/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 423 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C, D, E, F, G) -> Com_1(f2(H, B, I, D, E, F, G)) :|: A >= B
f0(A, B, C, D, E, F, G) -> Com_1(f0(H, B, C, I, 0, F, G)) :|: B >= A + 1
f0(A, B, C, D, E, F, G) -> Com_1(f0(H, B, C, I, J, K, G)) :|: 0 >= J + 1 && B >= A + 1
f0(A, B, C, D, E, F, G) -> Com_1(f0(H, B, C, I, J, K, G)) :|: J >= 1 && B >= A + 1
f1(A, B, C, D, E, F, G) -> Com_1(f0(A, B, C, D, E, F, H)) :|: TRUE

The start-symbols are:[f1_7]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f1

      0: f0 -> f2 : A'=free, C'=free_1, [ A>=B ], cost: 1

      1: f0 -> f0 : A'=free_2, D'=free_3, E'=0, [ B>=1+A ], cost: 1

      2: f0 -> f0 : A'=free_6, D'=free_7, E'=free_4, F'=free_5, [ 0>=1+free_4 && B>=1+A ], cost: 1

      3: f0 -> f0 : A'=free_10, D'=free_11, E'=free_8, F'=free_9, [ free_8>=1 && B>=1+A ], cost: 1

      4: f1 -> f0 : G'=free_12, [], cost: 1



Checking for constant complexity:

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

      4: f1 -> f0 : G'=free_12, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f1

      1: f0 -> f0 : A'=free_2, D'=free_3, E'=0, [ B>=1+A ], cost: 1

      2: f0 -> f0 : A'=free_6, D'=free_7, E'=free_4, F'=free_5, [ 0>=1+free_4 && B>=1+A ], cost: 1

      3: f0 -> f0 : A'=free_10, D'=free_11, E'=free_8, F'=free_9, [ free_8>=1 && B>=1+A ], cost: 1

      4: f1 -> f0 : G'=free_12, [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      1: f0 -> f0 : A'=free_2, D'=free_3, E'=0, [ B>=1+A ], cost: 1

      2: f0 -> f0 : A'=free_6, D'=free_7, E'=free_4, F'=free_5, [ 0>=1+free_4 && B>=1+A ], cost: 1

      3: f0 -> f0 : A'=free_10, D'=free_11, E'=free_8, F'=free_9, [ free_8>=1 && B>=1+A ], cost: 1



   Accelerated rule 1 with NONTERM (after strengthening guard), yielding the new rule 5.

   Accelerated rule 2 with NONTERM (after strengthening guard), yielding the new rule 6.

   Accelerated rule 3 with NONTERM (after strengthening guard), yielding the new rule 7.

   Removing the simple loops:.



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

   Start location: f1

      1: f0 -> f0 : A'=free_2, D'=free_3, E'=0, [ B>=1+A ], cost: 1

      2: f0 -> f0 : A'=free_6, D'=free_7, E'=free_4, F'=free_5, [ 0>=1+free_4 && B>=1+A ], cost: 1

      3: f0 -> f0 : A'=free_10, D'=free_11, E'=free_8, F'=free_9, [ free_8>=1 && B>=1+A ], cost: 1

      5: f0 -> [3] : [ B>=1+A && B>=1+free_2 ], cost: NONTERM

      6: f0 -> [3] : [ 0>=1+free_4 && B>=1+A && B>=1+free_6 ], cost: NONTERM

      7: f0 -> [3] : [ free_8>=1 && B>=1+A && B>=1+free_10 ], cost: NONTERM

      4: f1 -> f0 : G'=free_12, [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f1

      4: f1 -> f0 : G'=free_12, [], cost: 1

      8: f1 -> f0 : A'=free_2, D'=free_3, E'=0, G'=free_12, [ B>=1+A ], cost: 2

      9: f1 -> f0 : A'=free_6, D'=free_7, E'=free_4, F'=free_5, G'=free_12, [ 0>=1+free_4 && B>=1+A ], cost: 2

     10: f1 -> f0 : A'=free_10, D'=free_11, E'=free_8, F'=free_9, G'=free_12, [ free_8>=1 && B>=1+A ], cost: 2

     11: f1 -> [3] : G'=free_12, [ B>=1+A ], cost: NONTERM

     12: f1 -> [3] : G'=free_12, [ B>=1+A ], cost: NONTERM

     13: f1 -> [3] : G'=free_12, [ B>=1+A ], cost: NONTERM



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

   Start location: f1

     11: f1 -> [3] : G'=free_12, [ B>=1+A ], cost: NONTERM

     12: f1 -> [3] : G'=free_12, [ B>=1+A ], cost: NONTERM

     13: f1 -> [3] : G'=free_12, [ B>=1+A ], cost: NONTERM



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f1

     13: f1 -> [3] : G'=free_12, [ B>=1+A ], cost: NONTERM



Computing asymptotic complexity for rule 13

   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:  [ B>=1+A ]



NO


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(2)
BOUNDS(INF, INF)