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

YES