YES(?,O(1))
* Step 1: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f300(A,B,C,D,E) -> f300(1 + A,B,C,D,E) [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 100 >= A && B >= 1] (?,1)
          1. f300(A,B,C,D,E) -> f3(A,B,0,0,0)       [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 101]           (?,1)
          2. f2(A,B,C,D,E)   -> f300(1,B,C,D,E)     [B >= 1]                                                              (1,1)
          3. f2(A,B,C,D,E)   -> f3(0,B,0,0,0)       [0 >= B]                                                              (1,1)
        Signature:
          {(f2,5);(f3,5);(f300,5)}
        Flow Graph:
          [0->{0,1},1->{},2->{0,1},3->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(2,1)]
* Step 2: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f300(A,B,C,D,E) -> f300(1 + A,B,C,D,E) [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 100 >= A && B >= 1] (?,1)
          1. f300(A,B,C,D,E) -> f3(A,B,0,0,0)       [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 101]           (?,1)
          2. f2(A,B,C,D,E)   -> f300(1,B,C,D,E)     [B >= 1]                                                              (1,1)
          3. f2(A,B,C,D,E)   -> f3(0,B,0,0,0)       [0 >= B]                                                              (1,1)
        Signature:
          {(f2,5);(f3,5);(f300,5)}
        Flow Graph:
          [0->{0,1},1->{},2->{0},3->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 3: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f300(A,B,C,D,E) -> f300(1 + A,B,C,D,E) [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 100 >= A && B >= 1] (?,1)
          1. f300(A,B,C,D,E) -> f3(A,B,0,0,0)       [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 101]           (1,1)
          2. f2(A,B,C,D,E)   -> f300(1,B,C,D,E)     [B >= 1]                                                              (1,1)
          3. f2(A,B,C,D,E)   -> f3(0,B,0,0,0)       [0 >= B]                                                              (1,1)
        Signature:
          {(f2,5);(f3,5);(f300,5)}
        Flow Graph:
          [0->{0,1},1->{},2->{0},3->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
            p(f2) = 101        
            p(f3) = 101 + -1*x1
          p(f300) = 101 + -1*x1
        
        Following rules are strictly oriented:
        [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 100 >= A && B >= 1] ==>                    
                                                              f300(A,B,C,D,E)   = 101 + -1*A         
                                                                                > 100 + -1*A         
                                                                                = f300(1 + A,B,C,D,E)
        
                                                                     [B >= 1] ==>                    
                                                                f2(A,B,C,D,E)   = 101                
                                                                                > 100                
                                                                                = f300(1,B,C,D,E)    
        
        
        Following rules are weakly oriented:
        [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 101] ==>              
                                                    f300(A,B,C,D,E)   = 101 + -1*A   
                                                                     >= 101 + -1*A   
                                                                      = f3(A,B,0,0,0)
        
                                                           [0 >= B] ==>              
                                                      f2(A,B,C,D,E)   = 101          
                                                                     >= 101          
                                                                      = f3(0,B,0,0,0)
        
        
* Step 4: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f300(A,B,C,D,E) -> f300(1 + A,B,C,D,E) [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && 100 >= A && B >= 1] (101,1)
          1. f300(A,B,C,D,E) -> f3(A,B,0,0,0)       [-1 + B >= 0 && -2 + A + B >= 0 && -1 + A >= 0 && A >= 101]           (1,1)  
          2. f2(A,B,C,D,E)   -> f300(1,B,C,D,E)     [B >= 1]                                                              (1,1)  
          3. f2(A,B,C,D,E)   -> f3(0,B,0,0,0)       [0 >= B]                                                              (1,1)  
        Signature:
          {(f2,5);(f3,5);(f300,5)}
        Flow Graph:
          [0->{0,1},1->{},2->{0},3->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))