YES(?,O(1))
* Step 1: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J)         True                                                         (1,1)
          1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (?,1)
          2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J)     [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (?,1)
          3. f5(A,B,C,D,E,F,G,H,I,J) -> f27(A,B,C,D,B,B,K,L,L,L)        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 16] (?,1)
        Signature:
          {(f0,10);(f27,10);(f5,10)}
        Flow Graph:
          [0->{1,2,3},1->{1,2,3},2->{1,2,3},3->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,3)]
* Step 2: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J)         True                                                         (1,1)
          1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (?,1)
          2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J)     [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (?,1)
          3. f5(A,B,C,D,E,F,G,H,I,J) -> f27(A,B,C,D,B,B,K,L,L,L)        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 16] (?,1)
        Signature:
          {(f0,10);(f27,10);(f5,10)}
        Flow Graph:
          [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}]
        
    + 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,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J)         True                                                         (1,1)
          1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (?,1)
          2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J)     [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (?,1)
          3. f5(A,B,C,D,E,F,G,H,I,J) -> f27(A,B,C,D,B,B,K,L,L,L)        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 16] (1,1)
        Signature:
          {(f0,10);(f27,10);(f5,10)}
        Flow Graph:
          [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 16        
          p(f27) = 16 + -1*x3
           p(f5) = 16 + -1*x3
        
        Following rules are strictly oriented:
        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] ==>                            
                                             f5(A,B,C,D,E,F,G,H,I,J)   = 16 + -1*C                  
                                                                       > 15 + -1*C                  
                                                                       = f5(A,B,1 + C,0,E,F,G,H,I,J)
        
        
        Following rules are weakly oriented:
                                                                True ==>                                
                                             f0(A,B,C,D,E,F,G,H,I,J)   = 16                             
                                                                      >= 16                             
                                                                       = f5(K,0,0,D,E,F,G,H,I,J)        
        
        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] ==>                                
                                             f5(A,B,C,D,E,F,G,H,I,J)   = 16 + -1*C                      
                                                                      >= 15 + -1*C                      
                                                                       = f5(A,1 + B,1 + C,1,E,F,G,H,I,J)
        
        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 16] ==>                                
                                             f5(A,B,C,D,E,F,G,H,I,J)   = 16 + -1*C                      
                                                                      >= 16 + -1*C                      
                                                                       = f27(A,B,C,D,B,B,K,L,L,L)       
        
        
* Step 4: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J)         True                                                         (1,1) 
          1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (?,1) 
          2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J)     [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (16,1)
          3. f5(A,B,C,D,E,F,G,H,I,J) -> f27(A,B,C,D,B,B,K,L,L,L)        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 16] (1,1) 
        Signature:
          {(f0,10);(f27,10);(f5,10)}
        Flow Graph:
          [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 16        
          p(f27) = 16 + -1*x3
           p(f5) = 16 + -1*x3
        
        Following rules are strictly oriented:
        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] ==>                                
                                             f5(A,B,C,D,E,F,G,H,I,J)   = 16 + -1*C                      
                                                                       > 15 + -1*C                      
                                                                       = f5(A,1 + B,1 + C,1,E,F,G,H,I,J)
        
        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] ==>                                
                                             f5(A,B,C,D,E,F,G,H,I,J)   = 16 + -1*C                      
                                                                       > 15 + -1*C                      
                                                                       = f5(A,B,1 + C,0,E,F,G,H,I,J)    
        
        
        Following rules are weakly oriented:
                                                                True ==>                         
                                             f0(A,B,C,D,E,F,G,H,I,J)   = 16                      
                                                                      >= 16                      
                                                                       = f5(K,0,0,D,E,F,G,H,I,J) 
        
        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 16] ==>                         
                                             f5(A,B,C,D,E,F,G,H,I,J)   = 16 + -1*C               
                                                                      >= 16 + -1*C               
                                                                       = f27(A,B,C,D,B,B,K,L,L,L)
        
        
* Step 5: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E,F,G,H,I,J) -> f5(K,0,0,D,E,F,G,H,I,J)         True                                                         (1,1) 
          1. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,1 + B,1 + C,1,E,F,G,H,I,J) [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (16,1)
          2. f5(A,B,C,D,E,F,G,H,I,J) -> f5(A,B,1 + C,0,E,F,G,H,I,J)     [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && 15 >= C] (16,1)
          3. f5(A,B,C,D,E,F,G,H,I,J) -> f27(A,B,C,D,B,B,K,L,L,L)        [C >= 0 && B + C >= 0 && -1*B + C >= 0 && B >= 0 && C >= 16] (1,1) 
        Signature:
          {(f0,10);(f27,10);(f5,10)}
        Flow Graph:
          [0->{1,2},1->{1,2,3},2->{1,2,3},3->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))