YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0. f(A,B,C)  -> g(A,1,0)       True                                                                            (1,1)
          1. g(A,B,C)  -> g1(-1 + A,B,B) [-1 + B + -1*C >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + B >= 0 && -1 + A >= 0] (?,1)
          2. g1(A,B,C) -> g(A,B + C,C)   [B + -1*C >= 0                                                                  (?,1)
                                          && -1 + C >= 0                                                                      
                                          && -2 + B + C >= 0                                                                  
                                          && -1*B + C >= 0                                                                    
                                          && -1 + A + C >= 0                                                                  
                                          && -1 + B >= 0                                                                      
                                          && -1 + A + B >= 0                                                                  
                                          && A >= 0]                                                                          
          3. g(A,B,C)  -> h(A,B,C)       [-1 + B + -1*C >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + B >= 0 && 0 >= A]      (?,1)
          4. h(A,B,C)  -> h(A,-1 + B,C)  [C >= 0 && -1*A + C >= 0 && -1*A >= 0 && -1 + B >= 0]                           (?,1)
        Signature:
          {(f,3);(g,3);(g1,3);(h,3)}
        Flow Graph:
          [0->{1,3},1->{2},2->{1,3},3->{4},4->{4}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: Looptree YES
    + Considered Problem:
        Rules:
          0. f(A,B,C)  -> g(A,1,0)       True                                                                            (1,1)
          1. g(A,B,C)  -> g1(-1 + A,B,B) [-1 + B + -1*C >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + B >= 0 && -1 + A >= 0] (?,1)
          2. g1(A,B,C) -> g(A,B + C,C)   [B + -1*C >= 0                                                                  (?,1)
                                          && -1 + C >= 0                                                                      
                                          && -2 + B + C >= 0                                                                  
                                          && -1*B + C >= 0                                                                    
                                          && -1 + A + C >= 0                                                                  
                                          && -1 + B >= 0                                                                      
                                          && -1 + A + B >= 0                                                                  
                                          && A >= 0]                                                                          
          3. g(A,B,C)  -> h(A,B,C)       [-1 + B + -1*C >= 0 && C >= 0 && -1 + B + C >= 0 && -1 + B >= 0 && 0 >= A]      (1,1)
          4. h(A,B,C)  -> h(A,-1 + B,C)  [C >= 0 && -1*A + C >= 0 && -1*A >= 0 && -1 + B >= 0]                           (?,1)
        Signature:
          {(f,3);(g,3);(g1,3);(h,3)}
        Flow Graph:
          [0->{1,3},1->{2},2->{1,3},3->{4},4->{4}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4]
          |
          +- p:[1,2] c: [2]
          |
          `- p:[4] c: [4]

YES