/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 15 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) NonInfProof [EQUIVALENT, 23 ms]
        (27) AND
            (28) QDP
                (29) DependencyGraphProof [EQUIVALENT, 0 ms]
                (30) TRUE
            (31) QDP
                (32) DependencyGraphProof [EQUIVALENT, 0 ms]
                (33) TRUE
    (34) QDP
        (35) UsableRulesProof [EQUIVALENT, 0 ms]
        (36) QDP
        (37) QReductionProof [EQUIVALENT, 0 ms]
        (38) QDP
        (39) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (40) YES
    (41) QDP
        (42) UsableRulesProof [EQUIVALENT, 0 ms]
        (43) QDP
        (44) QReductionProof [EQUIVALENT, 0 ms]
        (45) QDP
        (46) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (47) YES


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

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

   app(x, y) -> helpa(0, plus(length(x), length(y)), x, y)
   plus(x, 0) -> x
   plus(x, s(y)) -> s(plus(x, y))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))
   helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if(true, c, l, ys, zs) -> nil
   if(false, c, l, ys, zs) -> helpb(c, l, ys, zs)
   take(0, cons(x, xs), ys) -> x
   take(0, nil, cons(y, ys)) -> y
   take(s(c), cons(x, xs), ys) -> take(c, xs, ys)
   take(s(c), nil, cons(y, ys)) -> take(c, nil, ys)
   helpb(c, l, ys, zs) -> cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

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:

   app(x, y) -> helpa(0, plus(length(x), length(y)), x, y)
   plus(x, 0) -> x
   plus(x, s(y)) -> s(plus(x, y))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))
   helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if(true, c, l, ys, zs) -> nil
   if(false, c, l, ys, zs) -> helpb(c, l, ys, zs)
   take(0, cons(x, xs), ys) -> x
   take(0, nil, cons(y, ys)) -> y
   take(s(c), cons(x, xs), ys) -> take(c, xs, ys)
   take(s(c), nil, cons(y, ys)) -> take(c, nil, ys)
   helpb(c, l, ys, zs) -> cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)


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

(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:

   APP(x, y) -> HELPA(0, plus(length(x), length(y)), x, y)
   APP(x, y) -> PLUS(length(x), length(y))
   APP(x, y) -> LENGTH(x)
   APP(x, y) -> LENGTH(y)
   PLUS(x, s(y)) -> PLUS(x, y)
   LENGTH(cons(x, y)) -> LENGTH(y)
   HELPA(c, l, ys, zs) -> IF(ge(c, l), c, l, ys, zs)
   HELPA(c, l, ys, zs) -> GE(c, l)
   GE(s(x), s(y)) -> GE(x, y)
   IF(false, c, l, ys, zs) -> HELPB(c, l, ys, zs)
   TAKE(s(c), cons(x, xs), ys) -> TAKE(c, xs, ys)
   TAKE(s(c), nil, cons(y, ys)) -> TAKE(c, nil, ys)
   HELPB(c, l, ys, zs) -> TAKE(c, ys, zs)
   HELPB(c, l, ys, zs) -> HELPA(s(c), l, ys, zs)

The TRS R consists of the following rules:

   app(x, y) -> helpa(0, plus(length(x), length(y)), x, y)
   plus(x, 0) -> x
   plus(x, s(y)) -> s(plus(x, y))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))
   helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if(true, c, l, ys, zs) -> nil
   if(false, c, l, ys, zs) -> helpb(c, l, ys, zs)
   take(0, cons(x, xs), ys) -> x
   take(0, nil, cons(y, ys)) -> y
   take(s(c), cons(x, xs), ys) -> take(c, xs, ys)
   take(s(c), nil, cons(y, ys)) -> take(c, nil, ys)
   helpb(c, l, ys, zs) -> cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 7 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

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

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

   TAKE(s(c), nil, cons(y, ys)) -> TAKE(c, nil, ys)

