MAYBE
* Step 1: TrivialSCCs MAYBE
    + Considered Problem:
        Rules:
          0. f0(A,B,C) -> f1(0,B,C)      True                                   (1,1)
          1. f1(A,B,C) -> f1(A,-1 + B,D) [-1*A >= 0 && A >= 0 && B >= 1]        (?,1)
          2. f1(A,B,C) -> f4(A,B,D)      [-1*A >= 0 && A >= 0 && 0 >= B]        (?,1)
          3. f4(A,B,C) -> f4(1,B,D)      [-1*B >= 0 && A + -1*B >= 0 && A >= 0] (?,1)
        Signature:
          {(f0,3);(f1,3);(f4,3)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{3},3->{3}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f0(A,B,C) -> f1(0,B,C)      True                                   (1,1)
          1. f1(A,B,C) -> f1(A,-1 + B,D) [-1*A >= 0 && A >= 0 && B >= 1]        (?,1)
          2. f1(A,B,C) -> f4(A,B,D)      [-1*A >= 0 && A >= 0 && 0 >= B]        (1,1)
          3. f4(A,B,C) -> f4(1,B,D)      [-1*B >= 0 && A + -1*B >= 0 && A >= 0] (?,1)
        Signature:
          {(f0,3);(f1,3);(f4,3)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{3},3->{3}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(f0) = x2
          p(f1) = x2
          p(f4) = x2
        
        Following rules are strictly oriented:
        [-1*A >= 0 && A >= 0 && B >= 1] ==>               
                              f1(A,B,C)   = B             
                                          > -1 + B        
                                          = f1(A,-1 + B,D)
        
        
        Following rules are weakly oriented:
                                          True ==>          
                                     f0(A,B,C)   = B        
                                                >= B        
                                                 = f1(0,B,C)
        
               [-1*A >= 0 && A >= 0 && 0 >= B] ==>          
                                     f1(A,B,C)   = B        
                                                >= B        
                                                 = f4(A,B,D)
        
        [-1*B >= 0 && A + -1*B >= 0 && A >= 0] ==>          
                                     f4(A,B,C)   = B        
                                                >= B        
                                                 = f4(1,B,D)
        
        
* Step 3: Failure MAYBE
  + Considered Problem:
      Rules:
        0. f0(A,B,C) -> f1(0,B,C)      True                                   (1,1)
        1. f1(A,B,C) -> f1(A,-1 + B,D) [-1*A >= 0 && A >= 0 && B >= 1]        (B,1)
        2. f1(A,B,C) -> f4(A,B,D)      [-1*A >= 0 && A >= 0 && 0 >= B]        (1,1)
        3. f4(A,B,C) -> f4(1,B,D)      [-1*B >= 0 && A + -1*B >= 0 && A >= 0] (?,1)
      Signature:
        {(f0,3);(f1,3);(f4,3)}
      Flow Graph:
        [0->{1,2},1->{1,2},2->{3},3->{3}]
      
  + Applied Processor:
      Failing "Open problems left."
  + Details:
      Open problems left.
MAYBE