/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 424 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f10(A, B, C, D, E) -> Com_1(f10(A - 1, B - 1, C + 1, F, E)) :|: A >= 1 && F >= 1
f10(A, B, C, D, E) -> Com_1(f10(A - 1, B, C, F, E)) :|: 0 >= F && A >= 1 && A >= B + 1
f20(A, B, C, D, E) -> Com_1(f20(A, B, C, D, E)) :|: TRUE
f22(A, B, C, D, E) -> Com_1(f25(A, B, C, D, E)) :|: TRUE
f10(A, B, C, D, E) -> Com_1(f20(A, B, C, D, E)) :|: 0 >= A
f0(A, B, C, D, E) -> Com_1(f10(8, F, 0, D, 8)) :|: F >= 1

The start-symbols are:[f0_5]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f10 -> f10 : A'=-1+A, B'=-1+B, C'=1+C, D'=free, [ A>=1 && free>=1 ], cost: 1

      1: f10 -> f10 : A'=-1+A, D'=free_1, [ 0>=free_1 && A>=1 && A>=1+B ], cost: 1

      4: f10 -> f20 : [ 0>=A ], cost: 1

      2: f20 -> f20 : [], cost: 1

      3: f22 -> f25 : [], cost: 1

      5: f0 -> f10 : A'=8, B'=free_2, C'=0, E'=8, [ free_2>=1 ], cost: 1



Checking for constant complexity:

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

      5: f0 -> f10 : A'=8, B'=free_2, C'=0, E'=8, [ free_2>=1 ], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      0: f10 -> f10 : A'=-1+A, B'=-1+B, C'=1+C, D'=free, [ A>=1 && free>=1 ], cost: 1

      1: f10 -> f10 : A'=-1+A, D'=free_1, [ 0>=free_1 && A>=1 && A>=1+B ], cost: 1

      4: f10 -> f20 : [ 0>=A ], cost: 1

      2: f20 -> f20 : [], cost: 1

      5: f0 -> f10 : A'=8, B'=free_2, C'=0, E'=8, [ free_2>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: f10 -> f10 : A'=-1+A, B'=-1+B, C'=1+C, D'=free, [ A>=1 && free>=1 ], cost: 1

      1: f10 -> f10 : A'=-1+A, D'=free_1, [ 0>=free_1 && A>=1 && A>=1+B ], cost: 1



   Accelerated rule 0 with metering function A, yielding the new rule 6.

   Found no metering function for rule 1.

   Removing the simple loops: 0.



Accelerating simple loops of location 1.

   Accelerating the following rules:

      2: f20 -> f20 : [], cost: 1



   Accelerated rule 2 with NONTERM, yielding the new rule 7.

   Removing the simple loops: 2.



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

   Start location: f0

      1: f10 -> f10 : A'=-1+A, D'=free_1, [ 0>=free_1 && A>=1 && A>=1+B ], cost: 1

      4: f10 -> f20 : [ 0>=A ], cost: 1

      6: f10 -> f10 : A'=0, B'=-A+B, C'=C+A, D'=free, [ A>=1 && free>=1 ], cost: A

      7: f20 -> [6] : [], cost: NONTERM

      5: f0 -> f10 : A'=8, B'=free_2, C'=0, E'=8, [ free_2>=1 ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f0

      4: f10 -> f20 : [ 0>=A ], cost: 1

     10: f10 -> [6] : [ 0>=A ], cost: NONTERM

      5: f0 -> f10 : A'=8, B'=free_2, C'=0, E'=8, [ free_2>=1 ], cost: 1

      8: f0 -> f10 : A'=7, B'=free_2, C'=0, D'=free_1, E'=8, [ free_2>=1 && 0>=free_1 && 8>=1+free_2 ], cost: 2

      9: f0 -> f10 : A'=0, B'=-8+free_2, C'=8, D'=free, E'=8, [ free_2>=1 && free>=1 ], cost: 9



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

   Start location: f0

     10: f10 -> [6] : [ 0>=A ], cost: NONTERM

      5: f0 -> f10 : A'=8, B'=free_2, C'=0, E'=8, [ free_2>=1 ], cost: 1

      8: f0 -> f10 : A'=7, B'=free_2, C'=0, D'=free_1, E'=8, [ free_2>=1 && 0>=free_1 && 8>=1+free_2 ], cost: 2

      9: f0 -> f10 : A'=0, B'=-8+free_2, C'=8, D'=free, E'=8, [ free_2>=1 && free>=1 ], cost: 9



Eliminated locations (on tree-shaped paths):

   Start location: f0

     11: f0 -> [6] : A'=0, B'=-8+free_2, C'=8, D'=free, E'=8, [ free_2>=1 && free>=1 ], cost: NONTERM



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

     11: f0 -> [6] : A'=0, B'=-8+free_2, C'=8, D'=free, E'=8, [ free_2>=1 && free>=1 ], cost: NONTERM



Computing asymptotic complexity for rule 11

   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:  [ free_2>=1 && free>=1 ]



NO


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(2)
BOUNDS(INF, INF)