YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f300(A,B,C) -> f300(-99 + A,0,C)   [0 >= 1 + A && 1 + B = 0]  (?,1)
          1. f300(A,B,C) -> f300(1 + A,1 + B,C) [0 >= 1 + A && B >= 0]     (?,1)
          2. f300(A,B,C) -> f300(1 + A,1 + B,C) [0 >= 1 + A && 0 >= 2 + B] (?,1)
          3. f300(A,B,C) -> f1(A,B,D)           [A >= 0]                   (?,1)
          4. f2(A,B,C)   -> f300(A,B,C)         True                       (1,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{0,1,2,3},1->{0,1,2,3},2->{0,1,2,3},3->{},4->{0,1,2,3}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,0),(0,2),(0,3),(1,0),(1,2),(2,1)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0. f300(A,B,C) -> f300(-99 + A,0,C)   [0 >= 1 + A && 1 + B = 0]  (?,1)
          1. f300(A,B,C) -> f300(1 + A,1 + B,C) [0 >= 1 + A && B >= 0]     (?,1)
          2. f300(A,B,C) -> f300(1 + A,1 + B,C) [0 >= 1 + A && 0 >= 2 + B] (?,1)
          3. f300(A,B,C) -> f1(A,B,D)           [A >= 0]                   (?,1)
          4. f2(A,B,C)   -> f300(A,B,C)         True                       (1,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1},1->{1,3},2->{0,2,3},3->{},4->{0,1,2,3}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          f300(A,B,C) -> f300(-99 + A,0,C)    [0 >= 1 + A && 1 + B = 0]
          f300(A,B,C) -> f300(1 + A,1 + B,C)  [0 >= 1 + A && B >= 0]
          f300(A,B,C) -> f300(1 + A,1 + B,C)  [0 >= 1 + A && 0 >= 2 + B]
          f300(A,B,C) -> f1(A,B,D)            [A >= 0]
          f2(A,B,C)   -> f300(A,B,C)          True
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Rule Graph:
          [0->{1},1->{1,3},2->{0,2,3},3->{},4->{0,1,2,3}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4]
          |
          +- p:[2] c: [2]
          |
          `- p:[1] c: [1]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          f300(A,B,C) -> f300(-99 + A,0,C)    [0 >= 1 + A && 1 + B = 0]
          f300(A,B,C) -> f300(1 + A,1 + B,C)  [0 >= 1 + A && B >= 0]
          f300(A,B,C) -> f300(1 + A,1 + B,C)  [0 >= 1 + A && 0 >= 2 + B]
          f300(A,B,C) -> f1(A,B,D)            [A >= 0]
          f2(A,B,C)   -> f300(A,B,C)          True
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Rule Graph:
          [0->{1},1->{1,3},2->{0,2,3},3->{},4->{0,1,2,3}]
        ,We construct a looptree:
          P: [0,1,2,3,4]
          |
          +- p:[2] c: [2]
          |
          `- p:[1] c: [1])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES