/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files


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WORST_CASE(NON_POLY, ?)
proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(INF, INF).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 222 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f3(A) -> Com_1(f2(A)) :|: TRUE
f2(A) -> Com_1(f2(-(1) + A)) :|: A >= 2
f2(A) -> Com_1(f2(-(1) + A)) :|: 1 >= A

The start-symbols are:[f3_1]


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(1) Loat Proof (FINISHED)


### Pre-processing the ITS problem ###



Initial linear ITS problem

   Start location: f3

      0: f3 -> f2 : [], cost: 1

      1: f2 -> f2 : A'=-1+A, [ A>=2 ], cost: 1

      2: f2 -> f2 : A'=-1+A, [ 1>=A ], cost: 1



Checking for constant complexity:

   The following rule is satisfiable with cost >= 1, yielding constant complexity:

      0: f3 -> f2 : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: f2 -> f2 : A'=-1+A, [ A>=2 ], cost: 1

      2: f2 -> f2 : A'=-1+A, [ 1>=A ], cost: 1



   Accelerated rule 1 with metering function -1+A, yielding the new rule 3.

   Accelerated rule 2 with NONTERM, yielding the new rule 4.

   Removing the simple loops: 1 2.



Accelerated all simple loops using metering functions (where possible):

   Start location: f3

      0: f3 -> f2 : [], cost: 1

      3: f2 -> f2 : A'=1, [ A>=2 ], cost: -1+A

      4: f2 -> [2] : [ 1>=A ], cost: NONTERM



Chained accelerated rules (with incoming rules):

   Start location: f3

      0: f3 -> f2 : [], cost: 1

      5: f3 -> f2 : A'=1, [ A>=2 ], cost: A

      6: f3 -> [2] : [ 1>=A ], cost: NONTERM



Removed unreachable locations (and leaf rules with constant cost):

   Start location: f3

      5: f3 -> f2 : A'=1, [ A>=2 ], cost: A

      6: f3 -> [2] : [ 1>=A ], cost: NONTERM



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f3

      5: f3 -> f2 : A'=1, [ A>=2 ], cost: A

      6: f3 -> [2] : [ 1>=A ], cost: NONTERM



Computing asymptotic complexity for rule 5

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -1+A (+/+!), A (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {A==n}

      resulting limit problem:

      [solved]



   Solution:

      A / n

   Resulting cost n has complexity: Poly(n^1)



Found new complexity Poly(n^1).



Computing asymptotic complexity for rule 6

   Guard is satisfiable, yielding nontermination

   Resulting cost NONTERM has complexity: Nonterm



Found new complexity Nonterm.



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Nonterm

   Cpx degree:  Nonterm

   Solved cost: NONTERM

   Rule cost:   NONTERM

   Rule guard:  [ 1>=A ]



NO


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(2)
BOUNDS(INF, INF)