MAYBE
* Step 1: TrivialSCCs MAYBE
    + Considered Problem:
        Rules:
          0. f2(A,B,C) -> f2(1 + A,B,C)  [-1 + C >= 0 && -1 + A + C >= 0 && A >= 0]                                        (?,1)
          1. f3(A,B,C) -> f3(A,-1 + B,C) [-1*C >= 0 && A + -1*C >= 0 && -1*A + -1*C >= 0 && -1*A >= 0 && A >= 0 && B >= 1] (?,1)
          2. f5(A,B,C) -> f5(A,B,1)      [-1*B >= 0 && A + -1*B >= 0 && -1*A + -1*B >= 0 && -1*A >= 0 && A >= 0]           (?,1)
          3. f0(A,B,C) -> f2(0,B,C)      [C >= 1]                                                                          (1,1)
          4. f0(A,B,C) -> f3(0,B,C)      [0 >= C]                                                                          (1,1)
          5. f3(A,B,C) -> f5(0,B,C)      [-1*C >= 0 && A + -1*C >= 0 && -1*A + -1*C >= 0 && -1*A >= 0 && A >= 0 && 0 >= B] (?,1)
        Signature:
          {(f0,3);(f2,3);(f3,3);(f5,3)}
        Flow Graph:
          [0->{0},1->{1,5},2->{2},3->{0},4->{1,5},5->{2}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: PolyRank MAYBE
    + Considered Problem:
        Rules:
          0. f2(A,B,C) -> f2(1 + A,B,C)  [-1 + C >= 0 && -1 + A + C >= 0 && A >= 0]                                        (?,1)
          1. f3(A,B,C) -> f3(A,-1 + B,C) [-1*C >= 0 && A + -1*C >= 0 && -1*A + -1*C >= 0 && -1*A >= 0 && A >= 0 && B >= 1] (?,1)
          2. f5(A,B,C) -> f5(A,B,1)      [-1*B >= 0 && A + -1*B >= 0 && -1*A + -1*B >= 0 && -1*A >= 0 && A >= 0]           (?,1)
          3. f0(A,B,C) -> f2(0,B,C)      [C >= 1]                                                                          (1,1)
          4. f0(A,B,C) -> f3(0,B,C)      [0 >= C]                                                                          (1,1)
          5. f3(A,B,C) -> f5(0,B,C)      [-1*C >= 0 && A + -1*C >= 0 && -1*A + -1*C >= 0 && -1*A >= 0 && A >= 0 && 0 >= B] (1,1)
        Signature:
          {(f0,3);(f2,3);(f3,3);(f5,3)}
        Flow Graph:
          [0->{0},1->{1,5},2->{2},3->{0},4->{1,5},5->{2}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(f0) = x2
          p(f2) = x2
          p(f3) = x2
          p(f5) = x2
        
        Following rules are strictly oriented:
        [-1*C >= 0 && A + -1*C >= 0 && -1*A + -1*C >= 0 && -1*A >= 0 && A >= 0 && B >= 1] ==>               
                                                                                f3(A,B,C)   = B             
                                                                                            > -1 + B        
                                                                                            = f3(A,-1 + B,C)
        
        
        Following rules are weakly oriented:
                                               [-1 + C >= 0 && -1 + A + C >= 0 && A >= 0] ==>              
                                                                                f2(A,B,C)   = B            
                                                                                           >= B            
                                                                                            = f2(1 + A,B,C)
        
                  [-1*B >= 0 && A + -1*B >= 0 && -1*A + -1*B >= 0 && -1*A >= 0 && A >= 0] ==>              
                                                                                f5(A,B,C)   = B            
                                                                                           >= B            
                                                                                            = f5(A,B,1)    
        
                                                                                 [C >= 1] ==>              
                                                                                f0(A,B,C)   = B            
                                                                                           >= B            
                                                                                            = f2(0,B,C)    
        
                                                                                 [0 >= C] ==>              
                                                                                f0(A,B,C)   = B            
                                                                                           >= B            
                                                                                            = f3(0,B,C)    
        
        [-1*C >= 0 && A + -1*C >= 0 && -1*A + -1*C >= 0 && -1*A >= 0 && A >= 0 && 0 >= B] ==>              
                                                                                f3(A,B,C)   = B            
                                                                                           >= B            
                                                                                            = f5(0,B,C)    
        
        
* Step 3: Failure MAYBE
  + Considered Problem:
      Rules:
        0. f2(A,B,C) -> f2(1 + A,B,C)  [-1 + C >= 0 && -1 + A + C >= 0 && A >= 0]                                        (?,1)
        1. f3(A,B,C) -> f3(A,-1 + B,C) [-1*C >= 0 && A + -1*C >= 0 && -1*A + -1*C >= 0 && -1*A >= 0 && A >= 0 && B >= 1] (B,1)
        2. f5(A,B,C) -> f5(A,B,1)      [-1*B >= 0 && A + -1*B >= 0 && -1*A + -1*B >= 0 && -1*A >= 0 && A >= 0]           (?,1)
        3. f0(A,B,C) -> f2(0,B,C)      [C >= 1]                                                                          (1,1)
        4. f0(A,B,C) -> f3(0,B,C)      [0 >= C]                                                                          (1,1)
        5. f3(A,B,C) -> f5(0,B,C)      [-1*C >= 0 && A + -1*C >= 0 && -1*A + -1*C >= 0 && -1*A >= 0 && A >= 0 && 0 >= B] (1,1)
      Signature:
        {(f0,3);(f2,3);(f3,3);(f5,3)}
      Flow Graph:
        [0->{0},1->{1,5},2->{2},3->{0},4->{1,5},5->{2}]
      
  + Applied Processor:
      Failing "Open problems left."
  + Details:
      Open problems left.
MAYBE