YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D)  -> f6(0,0,C,D)         True                                                            (1,1)
          1. f6(A,B,C,D)  -> f6(A,1 + B,C,D)     [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B]                  (?,1)
          2. f6(A,B,C,D)  -> f6(2 + A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B]                  (?,1)
          3. f15(A,B,C,D) -> f19(1 + C,B,C,1)    [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A = 1 + C]  (?,1)
          4. f15(A,B,C,D) -> f19(A,B,C,0)        [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && C >= A]     (?,1)
          5. f15(A,B,C,D) -> f19(A,B,C,0)        [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A >= 2 + C] (?,1)
          6. f6(A,B,C,D)  -> f15(A,B,C,D)        [B >= 0 && A + B >= 0 && A >= 0 && B >= C && C >= 1 + A]        (?,1)
          7. f6(A,B,C,D)  -> f15(A,B,C,D)        [B >= 0 && A + B >= 0 && A >= 0 && A >= 1 + C && B >= C]        (?,1)
          8. f6(A,B,C,D)  -> f19(A,B,A,1)        [B >= 0 && A + B >= 0 && A >= 0 && B >= C && A = C]             (?,1)
        Signature:
          {(f0,4);(f15,4);(f19,4);(f6,4)}
        Flow Graph:
          [0->{1,2,6,7,8},1->{1,2,6,7,8},2->{1,2,6,7,8},3->{},4->{},5->{},6->{3,4,5},7->{3,4,5},8->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,6),(6,3),(6,5),(7,4)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D)  -> f6(0,0,C,D)         True                                                            (1,1)
          1. f6(A,B,C,D)  -> f6(A,1 + B,C,D)     [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B]                  (?,1)
          2. f6(A,B,C,D)  -> f6(2 + A,1 + B,C,D) [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B]                  (?,1)
          3. f15(A,B,C,D) -> f19(1 + C,B,C,1)    [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A = 1 + C]  (?,1)
          4. f15(A,B,C,D) -> f19(A,B,C,0)        [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && C >= A]     (?,1)
          5. f15(A,B,C,D) -> f19(A,B,C,0)        [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A >= 2 + C] (?,1)
          6. f6(A,B,C,D)  -> f15(A,B,C,D)        [B >= 0 && A + B >= 0 && A >= 0 && B >= C && C >= 1 + A]        (?,1)
          7. f6(A,B,C,D)  -> f15(A,B,C,D)        [B >= 0 && A + B >= 0 && A >= 0 && A >= 1 + C && B >= C]        (?,1)
          8. f6(A,B,C,D)  -> f19(A,B,A,1)        [B >= 0 && A + B >= 0 && A >= 0 && B >= C && A = C]             (?,1)
        Signature:
          {(f0,4);(f15,4);(f19,4);(f6,4)}
        Flow Graph:
          [0->{1,2,7,8},1->{1,2,6,7,8},2->{1,2,6,7,8},3->{},4->{},5->{},6->{4},7->{3,5},8->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          f0(A,B,C,D)  -> f6(0,0,C,D)          True
          f6(A,B,C,D)  -> f6(A,1 + B,C,D)      [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B]
          f6(A,B,C,D)  -> f6(2 + A,1 + B,C,D)  [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B]
          f15(A,B,C,D) -> f19(1 + C,B,C,1)     [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A = 1 + C]
          f15(A,B,C,D) -> f19(A,B,C,0)         [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && C >= A]
          f15(A,B,C,D) -> f19(A,B,C,0)         [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A >= 2 + C]
          f6(A,B,C,D)  -> f15(A,B,C,D)         [B >= 0 && A + B >= 0 && A >= 0 && B >= C && C >= 1 + A]
          f6(A,B,C,D)  -> f15(A,B,C,D)         [B >= 0 && A + B >= 0 && A >= 0 && A >= 1 + C && B >= C]
          f6(A,B,C,D)  -> f19(A,B,A,1)         [B >= 0 && A + B >= 0 && A >= 0 && B >= C && A = C]
        Signature:
          {(f0,4);(f15,4);(f19,4);(f6,4)}
        Rule Graph:
          [0->{1,2,7,8},1->{1,2,6,7,8},2->{1,2,6,7,8},3->{},4->{},5->{},6->{4},7->{3,5},8->{}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8]
          |
          `- p:[1,2] c: [2]
             |
             `- p:[1] c: [1]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          f0(A,B,C,D)  -> f6(0,0,C,D)          True
          f6(A,B,C,D)  -> f6(A,1 + B,C,D)      [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B]
          f6(A,B,C,D)  -> f6(2 + A,1 + B,C,D)  [B >= 0 && A + B >= 0 && A >= 0 && C >= 1 + B]
          f15(A,B,C,D) -> f19(1 + C,B,C,1)     [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A = 1 + C]
          f15(A,B,C,D) -> f19(A,B,C,0)         [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && C >= A]
          f15(A,B,C,D) -> f19(A,B,C,0)         [B + -1*C >= 0 && B >= 0 && A + B >= 0 && A >= 0 && A >= 2 + C]
          f6(A,B,C,D)  -> f15(A,B,C,D)         [B >= 0 && A + B >= 0 && A >= 0 && B >= C && C >= 1 + A]
          f6(A,B,C,D)  -> f15(A,B,C,D)         [B >= 0 && A + B >= 0 && A >= 0 && A >= 1 + C && B >= C]
          f6(A,B,C,D)  -> f19(A,B,A,1)         [B >= 0 && A + B >= 0 && A >= 0 && B >= C && A = C]
        Signature:
          {(f0,4);(f15,4);(f19,4);(f6,4)}
        Rule Graph:
          [0->{1,2,7,8},1->{1,2,6,7,8},2->{1,2,6,7,8},3->{},4->{},5->{},6->{4},7->{3,5},8->{}]
        ,We construct a looptree:
          P: [0,1,2,3,4,5,6,7,8]
          |
          `- p:[1,2] c: [2]
             |
             `- p:[1] c: [1])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES