MAYBE
* Step 1: UnsatPaths MAYBE
    + Considered Problem:
        Rules:
          0. f1(A,B,C) -> f2(A,B,C)     True                   (1,1)
          1. f2(A,B,C) -> f300(A,B,D)   [A = B]                (?,1)
          2. f2(A,B,C) -> f2(1 + A,B,C) [A >= 1 + B && A >= B] (?,1)
          3. f2(A,B,C) -> f2(1 + A,B,C) [B >= 1 + A]           (?,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3},1->{},2->{1,2,3},3->{1,2,3}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(2,1),(2,3),(3,2)]
* Step 2: TrivialSCCs MAYBE
    + Considered Problem:
        Rules:
          0. f1(A,B,C) -> f2(A,B,C)     True                   (1,1)
          1. f2(A,B,C) -> f300(A,B,D)   [A = B]                (?,1)
          2. f2(A,B,C) -> f2(1 + A,B,C) [A >= 1 + B && A >= B] (?,1)
          3. f2(A,B,C) -> f2(1 + A,B,C) [B >= 1 + A]           (?,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3},1->{},2->{2},3->{1,3}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 3: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f1(A,B,C) -> f2(A,B,C)     True                   (1,1)
          1. f2(A,B,C) -> f300(A,B,D)   [A = B]                (1,1)
          2. f2(A,B,C) -> f2(1 + A,B,C) [A >= 1 + B && A >= B] (?,1)
          3. f2(A,B,C) -> f2(1 + A,B,C) [B >= 1 + A]           (?,1)
        Signature:
          {(f1,3);(f2,3);(f300,3)}
        Flow Graph:
          [0->{1,2,3},1->{},2->{2},3->{1,3}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
            p(f1) = -1*x1 + x2
            p(f2) = -1*x1 + x2
          p(f300) = -1*x1 + x2
        
        Following rules are strictly oriented:
        [B >= 1 + A] ==>              
           f2(A,B,C)   = -1*A + B     
                       > -1 + -1*A + B
                       = f2(1 + A,B,C)
        
        
        Following rules are weakly oriented:
                          True ==>              
                     f1(A,B,C)   = -1*A + B     
                                >= -1*A + B     
                                 = f2(A,B,C)    
        
                       [A = B] ==>              
                     f2(A,B,C)   = -1*A + B     
                                >= -1*A + B     
                                 = f300(A,B,D)  
        
        [A >= 1 + B && A >= B] ==>              
                     f2(A,B,C)   = -1*A + B     
                                >= -1 + -1*A + B
                                 = f2(1 + A,B,C)
        
        
* Step 4: Failure MAYBE
  + Considered Problem:
      Rules:
        0. f1(A,B,C) -> f2(A,B,C)     True                   (1,1)    
        1. f2(A,B,C) -> f300(A,B,D)   [A = B]                (1,1)    
        2. f2(A,B,C) -> f2(1 + A,B,C) [A >= 1 + B && A >= B] (?,1)    
        3. f2(A,B,C) -> f2(1 + A,B,C) [B >= 1 + A]           (A + B,1)
      Signature:
        {(f1,3);(f2,3);(f300,3)}
      Flow Graph:
        [0->{1,2,3},1->{},2->{2},3->{1,3}]
      
  + Applied Processor:
      Failing "Open problems left."
  + Details:
      Open problems left.
MAYBE