, 2 >= 2


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

(20)
YES

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

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(s(x), s(y)) -> GE(x, y)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   rev(x) -> if(x, eq(0, length(x)), nil, 0, length(x))
   if(x, true, z, c, l) -> z
   if(x, false, z, c, l) -> help(s(c), l, x, z)
   help(c, l, cons(x, y), z) -> if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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:

   GE(s(x), s(y)) -> GE(x, y)

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

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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].

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(x0, x1))


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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(s(x), s(y)) -> GE(x, y)

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

(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:
*GE(s(x), s(y)) -> GE(x, y)
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:

   HELP(c, l, cons(x, y), z) -> IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   IF(x, false, z, c, l) -> HELP(s(c), l, x, z)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   rev(x) -> if(x, eq(0, length(x)), nil, 0, length(x))
   if(x, true, z, c, l) -> z
   if(x, false, z, c, l) -> help(s(c), l, x, z)
   help(c, l, cons(x, y), z) -> if(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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:

   HELP(c, l, cons(x, y), z) -> IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   IF(x, false, z, c, l) -> HELP(s(c), l, x, z)

The TRS R consists of the following rules:

   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   append(nil, x0)
   append(cons(x0, x1), x2)
   length(nil)
   length(cons(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].

   rev(x0)
   if(x0, true, x1, x2, x3)
   if(x0, false, x1, x2, x3)
   help(x0, x1, cons(x2, x3), x4)
   length(nil)
   length(cons(x0, x1))


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

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

   HELP(c, l, cons(x, y), z) -> IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)
   IF(x, false, z, c, l) -> HELP(s(c), l, x, z)

The TRS R consists of the following rules:

   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, x0)
   append(cons(x0, x1), x2)

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

(33) NonInfProof (EQUIVALENT)
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:

Note that final constraints are written in bold face.



For Pair HELP(c, l, cons(x, y), z) -> IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l) the following chains were created:
*We consider the chain IF(x5, false, x6, x7, x8) -> HELP(s(x7), x8, x5, x6), HELP(x9, x10, cons(x11, x12), x13) -> IF(append(x12, cons(x11, nil)), ge(x9, x10), cons(x11, x13), x9, x10) which results in the following constraint:

(1)    (HELP(s(x7), x8, x5, x6)=HELP(x9, x10, cons(x11, x12), x13)  ==>  HELP(x9, x10, cons(x11, x12), x13)_>=_IF(append(x12, cons(x11, nil)), ge(x9, x10), cons(x11, x13), x9, x10))



We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:

(2)    (HELP(s(x7), x8, cons(x11, x12), x6)_>=_IF(append(x12, cons(x11, nil)), ge(s(x7), x8), cons(x11, x6), s(x7), x8))








For Pair IF(x, false, z, c, l) -> HELP(s(c), l, x, z) the following chains were created:
*We consider the chain HELP(x14, x15, cons(x16, x17), x18) -> IF(append(x17, cons(x16, nil)), ge(x14, x15), cons(x16, x18), x14, x15), IF(x19, false, x20, x21, x22) -> HELP(s(x21), x22, x19, x20) which results in the following constraint:

(1)    (IF(append(x17, cons(x16, nil)), ge(x14, x15), cons(x16, x18), x14, x15)=IF(x19, false, x20, x21, x22)  ==>  IF(x19, false, x20, x21, x22)_>=_HELP(s(x21), x22, x19, x20))



We simplified constraint (1) using rules (I), (II), (III), (VII) which results in the following new constraint:

(2)    (cons(x16, nil)=x27 & append(x17, x27)=x19 & ge(x14, x15)=false  ==>  IF(x19, false, cons(x16, x18), x14, x15)_>=_HELP(s(x14), x15, x19, cons(x16, x18)))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x14, x15)=false which results in the following new constraints:

(3)    (false=false & cons(x16, nil)=x27 & append(x17, x27)=x19  ==>  IF(x19, false, cons(x16, x18), 0, s(x29))_>=_HELP(s(0), s(x29), x19, cons(x16, x18)))

