/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files


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WORST_CASE(NON_POLY, ?)
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 110 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f2(A, B, C, D) -> Com_1(f2(1 + A, 1 + B, -(1) + C, D)) :|: TRUE
f1(A, B, C, D) -> Com_1(f2(A, B, C, D)) :|: A >= C && B >= A && C >= B
f1(A, B, C, D) -> Com_1(f300(A, B, C, E)) :|: C >= 1 + A && B >= A && C >= B
f1(A, B, C, D) -> Com_1(f300(A, B, C, E)) :|: B >= A && B >= C + 1
f1(A, B, C, D) -> Com_1(f300(A, B, C, E)) :|: A >= B + 1

The start-symbols are:[f1_4]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f1

      0: f2 -> f2 : A'=1+A, B'=1+B, C'=-1+C, [], cost: 1

      1: f1 -> f2 : [ A>=C && B>=A && C>=B ], cost: 1

      2: f1 -> f300 : D'=free, [ C>=1+A && B>=A && C>=B ], cost: 1

      3: f1 -> f300 : D'=free_1, [ B>=A && B>=1+C ], cost: 1

      4: f1 -> f300 : D'=free_2, [ A>=1+B ], cost: 1



Checking for constant complexity:

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

      1: f1 -> f2 : [ A>=C && B>=A && C>=B ], cost: 1



Removed unreachable and leaf rules:

   Start location: f1

      0: f2 -> f2 : A'=1+A, B'=1+B, C'=-1+C, [], cost: 1

      1: f1 -> f2 : [ A>=C && B>=A && C>=B ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: f2 -> f2 : A'=1+A, B'=1+B, C'=-1+C, [], cost: 1



   Accelerated rule 0 with NONTERM, yielding the new rule 5.

   Removing the simple loops: 0.



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

   Start location: f1

      5: f2 -> [3] : [], cost: NONTERM

      1: f1 -> f2 : [ A>=C && B>=A && C>=B ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f1

      1: f1 -> f2 : [ A>=C && B>=A && C>=B ], cost: 1

      6: f1 -> [3] : [ A>=C && B>=A && C>=B ], cost: NONTERM



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

   Start location: f1

      6: f1 -> [3] : [ A>=C && B>=A && C>=B ], cost: NONTERM



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f1

      6: f1 -> [3] : [ A>=C && B>=A && C>=B ], cost: NONTERM



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:  [ A>=C && B>=A && C>=B ]



NO


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(2)
BOUNDS(INF, INF)