WORST_CASE(INF,?)

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

Initial linear ITS problem
   Start location: __init
      0: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_1, arg2'=arg2P_1, arg3'=arg3P_1, [ arg1P_1<=arg1 && arg2>-1 && arg1>0 && arg1P_1>0 && 0==arg2P_1 && arg2==arg3P_1 ], cost: 1
      1: f66_0_main_GE -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg2P_2, arg3'=arg3P_2, [ arg3>arg2 && arg3>-1 && arg1P_2<=arg1 && arg1>0 && arg1P_2>0 && 1+arg2==arg2P_2 && arg3==arg3P_2 ], cost: 1
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_1, arg2'=0, arg3'=arg2, [ arg1P_1<=arg1 && arg2>-1 && arg1>0 && arg1P_1>0 ], cost: 1
      1: f66_0_main_GE -> f66_0_main_GE : arg1'=arg1P_2, arg2'=1+arg2, [ arg3>arg2 && arg3>-1 && arg1P_2<=arg1 && arg1>0 && arg1P_2>0 ], cost: 1
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

### Simplification by acceleration and chaining ###

Accelerating simple loops of location 1.
   Accelerating the following rules:
      1: f66_0_main_GE -> f66_0_main_GE : arg1'=arg1P_2, arg2'=1+arg2, [ arg3>arg2 && arg3>-1 && arg1P_2<=arg1 && arg1>0 && arg1P_2>0 ], cost: 1

   Accelerated rule 1 with metering function arg3-arg2, yielding the new rule 3.
   Removing the simple loops: 1.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
      0: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_1, arg2'=0, arg3'=arg2, [ arg1P_1<=arg1 && arg2>-1 && arg1>0 && arg1P_1>0 ], cost: 1
      3: f66_0_main_GE -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg3, [ arg3>arg2 && arg3>-1 && arg1P_2<=arg1 && arg1>0 && arg1P_2>0 ], cost: arg3-arg2
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

Chained accelerated rules (with incoming rules):
   Start location: __init
      0: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_1, arg2'=0, arg3'=arg2, [ arg1P_1<=arg1 && arg2>-1 && arg1>0 && arg1P_1>0 ], cost: 1
      4: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_2, arg3'=arg2, [ arg1>0 && arg2>0 && arg1P_2>0 && arg1P_2<=arg1 ], cost: 1+arg2
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
      4: f1_0_main_ConstantStackPush -> f66_0_main_GE : arg1'=arg1P_2, arg3'=arg2, [ arg1>0 && arg2>0 && arg1P_2>0 && arg1P_2<=arg1 ], cost: 1+arg2
      2: __init -> f1_0_main_ConstantStackPush : arg1'=arg1P_3, arg2'=arg2P_3, arg3'=arg3P_3, [], cost: 1

Eliminated locations (on linear paths):
   Start location: __init
      5: __init -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg2P_3, arg3'=arg2P_3, [ arg1P_3>0 && arg2P_3>0 && arg1P_2>0 && arg1P_2<=arg1P_3 ], cost: 2+arg2P_3

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
      5: __init -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg2P_3, arg3'=arg2P_3, [ arg1P_3>0 && arg2P_3>0 && arg1P_2>0 && arg1P_2<=arg1P_3 ], cost: 2+arg2P_3

Computing asymptotic complexity for rule 5
   Simplified the guard:
      5: __init -> f66_0_main_GE : arg1'=arg1P_2, arg2'=arg2P_3, arg3'=arg2P_3, [ arg2P_3>0 && arg1P_2>0 && arg1P_2<=arg1P_3 ], cost: 2+arg2P_3
   Solved the limit problem by the following transformations:
      Created initial limit problem:
      1+arg1P_3-arg1P_2 (+/+!), arg1P_2 (+/+!), 2+arg2P_3 (+), arg2P_3 (+/+!) [not solved]

      removing all constraints (solved by SMT)
      resulting limit problem: [solved]

      applying transformation rule (C) using substitution {arg1P_3==n,arg1P_2==1,arg2P_3==n}
      resulting limit problem:
      [solved]

   Solution:
      arg1P_3 / n
      arg1P_2 / 1
      arg2P_3 / n
   Resulting cost 2+n has complexity: Unbounded

Found new complexity Unbounded.

Obtained the following overall complexity (w.r.t. the length of the input n):
   Complexity:  Unbounded
   Cpx degree:  Unbounded
   Solved cost: 2+n
   Rule cost:   2+arg2P_3
   Rule guard:  [ arg2P_3>0 && arg1P_2>0 && arg1P_2<=arg1P_3 ]

WORST_CASE(INF,?)