YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f0(A) -> f3(0)     True                              (1,1)
          1. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A]               (?,1)
          2. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A && 0 >= 1 + B] (?,1)
          3. f3(A) -> f13(A)    [A >= 0 && A >= 42]               (?,1)
        Signature:
          {(f0,1);(f13,1);(f3,1)}
        Flow Graph:
          [0->{1,2,3},1->{1,2,3},2->{1,2,3},3->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,3)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0. f0(A) -> f3(0)     True                              (1,1)
          1. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A]               (?,1)
          2. f3(A) -> f3(1 + A) [A >= 0 && 41 >= A && 0 >= 1 + B] (?,1)
          3. f3(A) -> f13(A)    [A >= 0 && A >= 42]               (?,1)
        Signature:
          {(f0,1);(f13,1);(f3,1)}
        Flow Graph:
          [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          f0(A) -> f3(0)      True
          f3(A) -> f3(1 + A)  [A >= 0 && 41 >= A]
          f3(A) -> f3(1 + A)  [A >= 0 && 41 >= A && 0 >= 1 + B]
          f3(A) -> f13(A)     [A >= 0 && A >= 42]
        Signature:
          {(f0,1);(f13,1);(f3,1)}
        Rule Graph:
          [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[1,2] c: [2]
             |
             `- p:[1] c: [1]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          f0(A) -> f3(0)      True
          f3(A) -> f3(1 + A)  [A >= 0 && 41 >= A]
          f3(A) -> f3(1 + A)  [A >= 0 && 41 >= A && 0 >= 1 + B]
          f3(A) -> f13(A)     [A >= 0 && A >= 42]
        Signature:
          {(f0,1);(f13,1);(f3,1)}
        Rule Graph:
          [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}]
        ,We construct a looptree:
          P: [0,1,2,3]
          |
          `- p:[1,2] c: [2]
             |
             `- p:[1] c: [1])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES