YES(?,O(1))
* Step 1: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B)  -> f4(0,B)      True                   (1,1)
          1. f4(A,B)  -> f4(1 + A,B)  [1 >= A]               (?,1)
          2. f10(A,B) -> f10(A,1 + B) [1 >= B]               (?,1)
          3. f10(A,B) -> f18(A,B)     [B >= 2 && 0 >= 1 + C] (?,1)
          4. f10(A,B) -> f18(A,B)     [B >= 2]               (?,1)
          5. f4(A,B)  -> f10(A,0)     [A >= 2]               (?,1)
        Signature:
          {(f0,2);(f10,2);(f18,2);(f4,2)}
        Flow Graph:
          [0->{1,5},1->{1,5},2->{2,3,4},3->{},4->{},5->{2,3,4}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,5),(5,3),(5,4)]
* Step 2: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B)  -> f4(0,B)      True                   (1,1)
          1. f4(A,B)  -> f4(1 + A,B)  [1 >= A]               (?,1)
          2. f10(A,B) -> f10(A,1 + B) [1 >= B]               (?,1)
          3. f10(A,B) -> f18(A,B)     [B >= 2 && 0 >= 1 + C] (?,1)
          4. f10(A,B) -> f18(A,B)     [B >= 2]               (?,1)
          5. f4(A,B)  -> f10(A,0)     [A >= 2]               (?,1)
        Signature:
          {(f0,2);(f10,2);(f18,2);(f4,2)}
        Flow Graph:
          [0->{1},1->{1,5},2->{2,3,4},3->{},4->{},5->{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)  -> f4(0,B)      True                   (1,1)
          1. f4(A,B)  -> f4(1 + A,B)  [1 >= A]               (?,1)
          2. f10(A,B) -> f10(A,1 + B) [1 >= B]               (?,1)
          3. f10(A,B) -> f18(A,B)     [B >= 2 && 0 >= 1 + C] (1,1)
          4. f10(A,B) -> f18(A,B)     [B >= 2]               (1,1)
          5. f4(A,B)  -> f10(A,0)     [A >= 2]               (1,1)
        Signature:
          {(f0,2);(f10,2);(f18,2);(f4,2)}
        Flow Graph:
          [0->{1},1->{1,5},2->{2,3,4},3->{},4->{},5->{2}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 2        
          p(f10) = 2 + -1*x2
          p(f18) = 2 + -1*x2
           p(f4) = 2        
        
        Following rules are strictly oriented:
          [1 >= B] ==>             
          f10(A,B)   = 2 + -1*B    
                     > 1 + -1*B    
                     = f10(A,1 + B)
        
        
        Following rules are weakly oriented:
                          True ==>            
                       f0(A,B)   = 2          
                                >= 2          
                                 = f4(0,B)    
        
                      [1 >= A] ==>            
                       f4(A,B)   = 2          
                                >= 2          
                                 = f4(1 + A,B)
        
        [B >= 2 && 0 >= 1 + C] ==>            
                      f10(A,B)   = 2 + -1*B   
                                >= 2 + -1*B   
                                 = f18(A,B)   
        
                      [B >= 2] ==>            
                      f10(A,B)   = 2 + -1*B   
                                >= 2 + -1*B   
                                 = f18(A,B)   
        
                      [A >= 2] ==>            
                       f4(A,B)   = 2          
                                >= 2          
                                 = f10(A,0)   
        
        
* Step 4: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B)  -> f4(0,B)      True                   (1,1)
          1. f4(A,B)  -> f4(1 + A,B)  [1 >= A]               (?,1)
          2. f10(A,B) -> f10(A,1 + B) [1 >= B]               (2,1)
          3. f10(A,B) -> f18(A,B)     [B >= 2 && 0 >= 1 + C] (1,1)
          4. f10(A,B) -> f18(A,B)     [B >= 2]               (1,1)
          5. f4(A,B)  -> f10(A,0)     [A >= 2]               (1,1)
        Signature:
          {(f0,2);(f10,2);(f18,2);(f4,2)}
        Flow Graph:
          [0->{1},1->{1,5},2->{2,3,4},3->{},4->{},5->{2}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 2        
          p(f10) = -1*x1    
          p(f18) = -1*x1    
           p(f4) = 2 + -1*x1
        
        Following rules are strictly oriented:
         [1 >= A] ==>            
          f4(A,B)   = 2 + -1*A   
                    > 1 + -1*A   
                    = f4(1 + A,B)
        
        
        Following rules are weakly oriented:
                          True ==>             
                       f0(A,B)   = 2           
                                >= 2           
                                 = f4(0,B)     
        
                      [1 >= B] ==>             
                      f10(A,B)   = -1*A        
                                >= -1*A        
                                 = f10(A,1 + B)
        
        [B >= 2 && 0 >= 1 + C] ==>             
                      f10(A,B)   = -1*A        
                                >= -1*A        
                                 = f18(A,B)    
        
                      [B >= 2] ==>             
                      f10(A,B)   = -1*A        
                                >= -1*A        
                                 = f18(A,B)    
        
                      [A >= 2] ==>             
                       f4(A,B)   = 2 + -1*A    
                                >= -1*A        
                                 = f10(A,0)    
        
        
* Step 5: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B)  -> f4(0,B)      True                   (1,1)
          1. f4(A,B)  -> f4(1 + A,B)  [1 >= A]               (2,1)
          2. f10(A,B) -> f10(A,1 + B) [1 >= B]               (2,1)
          3. f10(A,B) -> f18(A,B)     [B >= 2 && 0 >= 1 + C] (1,1)
          4. f10(A,B) -> f18(A,B)     [B >= 2]               (1,1)
          5. f4(A,B)  -> f10(A,0)     [A >= 2]               (1,1)
        Signature:
          {(f0,2);(f10,2);(f18,2);(f4,2)}
        Flow Graph:
          [0->{1},1->{1,5},2->{2,3,4},3->{},4->{},5->{2}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))