The TRS R consists of the following rules:

   app(x, y) -> helpa(0, plus(length(x), length(y)), x, y)
   plus(x, 0) -> x
   plus(x, s(y)) -> s(plus(x, y))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))
   helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if(true, c, l, ys, zs) -> nil
   if(false, c, l, ys, zs) -> helpb(c, l, ys, zs)
   take(0, cons(x, xs), ys) -> x
   take(0, nil, cons(y, ys)) -> y
   take(s(c), cons(x, xs), ys) -> take(c, xs, ys)
   take(s(c), nil, cons(y, ys)) -> take(c, nil, ys)
   helpb(c, l, ys, zs) -> cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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:

   TAKE(s(c), nil, cons(y, ys)) -> TAKE(c, nil, ys)

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

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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].

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)


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

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

   TAKE(s(c), nil, cons(y, ys)) -> TAKE(c, nil, ys)

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:
*TAKE(s(c), nil, cons(y, ys)) -> TAKE(c, nil, ys)
The graph contains the following edges 1 > 1, 2 >= 2, 3 > 3


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

(13)
YES

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

(14)
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:

   app(x, y) -> helpa(0, plus(length(x), length(y)), x, y)
   plus(x, 0) -> x
   plus(x, s(y)) -> s(plus(x, y))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))
   helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if(true, c, l, ys, zs) -> nil
   if(false, c, l, ys, zs) -> helpb(c, l, ys, zs)
   take(0, cons(x, xs), ys) -> x
   take(0, nil, cons(y, ys)) -> y
   take(s(c), cons(x, xs), ys) -> take(c, xs, ys)
   take(s(c), nil, cons(y, ys)) -> take(c, nil, ys)
   helpb(c, l, ys, zs) -> cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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:

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

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

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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].

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)


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

(18)
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.
----------------------------------------

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

   IF(false, c, l, ys, zs) -> HELPB(c, l, ys, zs)
   HELPB(c, l, ys, zs) -> HELPA(s(c), l, ys, zs)
   HELPA(c, l, ys, zs) -> IF(ge(c, l), c, l, ys, zs)

The TRS R consists of the following rules:

   app(x, y) -> helpa(0, plus(length(x), length(y)), x, y)
   plus(x, 0) -> x
   plus(x, s(y)) -> s(plus(x, y))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))
   helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if(true, c, l, ys, zs) -> nil
   if(false, c, l, ys, zs) -> helpb(c, l, ys, zs)
   take(0, cons(x, xs), ys) -> x
   take(0, nil, cons(y, ys)) -> y
   take(s(c), cons(x, xs), ys) -> take(c, xs, ys)
   take(s(c), nil, cons(y, ys)) -> take(c, nil, ys)
   helpb(c, l, ys, zs) -> cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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:

   IF(false, c, l, ys, zs) -> HELPB(c, l, ys, zs)
   HELPB(c, l, ys, zs) -> HELPA(s(c), l, ys, zs)
   HELPA(c, l, ys, zs) -> IF(ge(c, l), c, l, ys, zs)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)

The set Q consists of the following terms:

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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].

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)


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

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

   IF(false, c, l, ys, zs) -> HELPB(c, l, ys, zs)
   HELPB(c, l, ys, zs) -> HELPA(s(c), l, ys, zs)
   HELPA(c, l, ys, zs) -> IF(ge(c, l), c, l, ys, zs)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(x)) -> 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))

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

(26) 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 IF(false, c, l, ys, zs) -> HELPB(c, l, ys, zs) the following chains were created:
*We consider the chain HELPA(x8, x9, x10, x11) -> IF(ge(x8, x9), x8, x9, x10, x11), IF(false, x12, x13, x14, x15) -> HELPB(x12, x13, x14, x15) which results in the following constraint:

(1)    (IF(ge(x8, x9), x8, x9, x10, x11)=IF(false, x12, x13, x14, x15)  ==>  IF(false, x12, x13, x14, x15)_>=_HELPB(x12, x13, x14, x15))



