YES(?,O(1))
* Step 1: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C)  -> f8(D,0,C)      True                                                    (1,1)
          1. f8(A,B,C)  -> f8(A,1 + B,C)  [B >= 0 && 9 >= B]                                      (?,1)
          2. f19(A,B,C) -> f19(A,B,1 + C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C]  (?,1)
          3. f19(A,B,C) -> f29(A,B,C)     [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] (?,1)
          4. f8(A,B,C)  -> f19(A,B,0)     [B >= 0 && B >= 10]                                     (?,1)
        Signature:
          {(f0,3);(f19,3);(f29,3);(f8,3)}
        Flow Graph:
          [0->{1,4},1->{1,4},2->{2,3},3->{},4->{2,3}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,4),(4,3)]
* Step 2: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C)  -> f8(D,0,C)      True                                                    (1,1)
          1. f8(A,B,C)  -> f8(A,1 + B,C)  [B >= 0 && 9 >= B]                                      (?,1)
          2. f19(A,B,C) -> f19(A,B,1 + C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C]  (?,1)
          3. f19(A,B,C) -> f29(A,B,C)     [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] (?,1)
          4. f8(A,B,C)  -> f19(A,B,0)     [B >= 0 && B >= 10]                                     (?,1)
        Signature:
          {(f0,3);(f19,3);(f29,3);(f8,3)}
        Flow Graph:
          [0->{1},1->{1,4},2->{2,3},3->{},4->{2}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 3: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C)  -> f8(D,0,C)      True                                                    (1,1)
          1. f8(A,B,C)  -> f8(A,1 + B,C)  [B >= 0 && 9 >= B]                                      (?,1)
          2. f19(A,B,C) -> f19(A,B,1 + C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C]  (?,1)
          3. f19(A,B,C) -> f29(A,B,C)     [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] (1,1)
          4. f8(A,B,C)  -> f19(A,B,0)     [B >= 0 && B >= 10]                                     (1,1)
        Signature:
          {(f0,3);(f19,3);(f29,3);(f8,3)}
        Flow Graph:
          [0->{1},1->{1,4},2->{2,3},3->{},4->{2}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 10        
          p(f19) = 10 + -1*x3
          p(f29) = 10 + -1*x3
           p(f8) = 10        
        
        Following rules are strictly oriented:
        [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C] ==>               
                                                    f19(A,B,C)   = 10 + -1*C     
                                                                 > 9 + -1*C      
                                                                 = f19(A,B,1 + C)
        
        
        Following rules are weakly oriented:
                                                           True ==>              
                                                      f0(A,B,C)   = 10           
                                                                 >= 10           
                                                                  = f8(D,0,C)    
        
                                             [B >= 0 && 9 >= B] ==>              
                                                      f8(A,B,C)   = 10           
                                                                 >= 10           
                                                                  = f8(A,1 + B,C)
        
        [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] ==>              
                                                     f19(A,B,C)   = 10 + -1*C    
                                                                 >= 10 + -1*C    
                                                                  = f29(A,B,C)   
        
                                            [B >= 0 && B >= 10] ==>              
                                                      f8(A,B,C)   = 10           
                                                                 >= 10           
                                                                  = f19(A,B,0)   
        
        
* Step 4: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C)  -> f8(D,0,C)      True                                                    (1,1) 
          1. f8(A,B,C)  -> f8(A,1 + B,C)  [B >= 0 && 9 >= B]                                      (?,1) 
          2. f19(A,B,C) -> f19(A,B,1 + C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C]  (10,1)
          3. f19(A,B,C) -> f29(A,B,C)     [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] (1,1) 
          4. f8(A,B,C)  -> f19(A,B,0)     [B >= 0 && B >= 10]                                     (1,1) 
        Signature:
          {(f0,3);(f19,3);(f29,3);(f8,3)}
        Flow Graph:
          [0->{1},1->{1,4},2->{2,3},3->{},4->{2}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 10        
          p(f19) = 10 + -1*x2
          p(f29) = 10 + -1*x2
           p(f8) = 10 + -1*x2
        
        Following rules are strictly oriented:
        [B >= 0 && 9 >= B] ==>              
                 f8(A,B,C)   = 10 + -1*B    
                             > 9 + -1*B     
                             = f8(A,1 + B,C)
        
        
        Following rules are weakly oriented:
                                                           True ==>               
                                                      f0(A,B,C)   = 10            
                                                                 >= 10            
                                                                  = f8(D,0,C)     
        
         [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C] ==>               
                                                     f19(A,B,C)   = 10 + -1*B     
                                                                 >= 10 + -1*B     
                                                                  = f19(A,B,1 + C)
        
        [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] ==>               
                                                     f19(A,B,C)   = 10 + -1*B     
                                                                 >= 10 + -1*B     
                                                                  = f29(A,B,C)    
        
                                            [B >= 0 && B >= 10] ==>               
                                                      f8(A,B,C)   = 10 + -1*B     
                                                                 >= 10 + -1*B     
                                                                  = f19(A,B,0)    
        
        
* Step 5: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C)  -> f8(D,0,C)      True                                                    (1,1) 
          1. f8(A,B,C)  -> f8(A,1 + B,C)  [B >= 0 && 9 >= B]                                      (10,1)
          2. f19(A,B,C) -> f19(A,B,1 + C) [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && 9 >= C]  (10,1)
          3. f19(A,B,C) -> f29(A,B,C)     [C >= 0 && -10 + B + C >= 0 && -10 + B >= 0 && C >= 10] (1,1) 
          4. f8(A,B,C)  -> f19(A,B,0)     [B >= 0 && B >= 10]                                     (1,1) 
        Signature:
          {(f0,3);(f19,3);(f29,3);(f8,3)}
        Flow Graph:
          [0->{1},1->{1,4},2->{2,3},3->{},4->{2}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))