YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0. f0(A,B)  -> f4(0,B)      True                                                               (1,1)
          1. f4(A,B)  -> f4(1 + A,B)  [A >= 0 && 1 >= A]                                                 (?,1)
          2. f10(A,B) -> f10(A,1 + B) [B >= 0 && -2 + A + B >= 0 && -2 + A >= 0 && 1 >= B]               (?,1)
          3. f10(A,B) -> f18(A,B)     [B >= 0 && -2 + A + B >= 0 && -2 + A >= 0 && B >= 2 && 0 >= 1 + C] (?,1)
          4. f10(A,B) -> f18(A,B)     [B >= 0 && -2 + A + B >= 0 && -2 + A >= 0 && B >= 2]               (?,1)
          5. f4(A,B)  -> f10(A,0)     [A >= 0 && A >= 2]                                                 (?,1)
        Signature:
          {(f0,2);(f10,2);(f18,2);(f4,2)}
        Flow Graph:
          [0->{1,5},1->{1,5},2->{2,3,4},3->{},4->{},5->{2,3,4}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f0(A,B)  -> f4(0,B)      True                                                               (1,1)
          1. f4(A,B)  -> f4(1 + A,B)  [A >= 0 && 1 >= A]                                                 (?,1)
          2. f10(A,B) -> f10(A,1 + B) [B >= 0 && -2 + A + B >= 0 && -2 + A >= 0 && 1 >= B]               (?,1)
          3. f10(A,B) -> f18(A,B)     [B >= 0 && -2 + A + B >= 0 && -2 + A >= 0 && B >= 2 && 0 >= 1 + C] (1,1)
          4. f10(A,B) -> f18(A,B)     [B >= 0 && -2 + A + B >= 0 && -2 + A >= 0 && B >= 2]               (1,1)
          5. f4(A,B)  -> f10(A,0)     [A >= 0 && A >= 2]                                                 (1,1)
        Signature:
          {(f0,2);(f10,2);(f18,2);(f4,2)}
        Flow Graph:
          [0->{1,5},1->{1,5},2->{2,3,4},3->{},4->{},5->{2,3,4}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,5),(5,3),(5,4)]
* Step 3: Looptree YES
    + Considered Problem:
        Rules:
          0. f0(A,B)  -> f4(0,B)      True                                                               (1,1)
          1. f4(A,B)  -> f4(1 + A,B)  [A >= 0 && 1 >= A]                                                 (?,1)
          2. f10(A,B) -> f10(A,1 + B) [B >= 0 && -2 + A + B >= 0 && -2 + A >= 0 && 1 >= B]               (?,1)
          3. f10(A,B) -> f18(A,B)     [B >= 0 && -2 + A + B >= 0 && -2 + A >= 0 && B >= 2 && 0 >= 1 + C] (1,1)
          4. f10(A,B) -> f18(A,B)     [B >= 0 && -2 + A + B >= 0 && -2 + A >= 0 && B >= 2]               (1,1)
          5. f4(A,B)  -> f10(A,0)     [A >= 0 && A >= 2]                                                 (1,1)
        Signature:
          {(f0,2);(f10,2);(f18,2);(f4,2)}
        Flow Graph:
          [0->{1},1->{1,5},2->{2,3,4},3->{},4->{},5->{2}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          +- p:[1] c: [1]
          |
          `- p:[2] c: [2]

YES