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


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NO
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 disproven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) NonTerminationLoopProof [COMPLETE, 5448 ms]
        (17) NO


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(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)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   id_inc(x) -> x
   id_inc(x) -> s(x)
   div(x, y) -> if(ge(y, s(0)), ge(x, y), x, y)
   if(false, b, x, y) -> div_by_zero
   if(true, false, x, y) -> 0
   if(true, true, x, y) -> id_inc(div(minus(x, y), y))

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   GE(s(x), s(y)) -> GE(x, y)
   MINUS(s(x), s(y)) -> MINUS(x, y)
   DIV(x, y) -> IF(ge(y, s(0)), ge(x, y), x, y)
   DIV(x, y) -> GE(y, s(0))
   DIV(x, y) -> GE(x, y)
   IF(true, true, x, y) -> ID_INC(div(minus(x, y), y))
   IF(true, true, x, y) -> DIV(minus(x, y), y)
   IF(true, true, x, y) -> MINUS(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)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   id_inc(x) -> x
   id_inc(x) -> s(x)
   div(x, y) -> if(ge(y, s(0)), ge(x, y), x, y)
   if(false, b, x, y) -> div_by_zero
   if(true, false, x, y) -> 0
   if(true, true, x, y) -> id_inc(div(minus(x, y), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS(s(x), s(y)) -> MINUS(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)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   id_inc(x) -> x
   id_inc(x) -> s(x)
   div(x, y) -> if(ge(y, s(0)), ge(x, y), x, y)
   if(false, b, x, y) -> div_by_zero
   if(true, false, x, y) -> 0
   if(true, true, x, y) -> id_inc(div(minus(x, y), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

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

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

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

(8) 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:
*MINUS(s(x), s(y)) -> MINUS(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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(9)
YES

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(10)
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)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   id_inc(x) -> x
   id_inc(x) -> s(x)
   div(x, y) -> if(ge(y, s(0)), ge(x, y), x, y)
   if(false, b, x, y) -> div_by_zero
   if(true, false, x, y) -> 0
   if(true, true, x, y) -> id_inc(div(minus(x, y), y))

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

(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(12)
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.
----------------------------------------

(13) 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


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(14)
YES

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(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DIV(x, y) -> IF(ge(y, s(0)), ge(x, y), x, y)
   IF(true, true, x, y) -> DIV(minus(x, y), 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)
   minus(x, 0) -> x
   minus(0, y) -> 0
   minus(s(x), s(y)) -> minus(x, y)
   id_inc(x) -> x
   id_inc(x) -> s(x)
   div(x, y) -> if(ge(y, s(0)), ge(x, y), x, y)
   if(false, b, x, y) -> div_by_zero
   if(true, false, x, y) -> 0
   if(true, true, x, y) -> id_inc(div(minus(x, y), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(16) NonTerminationLoopProof (COMPLETE)
We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = IF(ge(id_inc(x), s(0)), ge(x'', id_inc(0)), x', y') evaluates to  t =IF(ge(y', s(0)), ge(minus(x', y'), y'), minus(x', y'), y')

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
* Matcher: [x / 0, x'' / minus(x', id_inc(0)), x' / minus(x', id_inc(0))]
* Semiunifier: [y' / id_inc(0)]

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Rewriting sequence

IF(ge(id_inc(x), s(0)), ge(x'', id_inc(0)), x', id_inc(0)) -> IF(ge(id_inc(x), s(0)), ge(x'', 0), x', id_inc(0))
with rule id_inc(x1) -> x1 at position [1,1] and matcher [x1 / 0]

IF(ge(id_inc(x), s(0)), ge(x'', 0), x', id_inc(0)) -> IF(ge(id_inc(x), s(0)), true, x', id_inc(0))
with rule ge(x''', 0) -> true at position [1] and matcher [x''' / x'']

IF(ge(id_inc(x), s(0)), true, x', id_inc(0)) -> IF(ge(s(x), s(0)), true, x', id_inc(0))
with rule id_inc(x'') -> s(x'') at position [0,0] and matcher [x'' / x]

IF(ge(s(x), s(0)), true, x', id_inc(0)) -> IF(ge(x, 0), true, x', id_inc(0))
with rule ge(s(x''), s(y)) -> ge(x'', y) at position [0] and matcher [x'' / x, y / 0]

IF(ge(x, 0), true, x', id_inc(0)) -> IF(true, true, x', id_inc(0))
with rule ge(x'', 0) -> true at position [0] and matcher [x'' / x]

IF(true, true, x', id_inc(0)) -> DIV(minus(x', id_inc(0)), id_inc(0))
with rule IF(true, true, x'', y') -> DIV(minus(x'', y'), y') at position [] and matcher [x'' / x', y' / id_inc(0)]

DIV(minus(x', id_inc(0)), id_inc(0)) -> IF(ge(id_inc(0), s(0)), ge(minus(x', id_inc(0)), id_inc(0)), minus(x', id_inc(0)), id_inc(0))
with rule DIV(x, y) -> IF(ge(y, s(0)), ge(x, y), x, y)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.




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(17)
NO