WORST_CASE(?,O(n^1))
* Step 1: LocalSizeboundsProc WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval(A)  -> eval(2*B) [2*B >= 0 && A = 1 + 2*B] (?,1)
          1. start(A) -> eval(A)   True                      (1,1)
        Signature:
          {(eval,1);(start,1)}
        Flow Graph:
          [0->{0},1->{0}]
        
    + Applied Processor:
        LocalSizeboundsProc
    + Details:
        LocalSizebounds generated; rvgraph
          (<0,0,A>, 1 + A, .+ 1) 
          (<1,0,A>,     A, .= 0) 
* Step 2: SizeboundsProc WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval(A)  -> eval(2*B) [2*B >= 0 && A = 1 + 2*B] (?,1)
          1. start(A) -> eval(A)   True                      (1,1)
        Signature:
          {(eval,1);(start,1)}
        Flow Graph:
          [0->{0},1->{0}]
        Sizebounds:
          (<0,0,A>, ?) 
          (<1,0,A>, ?) 
    + Applied Processor:
        SizeboundsProc
    + Details:
        Sizebounds computed:
          (<0,0,A>, 2) 
          (<1,0,A>, A) 
* Step 3: LocationConstraintsProc WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval(A)  -> eval(2*B) [2*B >= 0 && A = 1 + 2*B] (?,1)
          1. start(A) -> eval(A)   True                      (1,1)
        Signature:
          {(eval,1);(start,1)}
        Flow Graph:
          [0->{0},1->{0}]
        Sizebounds:
          (<0,0,A>, 2) 
          (<1,0,A>, A) 
    + Applied Processor:
        LocationConstraintsProc
    + Details:
        We computed the location constraints  0 :  [2*B >= 0] 1 :  True .
* Step 4: PolyRank WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval(A)  -> eval(2*B) [2*B >= 0 && A = 1 + 2*B] (?,1)
          1. start(A) -> eval(A)   True                      (1,1)
        Signature:
          {(eval,1);(start,1)}
        Flow Graph:
          [0->{0},1->{0}]
        Sizebounds:
          (<0,0,A>, 2) 
          (<1,0,A>, A) 
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(eval) = x1
          p(start) = x1
        
        The following rules are strictly oriented:
        [2*B >= 0 && A = 1 + 2*B] ==>          
                          eval(A)   = A        
                                    > 2*B      
                                    = eval(2*B)
        
        
        The following rules are weakly oriented:
              True ==>        
          start(A)   = A      
                    >= A      
                     = eval(A)
        
        
* Step 5: LoopRecurrenceProcessor WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval(A)  -> eval(2*B) [2*B >= 0 && A = 1 + 2*B] (A,1)
          1. start(A) -> eval(A)   True                      (1,1)
        Signature:
          {(eval,1);(start,1)}
        Flow Graph:
          [0->{0},1->{0}]
        Sizebounds:
          (<0,0,A>, 2) 
          (<1,0,A>, A) 
    + Applied Processor:
        LoopRecurrenceProcessor [0]
    + Details:
        Applying the recurrence pattern linear * f to the expression A yields the solution A .
* Step 6: UnsatPaths WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval(A)  -> eval(2*B) [2*B >= 0 && A = 1 + 2*B] (A,1)
          1. start(A) -> eval(A)   True                      (1,1)
        Signature:
          {(eval,1);(start,1)}
        Flow Graph:
          [0->{0},1->{0}]
        Sizebounds:
          (<0,0,A>, 2) 
          (<1,0,A>, A) 
    + Applied Processor:
        UnsatPaths
    + Details:
        The problem is already solved.

WORST_CASE(?,O(n^1))