YES(?,O(1))
* Step 1: UnreachableRules WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f8(A,B,C,D)  -> f8(-1 + A,B,C,D)  [A >= 0]     (?,1)
          1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [B >= 0]     (?,1)
          2. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [C >= 0]     (?,1)
          3. f28(A,B,C,D) -> f36(A,B,C,D)      [0 >= 1 + C] (?,1)
          4. f19(A,B,C,D) -> f28(A,B,999,D)    [0 >= 1 + B] (?,1)
          5. f0(A,B,C,D)  -> f19(A,999,C,1)    True         (1,1)
          6. f8(A,B,C,D)  -> f19(A,999,C,D)    [0 >= 1 + A] (?,1)
        Signature:
          {(f0,4);(f19,4);(f28,4);(f36,4);(f8,4)}
        Flow Graph:
          [0->{0,6},1->{1,4},2->{2,3},3->{},4->{2,3},5->{1,4},6->{1,4}]
        
    + Applied Processor:
        UnreachableRules
    + Details:
        Following transitions are not reachable from the starting states and are revomed: [0,6]
* Step 2: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [B >= 0]     (?,1)
          2. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [C >= 0]     (?,1)
          3. f28(A,B,C,D) -> f36(A,B,C,D)      [0 >= 1 + C] (?,1)
          4. f19(A,B,C,D) -> f28(A,B,999,D)    [0 >= 1 + B] (?,1)
          5. f0(A,B,C,D)  -> f19(A,999,C,1)    True         (1,1)
        Signature:
          {(f0,4);(f19,4);(f28,4);(f36,4);(f8,4)}
        Flow Graph:
          [1->{1,4},2->{2,3},3->{},4->{2,3},5->{1,4}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(4,3),(5,4)]
* Step 3: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [B >= 0]     (?,1)
          2. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [C >= 0]     (?,1)
          3. f28(A,B,C,D) -> f36(A,B,C,D)      [0 >= 1 + C] (?,1)
          4. f19(A,B,C,D) -> f28(A,B,999,D)    [0 >= 1 + B] (?,1)
          5. f0(A,B,C,D)  -> f19(A,999,C,1)    True         (1,1)
        Signature:
          {(f0,4);(f19,4);(f28,4);(f36,4);(f8,4)}
        Flow Graph:
          [1->{1,4},2->{2,3},3->{},4->{2},5->{1}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 4: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [B >= 0]     (?,1)
          2. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [C >= 0]     (?,1)
          3. f28(A,B,C,D) -> f36(A,B,C,D)      [0 >= 1 + C] (1,1)
          4. f19(A,B,C,D) -> f28(A,B,999,D)    [0 >= 1 + B] (1,1)
          5. f0(A,B,C,D)  -> f19(A,999,C,1)    True         (1,1)
        Signature:
          {(f0,4);(f19,4);(f28,4);(f36,4);(f8,4)}
        Flow Graph:
          [1->{1,4},2->{2,3},3->{},4->{2},5->{1}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 1000  
          p(f19) = 1000  
          p(f28) = 1 + x3
          p(f36) = 1 + x3
        
        Following rules are strictly oriented:
              [C >= 0] ==>                  
          f28(A,B,C,D)   = 1 + C            
                         > C                
                         = f28(A,B,-1 + C,D)
        
        
        Following rules are weakly oriented:
              [B >= 0] ==>                  
          f19(A,B,C,D)   = 1000             
                        >= 1000             
                         = f19(A,-1 + B,C,D)
        
          [0 >= 1 + C] ==>                  
          f28(A,B,C,D)   = 1 + C            
                        >= 1 + C            
                         = f36(A,B,C,D)     
        
          [0 >= 1 + B] ==>                  
          f19(A,B,C,D)   = 1000             
                        >= 1000             
                         = f28(A,B,999,D)   
        
                  True ==>                  
           f0(A,B,C,D)   = 1000             
                        >= 1000             
                         = f19(A,999,C,1)   
        
        
* Step 5: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [B >= 0]     (?,1)   
          2. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [C >= 0]     (1000,1)
          3. f28(A,B,C,D) -> f36(A,B,C,D)      [0 >= 1 + C] (1,1)   
          4. f19(A,B,C,D) -> f28(A,B,999,D)    [0 >= 1 + B] (1,1)   
          5. f0(A,B,C,D)  -> f19(A,999,C,1)    True         (1,1)   
        Signature:
          {(f0,4);(f19,4);(f28,4);(f36,4);(f8,4)}
        Flow Graph:
          [1->{1,4},2->{2,3},3->{},4->{2},5->{1}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 1000  
          p(f19) = 1 + x2
          p(f28) = 1 + x2
          p(f36) = 1 + x2
        
        Following rules are strictly oriented:
              [B >= 0] ==>                  
          f19(A,B,C,D)   = 1 + B            
                         > B                
                         = f19(A,-1 + B,C,D)
        
        
        Following rules are weakly oriented:
              [C >= 0] ==>                  
          f28(A,B,C,D)   = 1 + B            
                        >= 1 + B            
                         = f28(A,B,-1 + C,D)
        
          [0 >= 1 + C] ==>                  
          f28(A,B,C,D)   = 1 + B            
                        >= 1 + B            
                         = f36(A,B,C,D)     
        
          [0 >= 1 + B] ==>                  
          f19(A,B,C,D)   = 1 + B            
                        >= 1 + B            
                         = f28(A,B,999,D)   
        
                  True ==>                  
           f0(A,B,C,D)   = 1000             
                        >= 1000             
                         = f19(A,999,C,1)   
        
        
* Step 6: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          1. f19(A,B,C,D) -> f19(A,-1 + B,C,D) [B >= 0]     (1000,1)
          2. f28(A,B,C,D) -> f28(A,B,-1 + C,D) [C >= 0]     (1000,1)
          3. f28(A,B,C,D) -> f36(A,B,C,D)      [0 >= 1 + C] (1,1)   
          4. f19(A,B,C,D) -> f28(A,B,999,D)    [0 >= 1 + B] (1,1)   
          5. f0(A,B,C,D)  -> f19(A,999,C,1)    True         (1,1)   
        Signature:
          {(f0,4);(f19,4);(f28,4);(f36,4);(f8,4)}
        Flow Graph:
          [1->{1,4},2->{2,3},3->{},4->{2},5->{1}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))