(4)    (ge(x31, x30)=false & cons(x16, nil)=x27 & append(x17, x27)=x19 & (\/x32,x33,x34,x35,x36:ge(x31, x30)=false & cons(x32, nil)=x33 & append(x34, x33)=x35  ==>  IF(x35, false, cons(x32, x36), x31, x30)_>=_HELP(s(x31), x30, x35, cons(x32, x36)))  ==>  IF(x19, false, cons(x16, x18), s(x31), s(x30))_>=_HELP(s(s(x31)), s(x30), x19, cons(x16, x18)))



We simplified constraint (3) using rules (I), (II), (III), (IV) which results in the following new constraint:

(5)    (IF(x19, false, cons(x16, x18), 0, s(x29))_>=_HELP(s(0), s(x29), x19, cons(x16, x18)))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (\/x32,x33,x34,x35,x36:ge(x31, x30)=false & cons(x32, nil)=x33 & append(x34, x33)=x35  ==>  IF(x35, false, cons(x32, x36), x31, x30)_>=_HELP(s(x31), x30, x35, cons(x32, x36))) with sigma = [x32 / x16, x33 / x27, x34 / x17, x35 / x19, x36 / x18] which results in the following new constraint:

(6)    (IF(x19, false, cons(x16, x18), x31, x30)_>=_HELP(s(x31), x30, x19, cons(x16, x18))  ==>  IF(x19, false, cons(x16, x18), s(x31), s(x30))_>=_HELP(s(s(x31)), s(x30), x19, cons(x16, x18)))








To summarize, we get the following constraints P__>=_ for the following pairs.

*HELP(c, l, cons(x, y), z) -> IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)

*(HELP(s(x7), x8, cons(x11, x12), x6)_>=_IF(append(x12, cons(x11, nil)), ge(s(x7), x8), cons(x11, x6), s(x7), x8))




*IF(x, false, z, c, l) -> HELP(s(c), l, x, z)

*(IF(x19, false, cons(x16, x18), x31, x30)_>=_HELP(s(x31), x30, x19, cons(x16, x18))  ==>  IF(x19, false, cons(x16, x18), s(x31), s(x30))_>=_HELP(s(s(x31)), s(x30), x19, cons(x16, x18)))


*(IF(x19, false, cons(x16, x18), 0, s(x29))_>=_HELP(s(0), s(x29), x19, cons(x16, x18)))








The constraints for P_> respective P_bound are constructed from P__>=_ where we just replace every occurence of "t _>=_ s" in P__>=_ by  "t > s" respective "t _>=_ c". Here c stands for the fresh constant used for P_bound. 

Using the following integer polynomial ordering the  resulting constraints can be solved 

Polynomial interpretation [NONINF]:

   POL(0) = 0
   POL(HELP(x_1, x_2, x_3, x_4)) = -1 - x_1 + x_2 - x_3
   POL(IF(x_1, x_2, x_3, x_4, x_5)) = -1 + x_1 - x_2 - x_3 - x_4 + x_5
   POL(append(x_1, x_2)) = x_2
   POL(c) = -2
   POL(cons(x_1, x_2)) = 0
   POL(false) = 0
   POL(ge(x_1, x_2)) = 0
   POL(nil) = 0
   POL(s(x_1)) = 1 + x_1
   POL(true) = 0


The following pairs  are in P_>:
   IF(x, false, z, c, l) -> HELP(s(c), l, x, z)
The following pairs are in P_bound:
   IF(x, false, z, c, l) -> HELP(s(c), l, x, z)
The following rules are usable:
   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   true -> ge(x, 0)
   false -> ge(0, s(y))
   ge(x, y) -> ge(s(x), s(y))

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

   HELP(c, l, cons(x, y), z) -> IF(append(y, cons(x, nil)), ge(c, l), cons(x, z), c, l)

The TRS R consists of the following rules:

   append(nil, y) -> y
   append(cons(x, y), z) -> cons(x, append(y, z))
   ge(x, 0) -> true
   ge(0, s(y)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   append(nil, x0)
   append(cons(x0, x1), x2)

We have to consider all minimal (P,Q,R)-chains.
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(35) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(36)
TRUE