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

(2)    (ge(x8, x9)=false  ==>  IF(false, x8, x9, x10, x11)_>=_HELPB(x8, x9, x10, x11))



We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on ge(x8, x9)=false which results in the following new constraints:

(3)    (false=false  ==>  IF(false, 0, s(x49), x10, x11)_>=_HELPB(0, s(x49), x10, x11))

(4)    (ge(x51, x50)=false & (\/x52,x53:ge(x51, x50)=false  ==>  IF(false, x51, x50, x52, x53)_>=_HELPB(x51, x50, x52, x53))  ==>  IF(false, s(x51), s(x50), x10, x11)_>=_HELPB(s(x51), s(x50), x10, x11))



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

(5)    (IF(false, 0, s(x49), x10, x11)_>=_HELPB(0, s(x49), x10, x11))



We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (\/x52,x53:ge(x51, x50)=false  ==>  IF(false, x51, x50, x52, x53)_>=_HELPB(x51, x50, x52, x53)) with sigma = [x52 / x10, x53 / x11] which results in the following new constraint:

(6)    (IF(false, x51, x50, x10, x11)_>=_HELPB(x51, x50, x10, x11)  ==>  IF(false, s(x51), s(x50), x10, x11)_>=_HELPB(s(x51), s(x50), x10, x11))








For Pair HELPB(c, l, ys, zs) -> HELPA(s(c), l, ys, zs) the following chains were created:
*We consider the chain IF(false, x16, x17, x18, x19) -> HELPB(x16, x17, x18, x19), HELPB(x20, x21, x22, x23) -> HELPA(s(x20), x21, x22, x23) which results in the following constraint:

(1)    (HELPB(x16, x17, x18, x19)=HELPB(x20, x21, x22, x23)  ==>  HELPB(x20, x21, x22, x23)_>=_HELPA(s(x20), x21, x22, x23))



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

(2)    (HELPB(x16, x17, x18, x19)_>=_HELPA(s(x16), x17, x18, x19))








For Pair HELPA(c, l, ys, zs) -> IF(ge(c, l), c, l, ys, zs) the following chains were created:
*We consider the chain HELPB(x36, x37, x38, x39) -> HELPA(s(x36), x37, x38, x39), HELPA(x40, x41, x42, x43) -> IF(ge(x40, x41), x40, x41, x42, x43) which results in the following constraint:

(1)    (HELPA(s(x36), x37, x38, x39)=HELPA(x40, x41, x42, x43)  ==>  HELPA(x40, x41, x42, x43)_>=_IF(ge(x40, x41), x40, x41, x42, x43))



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

(2)    (HELPA(s(x36), x37, x38, x39)_>=_IF(ge(s(x36), x37), s(x36), x37, x38, x39))








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

*IF(false, c, l, ys, zs) -> HELPB(c, l, ys, zs)

*(IF(false, 0, s(x49), x10, x11)_>=_HELPB(0, s(x49), x10, x11))


*(IF(false, x51, x50, x10, x11)_>=_HELPB(x51, x50, x10, x11)  ==>  IF(false, s(x51), s(x50), x10, x11)_>=_HELPB(s(x51), s(x50), x10, x11))




*HELPB(c, l, ys, zs) -> HELPA(s(c), l, ys, zs)

*(HELPB(x16, x17, x18, x19)_>=_HELPA(s(x16), x17, x18, x19))




*HELPA(c, l, ys, zs) -> IF(ge(c, l), c, l, ys, zs)

*(HELPA(s(x36), x37, x38, x39)_>=_IF(ge(s(x36), x37), s(x36), x37, x38, x39))








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(HELPA(x_1, x_2, x_3, x_4)) = -1 - x_1 + x_2 + x_3 + x_4
   POL(HELPB(x_1, x_2, x_3, x_4)) = -1 - x_1 + x_2 + x_3 + x_4
   POL(IF(x_1, x_2, x_3, x_4, x_5)) = -1 - x_1 - x_2 + x_3 + x_4 + x_5
   POL(c) = -2
   POL(false) = 0
   POL(ge(x_1, x_2)) = 0
   POL(s(x_1)) = 1 + x_1
   POL(true) = 0


