YES(?,O(1))
* Step 1: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D) -> f8(0,B,C,D)         True                             (1,1)
          1. f8(A,B,C,D) -> f8(1 + A,B,C,D)     [A >= 0 && 3 >= A]               (?,1)
          2. f8(A,B,C,D) -> f8(1 + A,A,1 + A,E) [A >= 0 && 3 >= A]               (?,1)
          3. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4 && 0 >= 1 + E] (?,1)
          4. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4]               (?,1)
        Signature:
          {(f0,4);(f23,4);(f8,4)}
        Flow Graph:
          [0->{1,2,3,4},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,3),(0,4)]
* Step 2: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D) -> f8(0,B,C,D)         True                             (1,1)
          1. f8(A,B,C,D) -> f8(1 + A,B,C,D)     [A >= 0 && 3 >= A]               (?,1)
          2. f8(A,B,C,D) -> f8(1 + A,A,1 + A,E) [A >= 0 && 3 >= A]               (?,1)
          3. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4 && 0 >= 1 + E] (?,1)
          4. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4]               (?,1)
        Signature:
          {(f0,4);(f23,4);(f8,4)}
        Flow Graph:
          [0->{1,2},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        
    + 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) -> f8(0,B,C,D)         True                             (1,1)
          1. f8(A,B,C,D) -> f8(1 + A,B,C,D)     [A >= 0 && 3 >= A]               (?,1)
          2. f8(A,B,C,D) -> f8(1 + A,A,1 + A,E) [A >= 0 && 3 >= A]               (?,1)
          3. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4 && 0 >= 1 + E] (1,1)
          4. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4]               (1,1)
        Signature:
          {(f0,4);(f23,4);(f8,4)}
        Flow Graph:
          [0->{1,2},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 10        
          p(f23) = 10 + -3*x1
           p(f8) = 10 + -3*x1
        
        Following rules are strictly oriented:
        [A >= 0 && 3 >= A] ==>                    
               f8(A,B,C,D)   = 10 + -3*A          
                             > 7 + -3*A           
                             = f8(1 + A,A,1 + A,E)
        
        
        Following rules are weakly oriented:
                                    True ==>                
                             f0(A,B,C,D)   = 10             
                                          >= 10             
                                           = f8(0,B,C,D)    
        
                      [A >= 0 && 3 >= A] ==>                
                             f8(A,B,C,D)   = 10 + -3*A      
                                          >= 7 + -3*A       
                                           = f8(1 + A,B,C,D)
        
        [A >= 0 && A >= 4 && 0 >= 1 + E] ==>                
                             f8(A,B,C,D)   = 10 + -3*A      
                                          >= 10 + -3*A      
                                           = f23(A,B,C,D)   
        
                      [A >= 0 && A >= 4] ==>                
                             f8(A,B,C,D)   = 10 + -3*A      
                                          >= 10 + -3*A      
                                           = f23(A,B,C,D)   
        
        
* Step 4: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D) -> f8(0,B,C,D)         True                             (1,1) 
          1. f8(A,B,C,D) -> f8(1 + A,B,C,D)     [A >= 0 && 3 >= A]               (?,1) 
          2. f8(A,B,C,D) -> f8(1 + A,A,1 + A,E) [A >= 0 && 3 >= A]               (10,1)
          3. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4 && 0 >= 1 + E] (1,1) 
          4. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4]               (1,1) 
        Signature:
          {(f0,4);(f23,4);(f8,4)}
        Flow Graph:
          [0->{1,2},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 4        
          p(f23) = 4 + -1*x1
           p(f8) = 4 + -1*x1
        
        Following rules are strictly oriented:
        [A >= 0 && 3 >= A] ==>                    
               f8(A,B,C,D)   = 4 + -1*A           
                             > 3 + -1*A           
                             = f8(1 + A,B,C,D)    
        
        [A >= 0 && 3 >= A] ==>                    
               f8(A,B,C,D)   = 4 + -1*A           
                             > 3 + -1*A           
                             = f8(1 + A,A,1 + A,E)
        
        
        Following rules are weakly oriented:
                                    True ==>             
                             f0(A,B,C,D)   = 4           
                                          >= 4           
                                           = f8(0,B,C,D) 
        
        [A >= 0 && A >= 4 && 0 >= 1 + E] ==>             
                             f8(A,B,C,D)   = 4 + -1*A    
                                          >= 4 + -1*A    
                                           = f23(A,B,C,D)
        
                      [A >= 0 && A >= 4] ==>             
                             f8(A,B,C,D)   = 4 + -1*A    
                                          >= 4 + -1*A    
                                           = f23(A,B,C,D)
        
        
* Step 5: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B,C,D) -> f8(0,B,C,D)         True                             (1,1) 
          1. f8(A,B,C,D) -> f8(1 + A,B,C,D)     [A >= 0 && 3 >= A]               (4,1) 
          2. f8(A,B,C,D) -> f8(1 + A,A,1 + A,E) [A >= 0 && 3 >= A]               (10,1)
          3. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4 && 0 >= 1 + E] (1,1) 
          4. f8(A,B,C,D) -> f23(A,B,C,D)        [A >= 0 && A >= 4]               (1,1) 
        Signature:
          {(f0,4);(f23,4);(f8,4)}
        Flow Graph:
          [0->{1,2},1->{1,2,3,4},2->{1,2,3,4},3->{},4->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))