Trying to load file: main.koat

Initial Control flow graph problem:
  Start location: start0
      0: start -> stop : [ 0>=1+A && B==C && D==A ], cost: 1
      1: start -> stop : [ 0>=1+C && B==C && D==A ], cost: 1
      2: start -> stop : [ A>=0 && 2+A>=C && C>=0 && 2+C>=A && B==C && D==A ], cost: 1
      3: start -> lbl81 : D'=1+D, [ A>=0 && C>=3+A && B==C && D==A ], cost: 1
      4: start -> lbl91 : B'=1+B, [ A>=3+C && C>=0 && B==C && D==A ], cost: 1
      5: lbl81 -> stop : [ C>=3+A && A>=0 && 2+D==C && B==C ], cost: 1
      6: lbl81 -> lbl81 : D'=1+D, [ C>=3+D && D>=1+A && A>=0 && C>=2+D && B==C ], cost: 1
      7: lbl81 -> lbl91 : B'=1+B, [ D>=3+C && D>=1+A && A>=0 && C>=2+D && B==C ], cost: 1
      8: lbl91 -> stop : [ A>=3+C && C>=0 && 2+B==A && D==A ], cost: 1
      9: lbl91 -> lbl81 : D'=1+D, [ B>=3+A && A>=2+B && B>=1+C && C>=0 && D==A ], cost: 1
     10: lbl91 -> lbl91 : B'=1+B, [ A>=3+B && A>=2+B && B>=1+C && C>=0 && D==A ], cost: 1
     11: start0 -> start : B'=C, D'=A, [], cost: 1


Simplified the transitions:
  Start location: start0
      3: start -> lbl81 : D'=1+D, [ A>=0 && C>=3+A && B==C && D==A ], cost: 1
      4: start -> lbl91 : B'=1+B, [ A>=3+C && C>=0 && B==C && D==A ], cost: 1
      6: lbl81 -> lbl81 : D'=1+D, [ C>=3+D && D>=1+A && A>=0 && B==C ], cost: 1
      7: lbl81 -> lbl91 : B'=1+B, [ D>=3+C && D>=1+A && A>=0 && C>=2+D && B==C ], cost: 1
      9: lbl91 -> lbl81 : D'=1+D, [ B>=3+A && A>=2+B && B>=1+C && C>=0 && D==A ], cost: 1
     10: lbl91 -> lbl91 : B'=1+B, [ A>=3+B && B>=1+C && C>=0 && D==A ], cost: 1
     11: start0 -> start : B'=C, D'=A, [], cost: 1

Eliminating 1 self-loops for location lbl81
  Self-Loop 6 has the metering function: -2+C-D, resulting in the new transition 12.
  Removing the self-loops: 6.
Eliminating 1 self-loops for location lbl91
  Self-Loop 10 has the metering function: -2-B+A, resulting in the new transition 13.
  Removing the self-loops: 10.

Removed all Self-loops using metering functions (where possible):
  Start location: start0
      3: start -> lbl81 : D'=1+D, [ A>=0 && C>=3+A && B==C && D==A ], cost: 1
      4: start -> lbl91 : B'=1+B, [ A>=3+C && C>=0 && B==C && D==A ], cost: 1
     12: lbl81 -> [5] : D'=-2+C, [ C>=3+D && D>=1+A && A>=0 && B==C ], cost: -2+C-D
     13: lbl91 -> [6] : B'=-2+A, [ A>=3+B && B>=1+C && C>=0 && D==A ], cost: -2-B+A
     11: start0 -> start : B'=C, D'=A, [], cost: 1
      7: [5] -> lbl91 : B'=1+B, [ D>=3+C && D>=1+A && A>=0 && C>=2+D && B==C ], cost: 1
      9: [6] -> lbl81 : D'=1+D, [ B>=3+A && A>=2+B && B>=1+C && C>=0 && D==A ], cost: 1


Applied chaining over branches and pruning:
  Start location: start0
     16: lbl81 -> [7] : D'=-2+C, [ C>=3+D && D>=1+A && A>=0 && B==C ], cost: -2+C-D
     17: lbl91 -> [8] : B'=-2+A, [ A>=3+B && B>=1+C && C>=0 && D==A ], cost: -2-B+A
     14: start0 -> lbl81 : B'=C, D'=1+A, [ A>=0 && C>=3+A && C==C && A==A ], cost: 2
     15: start0 -> lbl91 : B'=1+C, D'=A, [ A>=3+C && C>=0 && C==C && A==A ], cost: 2


Applied simple chaining:
  Start location: start0
     14: start0 -> [7] : B'=C, D'=-2+C, [ A>=0 && C>=3+A && C==C && A==A && C>=4+A && 1+A>=1+A && A>=0 && C==C ], cost: -1+C-A
     15: start0 -> [8] : B'=-2+A, D'=A, [ A>=3+C && C>=0 && C==C && A==A && A>=4+C && 1+C>=1+C && C>=0 && A==A ], cost: -1-C+A


Final control flow graph problem, now checking costs for infinitely many models:
  Start location: start0
     14: start0 -> [7] : B'=C, D'=-2+C, [ A>=0 && C>=3+A && C==C && A==A && C>=4+A && 1+A>=1+A && A>=0 && C==C ], cost: -1+C-A
     15: start0 -> [8] : B'=-2+A, D'=A, [ A>=3+C && C>=0 && C==C && A==A && A>=4+C && 1+C>=1+C && C>=0 && A==A ], cost: -1-C+A


Computing complexity for remaining 2 transitions.

  Found configuration with infinitely models for cost: -1+C-A
  and guard: A>=0 && C>=3+A && C==C && A==A && C>=4+A && 1+A>=1+A && A>=0 && C==C:
  C: Pos, A: Pos, where: C > A

Found new complexity n^1, because: Found infinity configuration.


The final runtime is determined by this resulting transition:
  Final Guard: A>=0 && C>=3+A && C==C && A==A && C>=4+A && 1+A>=1+A && A>=0 && C==C
  Final Cost:  -1+C-A

Obtained the following complexity w.r.t. the length of the input n:
  Complexity class: n^1
  Complexity value: 1

WORST_CASE(Omega(n^1),?)