The following pairs  are in P_>:
   HELPB(c, l, ys, zs) -> HELPA(s(c), l, ys, zs)
The following pairs are in P_bound:
   IF(false, c, l, ys, zs) -> HELPB(c, l, ys, zs)
The following rules are usable:
   true -> ge(x, 0)
   false -> ge(0, s(x))
   ge(x, y) -> ge(s(x), s(y))

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

(27)
Complex Obligation (AND)

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

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

   IF(false, c, l, ys, zs) -> HELPB(c, l, ys, zs)
   HELPA(c, l, ys, zs) -> IF(ge(c, l), c, l, ys, zs)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(x)) -> 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))

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

(29) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(30)
TRUE

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

(31)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   HELPB(c, l, ys, zs) -> HELPA(s(c), l, ys, zs)
   HELPA(c, l, ys, zs) -> IF(ge(c, l), c, l, ys, zs)

The TRS R consists of the following rules:

   ge(x, 0) -> true
   ge(0, s(x)) -> 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))

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

(32) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(33)
TRUE

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

(34)
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:

   app(x, y) -> helpa(0, plus(length(x), length(y)), x, y)
   plus(x, 0) -> x
   plus(x, s(y)) -> s(plus(x, y))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))
   helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if(true, c, l, ys, zs) -> nil
   if(false, c, l, ys, zs) -> helpb(c, l, ys, zs)
   take(0, cons(x, xs), ys) -> x
   take(0, nil, cons(y, ys)) -> y
   take(s(c), cons(x, xs), ys) -> take(c, xs, ys)
   take(s(c), nil, cons(y, ys)) -> take(c, nil, ys)
   helpb(c, l, ys, zs) -> cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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

(35) 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.
----------------------------------------

(36)
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:

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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

(37) 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].

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)


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

(38)
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.
----------------------------------------

(39) 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


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

(40)
YES

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

(41)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

   app(x, y) -> helpa(0, plus(length(x), length(y)), x, y)
   plus(x, 0) -> x
   plus(x, s(y)) -> s(plus(x, y))
   length(nil) -> 0
   length(cons(x, y)) -> s(length(y))
   helpa(c, l, ys, zs) -> if(ge(c, l), c, l, ys, zs)
   ge(x, 0) -> true
   ge(0, s(x)) -> false
   ge(s(x), s(y)) -> ge(x, y)
   if(true, c, l, ys, zs) -> nil
   if(false, c, l, ys, zs) -> helpb(c, l, ys, zs)
   take(0, cons(x, xs), ys) -> x
   take(0, nil, cons(y, ys)) -> y
   take(s(c), cons(x, xs), ys) -> take(c, xs, ys)
   take(s(c), nil, cons(y, ys)) -> take(c, nil, ys)
   helpb(c, l, ys, zs) -> cons(take(c, ys, zs), helpa(s(c), l, ys, zs))

The set Q consists of the following terms:

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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

(42) 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.
----------------------------------------

(43)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)

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

(44) 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].

   app(x0, x1)
   plus(x0, 0)
   plus(x0, s(x1))
   length(nil)
   length(cons(x0, x1))
   helpa(x0, x1, x2, x3)
   ge(x0, 0)
   ge(0, s(x0))
   ge(s(x0), s(x1))
   if(true, x0, x1, x2, x3)
   if(false, x0, x1, x2, x3)
   take(0, cons(x0, xs), x1)
   take(0, nil, cons(x0, x1))
   take(s(x0), cons(x1, xs), x2)
   take(s(x0), nil, cons(x1, x2))
   helpb(x0, x1, x2, x3)


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

(45)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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

(46) 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:
*PLUS(x, s(y)) -> PLUS(x, y)
The graph contains the following edges 1 >= 1, 2 > 2


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

(47)
YES