YES
* Step 1: TrivialSCCs YES
    + Considered Problem:
        Rules:
          0. f4(A)  -> f5(A)     [A >= 0 && 0 >= 1 + B] (?,1)
          1. f4(A)  -> f5(A)     [A >= 0]               (?,1)
          2. f0(A)  -> f4(0)     True                   (1,1)
          3. f5(A)  -> f11(A)    [A >= 0 && A >= 3]     (?,1)
          4. f4(A)  -> f11(A)    [A >= 0]               (?,1)
          5. f5(A)  -> f4(1 + A) [A >= 0 && 2 >= A]     (?,1)
          6. f11(A) -> f14(A)    [A >= 0 && 1 >= A]     (?,1)
          7. f11(A) -> f14(A)    [A >= 0 && A >= 2]     (?,1)
        Signature:
          {(f0,1);(f11,1);(f14,1);(f4,1);(f5,1)}
        Flow Graph:
          [0->{3,5},1->{3,5},2->{0,1,4},3->{6,7},4->{6,7},5->{0,1,4},6->{},7->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f4(A)  -> f5(A)     [A >= 0 && 0 >= 1 + B] (?,1)
          1. f4(A)  -> f5(A)     [A >= 0]               (?,1)
          2. f0(A)  -> f4(0)     True                   (1,1)
          3. f5(A)  -> f11(A)    [A >= 0 && A >= 3]     (1,1)
          4. f4(A)  -> f11(A)    [A >= 0]               (1,1)
          5. f5(A)  -> f4(1 + A) [A >= 0 && 2 >= A]     (?,1)
          6. f11(A) -> f14(A)    [A >= 0 && 1 >= A]     (1,1)
          7. f11(A) -> f14(A)    [A >= 0 && A >= 2]     (1,1)
        Signature:
          {(f0,1);(f11,1);(f14,1);(f4,1);(f5,1)}
        Flow Graph:
          [0->{3,5},1->{3,5},2->{0,1,4},3->{6,7},4->{6,7},5->{0,1,4},6->{},7->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(3,6)]
* Step 3: Looptree YES
    + Considered Problem:
        Rules:
          0. f4(A)  -> f5(A)     [A >= 0 && 0 >= 1 + B] (?,1)
          1. f4(A)  -> f5(A)     [A >= 0]               (?,1)
          2. f0(A)  -> f4(0)     True                   (1,1)
          3. f5(A)  -> f11(A)    [A >= 0 && A >= 3]     (1,1)
          4. f4(A)  -> f11(A)    [A >= 0]               (1,1)
          5. f5(A)  -> f4(1 + A) [A >= 0 && 2 >= A]     (?,1)
          6. f11(A) -> f14(A)    [A >= 0 && 1 >= A]     (1,1)
          7. f11(A) -> f14(A)    [A >= 0 && A >= 2]     (1,1)
        Signature:
          {(f0,1);(f11,1);(f14,1);(f4,1);(f5,1)}
        Flow Graph:
          [0->{3,5},1->{3,5},2->{0,1,4},3->{7},4->{6,7},5->{0,1,4},6->{},7->{}]
        
    + Applied Processor:
        Looptree
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4,5,6,7]
          |
          `- p:[0,5,1] c: [5]

YES