YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F)      True                   (1,1)
          1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1]     (?,1)
          2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (?,1)
          3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3]     (?,1)
          4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F)      [0 >= A]               (?,1)
        Signature:
          {(f0,6);(f1,6);(f2,6)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3,4},4->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(3,4)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0. f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F)      True                   (1,1)
          1. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [H >= 1 && A >= 1]     (?,1)
          2. f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F) [0 >= 1 + H && A >= 1] (?,1)
          3. f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H) [A >= 1 && C >= 3]     (?,1)
          4. f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F)      [0 >= A]               (?,1)
        Signature:
          {(f0,6);(f1,6);(f2,6)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3},4->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F)       True
          f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F)  [H >= 1 && A >= 1]
          f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F)  [0 >= 1 + H && A >= 1]
          f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H)  [A >= 1 && C >= 3]
          f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F)       [0 >= A]
        Signature:
          {(f0,6);(f1,6);(f2,6)}
        Rule Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3},4->{}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[1,2,3] c: [2]
             |
             `- p:[1,3] c: [1]
                |
                `- p:[3] c: [3]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          f1(A,B,C,D,E,F) -> f0(A,B,C,D,E,F)       True
          f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F)  [H >= 1 && A >= 1]
          f0(A,B,C,D,E,F) -> f0(-1 + A,B,I,G,H,F)  [0 >= 1 + H && A >= 1]
          f0(A,B,C,D,E,F) -> f0(A,B,-1 + C,G,0,H)  [A >= 1 && C >= 3]
          f0(A,B,C,D,E,F) -> f2(A,G,C,D,E,F)       [0 >= A]
        Signature:
          {(f0,6);(f1,6);(f2,6)}
        Rule Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{1,2,3},4->{}]
        ,We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[1,2,3] c: [2]
             |
             `- p:[1,3] c: [1]
                |
                `- p:[3] c: [3])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES