YES(?,O(1))
* Step 1: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E) -> f7(F,0,0,D,E)         [0 >= 1 + F]                        (1,1)
          1. f0(A,B,C,D,E) -> f7(F,0,0,D,E)         [F >= 1]                            (1,1)
          2. f0(A,B,C,D,E) -> f7(0,1023,0,D,E)      True                                (1,1)
          3. f7(A,B,C,D,E) -> f7(A,B,1 + C,2 + D,E) [B >= C]                            (?,1)
          4. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [E >= 0 && C >= 1 + B && 1022 >= E] (?,1)
          5. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [C >= 1 + B && E >= 1023]           (?,1)
          6. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [C >= 1 + B && 0 >= 1 + E]          (?,1)
        Signature:
          {(f0,5);(f21,5);(f7,5)}
        Flow Graph:
          [0->{3,4,5,6},1->{3,4,5,6},2->{3,4,5,6},3->{3,4,5,6},4->{},5->{},6->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,4),(0,5),(0,6),(1,4),(1,5),(1,6),(2,4),(2,5),(2,6)]
* Step 2: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E) -> f7(F,0,0,D,E)         [0 >= 1 + F]                        (1,1)
          1. f0(A,B,C,D,E) -> f7(F,0,0,D,E)         [F >= 1]                            (1,1)
          2. f0(A,B,C,D,E) -> f7(0,1023,0,D,E)      True                                (1,1)
          3. f7(A,B,C,D,E) -> f7(A,B,1 + C,2 + D,E) [B >= C]                            (?,1)
          4. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [E >= 0 && C >= 1 + B && 1022 >= E] (?,1)
          5. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [C >= 1 + B && E >= 1023]           (?,1)
          6. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [C >= 1 + B && 0 >= 1 + E]          (?,1)
        Signature:
          {(f0,5);(f21,5);(f7,5)}
        Flow Graph:
          [0->{3},1->{3},2->{3},3->{3,4,5,6},4->{},5->{},6->{}]
        
    + 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) -> f7(F,0,0,D,E)         [0 >= 1 + F]                        (1,1)
          1. f0(A,B,C,D,E) -> f7(F,0,0,D,E)         [F >= 1]                            (1,1)
          2. f0(A,B,C,D,E) -> f7(0,1023,0,D,E)      True                                (1,1)
          3. f7(A,B,C,D,E) -> f7(A,B,1 + C,2 + D,E) [B >= C]                            (?,1)
          4. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [E >= 0 && C >= 1 + B && 1022 >= E] (1,1)
          5. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [C >= 1 + B && E >= 1023]           (1,1)
          6. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [C >= 1 + B && 0 >= 1 + E]          (1,1)
        Signature:
          {(f0,5);(f21,5);(f7,5)}
        Flow Graph:
          [0->{3},1->{3},2->{3},3->{3,4,5,6},4->{},5->{},6->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 1024          
          p(f21) = 1 + x2 + -1*x3
           p(f7) = 1 + x2 + -1*x3
        
        Following rules are strictly oriented:
               [B >= C] ==>                      
          f7(A,B,C,D,E)   = 1 + B + -1*C         
                          > B + -1*C             
                          = f7(A,B,1 + C,2 + D,E)
        
        
        Following rules are weakly oriented:
                               [0 >= 1 + F] ==>                 
                              f0(A,B,C,D,E)   = 1024            
                                             >= 1               
                                              = f7(F,0,0,D,E)   
        
                                   [F >= 1] ==>                 
                              f0(A,B,C,D,E)   = 1024            
                                             >= 1               
                                              = f7(F,0,0,D,E)   
        
                                       True ==>                 
                              f0(A,B,C,D,E)   = 1024            
                                             >= 1024            
                                              = f7(0,1023,0,D,E)
        
        [E >= 0 && C >= 1 + B && 1022 >= E] ==>                 
                              f7(A,B,C,D,E)   = 1 + B + -1*C    
                                             >= 1 + B + -1*C    
                                              = f21(A,B,C,D,E)  
        
                  [C >= 1 + B && E >= 1023] ==>                 
                              f7(A,B,C,D,E)   = 1 + B + -1*C    
                                             >= 1 + B + -1*C    
                                              = f21(A,B,C,D,E)  
        
                 [C >= 1 + B && 0 >= 1 + E] ==>                 
                              f7(A,B,C,D,E)   = 1 + B + -1*C    
                                             >= 1 + B + -1*C    
                                              = f21(A,B,C,D,E)  
        
        
* Step 4: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D,E) -> f7(F,0,0,D,E)         [0 >= 1 + F]                        (1,1)   
          1. f0(A,B,C,D,E) -> f7(F,0,0,D,E)         [F >= 1]                            (1,1)   
          2. f0(A,B,C,D,E) -> f7(0,1023,0,D,E)      True                                (1,1)   
          3. f7(A,B,C,D,E) -> f7(A,B,1 + C,2 + D,E) [B >= C]                            (1024,1)
          4. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [E >= 0 && C >= 1 + B && 1022 >= E] (1,1)   
          5. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [C >= 1 + B && E >= 1023]           (1,1)   
          6. f7(A,B,C,D,E) -> f21(A,B,C,D,E)        [C >= 1 + B && 0 >= 1 + E]          (1,1)   
        Signature:
          {(f0,5);(f21,5);(f7,5)}
        Flow Graph:
          [0->{3},1->{3},2->{3},3->{3,4,5,6},4->{},5->{},6->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))