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

YES