YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f(A,B,C) -> g(A,1,1)        True                                                                                             (1,1)
          1. g(A,B,C) -> g(-1 + A,2*B,C) [1 + -1*C >= 0 && B + -1*C >= 0 && -1 + C >= 0 && -2 + B + C >= 0 && -1 + B >= 0 && -1 + A >= 0] (?,1)
          2. g(A,B,C) -> h(A,B,C)        [1 + -1*C >= 0 && B + -1*C >= 0 && -1 + C >= 0 && -2 + B + C >= 0 && -1 + B >= 0 && 0 >= A]      (?,1)
          3. h(A,B,C) -> h(A,-1 + B,2*C) [-1 + C >= 0 && -2 + B + C >= 0 && -1 + -1*A + C >= 0 && -1*A >= 0 && -1 + B >= 0]               (?,1)
          4. h(A,B,C) -> i(A,B,C)        [-1 + C >= 0 && -2 + B + C >= 0 && -1 + -1*A + C >= 0 && -1*A >= 0 && 0 >= B]                    (?,1)
          5. i(A,B,C) -> i(A,B,-1 + C)   [-1*B >= 0 && -1*A + -1*B >= 0 && -1*A >= 0 && -1 + C >= 0]                                      (?,1)
        Signature:
          {(f,3);(g,3);(h,3);(i,3)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{3,4},3->{3,4},4->{5},5->{5}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(2,4)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0. f(A,B,C) -> g(A,1,1)        True                                                                                             (1,1)
          1. g(A,B,C) -> g(-1 + A,2*B,C) [1 + -1*C >= 0 && B + -1*C >= 0 && -1 + C >= 0 && -2 + B + C >= 0 && -1 + B >= 0 && -1 + A >= 0] (?,1)
          2. g(A,B,C) -> h(A,B,C)        [1 + -1*C >= 0 && B + -1*C >= 0 && -1 + C >= 0 && -2 + B + C >= 0 && -1 + B >= 0 && 0 >= A]      (?,1)
          3. h(A,B,C) -> h(A,-1 + B,2*C) [-1 + C >= 0 && -2 + B + C >= 0 && -1 + -1*A + C >= 0 && -1*A >= 0 && -1 + B >= 0]               (?,1)
          4. h(A,B,C) -> i(A,B,C)        [-1 + C >= 0 && -2 + B + C >= 0 && -1 + -1*A + C >= 0 && -1*A >= 0 && 0 >= B]                    (?,1)
          5. i(A,B,C) -> i(A,B,-1 + C)   [-1*B >= 0 && -1*A + -1*B >= 0 && -1*A >= 0 && -1 + C >= 0]                                      (?,1)
        Signature:
          {(f,3);(g,3);(h,3);(i,3)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{3},3->{3,4},4->{5},5->{5}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          f(A,B,C) -> g(A,1,1)         True
          g(A,B,C) -> g(-1 + A,2*B,C)  [1 + -1*C >= 0
                                        && B + -1*C >= 0
                                        && -1 + C >= 0
                                        && -2 + B + C >= 0
                                        && -1 + B >= 0
                                        && -1 + A >= 0]
          g(A,B,C) -> h(A,B,C)         [1 + -1*C >= 0
                                        && B + -1*C >= 0
                                        && -1 + C >= 0
                                        && -2 + B + C >= 0
                                        && -1 + B >= 0
                                        && 0 >= A]
          h(A,B,C) -> h(A,-1 + B,2*C)  [-1 + C >= 0
                                        && -2 + B + C >= 0
                                        && -1 + -1*A + C >= 0
                                        && -1*A >= 0
                                        && -1 + B >= 0]
          h(A,B,C) -> i(A,B,C)         [-1 + C >= 0 && -2 + B + C >= 0 && -1 + -1*A + C >= 0 && -1*A >= 0 && 0 >= B]
          i(A,B,C) -> i(A,B,-1 + C)    [-1*B >= 0 && -1*A + -1*B >= 0 && -1*A >= 0 && -1 + C >= 0]
        Signature:
          {(f,3);(g,3);(h,3);(i,3)}
        Rule Graph:
          [0->{1,2},1->{1,2},2->{3},3->{3,4},4->{5},5->{5}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          +- p:[1] c: [1]
          |
          +- p:[3] c: [3]
          |
          `- p:[5] c: [5]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          f(A,B,C) -> g(A,1,1)         True
          g(A,B,C) -> g(-1 + A,2*B,C)  [1 + -1*C >= 0
                                        && B + -1*C >= 0
                                        && -1 + C >= 0
                                        && -2 + B + C >= 0
                                        && -1 + B >= 0
                                        && -1 + A >= 0]
          g(A,B,C) -> h(A,B,C)         [1 + -1*C >= 0
                                        && B + -1*C >= 0
                                        && -1 + C >= 0
                                        && -2 + B + C >= 0
                                        && -1 + B >= 0
                                        && 0 >= A]
          h(A,B,C) -> h(A,-1 + B,2*C)  [-1 + C >= 0
                                        && -2 + B + C >= 0
                                        && -1 + -1*A + C >= 0
                                        && -1*A >= 0
                                        && -1 + B >= 0]
          h(A,B,C) -> i(A,B,C)         [-1 + C >= 0 && -2 + B + C >= 0 && -1 + -1*A + C >= 0 && -1*A >= 0 && 0 >= B]
          i(A,B,C) -> i(A,B,-1 + C)    [-1*B >= 0 && -1*A + -1*B >= 0 && -1*A >= 0 && -1 + C >= 0]
        Signature:
          {(f,3);(g,3);(h,3);(i,3)}
        Rule Graph:
          [0->{1,2},1->{1,2},2->{3},3->{3,4},4->{5},5->{5}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5]
          |
          +- p:[1] c: [1]
          |
          +- p:[3] c: [3]
          |
          `- p:[5] c: [5])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES