MAYBE
* Step 1: TrivialSCCs MAYBE
    + Considered Problem:
        Rules:
          0. f3(A,B) -> f0(0,B)      True                                                                                      (1,1)
          1. f0(A,B) -> f0(A,-1 + B) [-1*A >= 0 && A >= 0 && B >= 1]                                                           (?,1)
          2. f4(A,B) -> f4(A,B)      [-1*B >= 0 && 1 + A + -1*B >= 0 && -1 + -1*A + -1*B >= 0 && -1 + -1*A >= 0 && 1 + A >= 0] (?,1)
          3. f0(A,B) -> f4(-1,B)     [-1*A >= 0 && A >= 0 && 0 >= B]                                                           (?,1)
        Signature:
          {(f0,2);(f3,2);(f4,2)}
        Flow Graph:
          [0->{1,3},1->{1,3},2->{2},3->{2}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f3(A,B) -> f0(0,B)      True                                                                                      (1,1)
          1. f0(A,B) -> f0(A,-1 + B) [-1*A >= 0 && A >= 0 && B >= 1]                                                           (?,1)
          2. f4(A,B) -> f4(A,B)      [-1*B >= 0 && 1 + A + -1*B >= 0 && -1 + -1*A + -1*B >= 0 && -1 + -1*A >= 0 && 1 + A >= 0] (?,1)
          3. f0(A,B) -> f4(-1,B)     [-1*A >= 0 && A >= 0 && 0 >= B]                                                           (1,1)
        Signature:
          {(f0,2);(f3,2);(f4,2)}
        Flow Graph:
          [0->{1,3},1->{1,3},2->{2},3->{2}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(f0) = x2
          p(f3) = x2
          p(f4) = x2
        
        Following rules are strictly oriented:
        [-1*A >= 0 && A >= 0 && B >= 1] ==>             
                                f0(A,B)   = B           
                                          > -1 + B      
                                          = f0(A,-1 + B)
        
        
        Following rules are weakly oriented:
                                                                                             True ==>         
                                                                                          f3(A,B)   = B       
                                                                                                   >= B       
                                                                                                    = f0(0,B) 
        
        [-1*B >= 0 && 1 + A + -1*B >= 0 && -1 + -1*A + -1*B >= 0 && -1 + -1*A >= 0 && 1 + A >= 0] ==>         
                                                                                          f4(A,B)   = B       
                                                                                                   >= B       
                                                                                                    = f4(A,B) 
        
                                                                  [-1*A >= 0 && A >= 0 && 0 >= B] ==>         
                                                                                          f0(A,B)   = B       
                                                                                                   >= B       
                                                                                                    = f4(-1,B)
        
        
* Step 3: Failure MAYBE
  + Considered Problem:
      Rules:
        0. f3(A,B) -> f0(0,B)      True                                                                                      (1,1)
        1. f0(A,B) -> f0(A,-1 + B) [-1*A >= 0 && A >= 0 && B >= 1]                                                           (B,1)
        2. f4(A,B) -> f4(A,B)      [-1*B >= 0 && 1 + A + -1*B >= 0 && -1 + -1*A + -1*B >= 0 && -1 + -1*A >= 0 && 1 + A >= 0] (?,1)
        3. f0(A,B) -> f4(-1,B)     [-1*A >= 0 && A >= 0 && 0 >= B]                                                           (1,1)
      Signature:
        {(f0,2);(f3,2);(f4,2)}
      Flow Graph:
        [0->{1,3},1->{1,3},2->{2},3->{2}]
      
  + Applied Processor:
      Failing "Open problems left."
  + Details:
      Open problems left.
MAYBE