YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f12(A,B,C,D,E,F) -> f5(4,0,0,D,E,F)     True                  (1,1)
          1. f5(A,B,C,D,E,F)  -> f11(A,B,0,0,0,F)    [B >= A && C = 0]     (?,1)
          2. f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)    [C >= 1]              (?,1)
          3. f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)    [0 >= 1 + C]          (?,1)
          4. f7(A,B,C,D,E,F)  -> f5(A,1 + B,G,H,E,G) True                  (?,1)
          5. f8(A,B,C,D,E,F)  -> f5(A,1 + B,G,H,E,G) True                  (?,1)
          6. f5(A,B,C,D,E,F)  -> f5(A,1 + B,G,H,E,G) [A >= 1 + B && C = 0] (?,1)
        Signature:
          {(f10,6);(f11,6);(f12,6);(f5,6);(f7,6);(f8,6)}
        Flow Graph:
          [0->{1,2,3,6},1->{},2->{},3->{},4->{1,2,3,6},5->{1,2,3,6},6->{1,2,3,6}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,1),(0,2),(0,3)]
* Step 2: UnreachableRules YES
    + Considered Problem:
        Rules:
          0. f12(A,B,C,D,E,F) -> f5(4,0,0,D,E,F)     True                  (1,1)
          1. f5(A,B,C,D,E,F)  -> f11(A,B,0,0,0,F)    [B >= A && C = 0]     (?,1)
          2. f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)    [C >= 1]              (?,1)
          3. f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)    [0 >= 1 + C]          (?,1)
          4. f7(A,B,C,D,E,F)  -> f5(A,1 + B,G,H,E,G) True                  (?,1)
          5. f8(A,B,C,D,E,F)  -> f5(A,1 + B,G,H,E,G) True                  (?,1)
          6. f5(A,B,C,D,E,F)  -> f5(A,1 + B,G,H,E,G) [A >= 1 + B && C = 0] (?,1)
        Signature:
          {(f10,6);(f11,6);(f12,6);(f5,6);(f7,6);(f8,6)}
        Flow Graph:
          [0->{6},1->{},2->{},3->{},4->{1,2,3,6},5->{1,2,3,6},6->{1,2,3,6}]
        
    + Applied Processor:
        UnreachableRules
    + Details:
        Following transitions are not reachable from the starting states and are revomed: [4,5]
* Step 3: FromIts YES
    + Considered Problem:
        Rules:
          0. f12(A,B,C,D,E,F) -> f5(4,0,0,D,E,F)     True                  (1,1)
          1. f5(A,B,C,D,E,F)  -> f11(A,B,0,0,0,F)    [B >= A && C = 0]     (?,1)
          2. f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)    [C >= 1]              (?,1)
          3. f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)    [0 >= 1 + C]          (?,1)
          6. f5(A,B,C,D,E,F)  -> f5(A,1 + B,G,H,E,G) [A >= 1 + B && C = 0] (?,1)
        Signature:
          {(f10,6);(f11,6);(f12,6);(f5,6);(f7,6);(f8,6)}
        Flow Graph:
          [0->{6},1->{},2->{},3->{},6->{1,2,3,6}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 4: Decompose YES
    + Considered Problem:
        Rules:
          f12(A,B,C,D,E,F) -> f5(4,0,0,D,E,F)      True
          f5(A,B,C,D,E,F)  -> f11(A,B,0,0,0,F)     [B >= A && C = 0]
          f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)     [C >= 1]
          f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)     [0 >= 1 + C]
          f5(A,B,C,D,E,F)  -> f5(A,1 + B,G,H,E,G)  [A >= 1 + B && C = 0]
        Signature:
          {(f10,6);(f11,6);(f12,6);(f5,6);(f7,6);(f8,6)}
        Rule Graph:
          [0->{6},1->{},2->{},3->{},6->{1,2,3,6}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,6]
          |
          `- p:[6] c: [6]
* Step 5: CloseWith YES
    + Considered Problem:
        (Rules:
          f12(A,B,C,D,E,F) -> f5(4,0,0,D,E,F)      True
          f5(A,B,C,D,E,F)  -> f11(A,B,0,0,0,F)     [B >= A && C = 0]
          f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)     [C >= 1]
          f5(A,B,C,D,E,F)  -> f10(A,B,C,C,C,F)     [0 >= 1 + C]
          f5(A,B,C,D,E,F)  -> f5(A,1 + B,G,H,E,G)  [A >= 1 + B && C = 0]
        Signature:
          {(f10,6);(f11,6);(f12,6);(f5,6);(f7,6);(f8,6)}
        Rule Graph:
          [0->{6},1->{},2->{},3->{},6->{1,2,3,6}]
        ,We construct a looptree:
          P: [0,1,2,3,6]
          |
          `- p:[6] c: [6])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES