MAYBE
* Step 1: UnsatPaths MAYBE
    + Considered Problem:
        Rules:
          0. f300(A,B,C,D) -> f1(A,B,C,E)       [A >= B]                   (?,1)
          1. f300(A,B,C,D) -> f300(A,B,0,D)     [B >= 1 + A]               (?,1)
          2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1)
          3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A]     (?,1)
          4. f2(A,B,C,D)   -> f300(A,B,C,D)     True                       (1,1)
        Signature:
          {(f1,4);(f2,4);(f300,4)}
        Flow Graph:
          [0->{},1->{0,1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(1,0)]
* Step 2: TrivialSCCs MAYBE
    + Considered Problem:
        Rules:
          0. f300(A,B,C,D) -> f1(A,B,C,E)       [A >= B]                   (?,1)
          1. f300(A,B,C,D) -> f300(A,B,0,D)     [B >= 1 + A]               (?,1)
          2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1)
          3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A]     (?,1)
          4. f2(A,B,C,D)   -> f300(A,B,C,D)     True                       (1,1)
        Signature:
          {(f1,4);(f2,4);(f300,4)}
        Flow Graph:
          [0->{},1->{1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 3: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f300(A,B,C,D) -> f1(A,B,C,E)       [A >= B]                   (1,1)
          1. f300(A,B,C,D) -> f300(A,B,0,D)     [B >= 1 + A]               (?,1)
          2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1)
          3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A]     (?,1)
          4. f2(A,B,C,D)   -> f300(A,B,C,D)     True                       (1,1)
        Signature:
          {(f1,4);(f2,4);(f300,4)}
        Flow Graph:
          [0->{},1->{1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}]
        
    + 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:
        [E >= 1 && B >= 1 + A] ==>                  
                 f300(A,B,C,D)   = -1*A + B         
                                 > -1 + -1*A + B    
                                 = f300(1 + A,B,E,D)
        
        
        Following rules are weakly oriented:
                          [A >= B] ==>                  
                     f300(A,B,C,D)   = -1*A + B         
                                    >= -1*A + B         
                                     = f1(A,B,C,E)      
        
                      [B >= 1 + A] ==>                  
                     f300(A,B,C,D)   = -1*A + B         
                                    >= -1*A + B         
                                     = f300(A,B,0,D)    
        
        [0 >= 1 + E && B >= 1 + A] ==>                  
                     f300(A,B,C,D)   = -1*A + B         
                                    >= -1 + -1*A + B    
                                     = f300(1 + A,B,E,D)
        
                              True ==>                  
                       f2(A,B,C,D)   = -1*A + B         
                                    >= -1*A + B         
                                     = f300(A,B,C,D)    
        
        
* Step 4: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f300(A,B,C,D) -> f1(A,B,C,E)       [A >= B]                   (1,1)    
          1. f300(A,B,C,D) -> f300(A,B,0,D)     [B >= 1 + A]               (?,1)    
          2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (?,1)    
          3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A]     (A + B,1)
          4. f2(A,B,C,D)   -> f300(A,B,C,D)     True                       (1,1)    
        Signature:
          {(f1,4);(f2,4);(f300,4)}
        Flow Graph:
          [0->{},1->{1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}]
        
    + 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:
        [0 >= 1 + E && B >= 1 + A] ==>                  
                     f300(A,B,C,D)   = -1*A + B         
                                     > -1 + -1*A + B    
                                     = f300(1 + A,B,E,D)
        
            [E >= 1 && B >= 1 + A] ==>                  
                     f300(A,B,C,D)   = -1*A + B         
                                     > -1 + -1*A + B    
                                     = f300(1 + A,B,E,D)
        
        
        Following rules are weakly oriented:
               [A >= B] ==>              
          f300(A,B,C,D)   = -1*A + B     
                         >= -1*A + B     
                          = f1(A,B,C,E)  
        
           [B >= 1 + A] ==>              
          f300(A,B,C,D)   = -1*A + B     
                         >= -1*A + B     
                          = f300(A,B,0,D)
        
                   True ==>              
            f2(A,B,C,D)   = -1*A + B     
                         >= -1*A + B     
                          = f300(A,B,C,D)
        
        
* Step 5: Failure MAYBE
  + Considered Problem:
      Rules:
        0. f300(A,B,C,D) -> f1(A,B,C,E)       [A >= B]                   (1,1)    
        1. f300(A,B,C,D) -> f300(A,B,0,D)     [B >= 1 + A]               (?,1)    
        2. f300(A,B,C,D) -> f300(1 + A,B,E,D) [0 >= 1 + E && B >= 1 + A] (A + B,1)
        3. f300(A,B,C,D) -> f300(1 + A,B,E,D) [E >= 1 && B >= 1 + A]     (A + B,1)
        4. f2(A,B,C,D)   -> f300(A,B,C,D)     True                       (1,1)    
      Signature:
        {(f1,4);(f2,4);(f300,4)}
      Flow Graph:
        [0->{},1->{1,2,3},2->{0,1,2,3},3->{0,1,2,3},4->{0,1,2,3}]
      
  + Applied Processor:
      Failing "Open problems left."
  + Details:
      Open problems left.
MAYBE