YES(?,O(1))
* Step 1: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f3(0)     True                              (1,1)
          1. f3(A) -> f3(1 + A) [A >= 0 && 9 >= A]                (?,1)
          2. f3(A) -> f11(A)    [A >= 0 && A >= 10 && 0 >= 1 + B] (?,1)
          3. f3(A) -> f11(A)    [A >= 0 && A >= 10]               (?,1)
        Signature:
          {(f0,1);(f11,1);(f3,1)}
        Flow Graph:
          [0->{1,2,3},1->{1,2,3},2->{},3->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,2),(0,3)]
* Step 2: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f3(0)     True                              (1,1)
          1. f3(A) -> f3(1 + A) [A >= 0 && 9 >= A]                (?,1)
          2. f3(A) -> f11(A)    [A >= 0 && A >= 10 && 0 >= 1 + B] (?,1)
          3. f3(A) -> f11(A)    [A >= 0 && A >= 10]               (?,1)
        Signature:
          {(f0,1);(f11,1);(f3,1)}
        Flow Graph:
          [0->{1},1->{1,2,3},2->{},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) -> f3(0)     True                              (1,1)
          1. f3(A) -> f3(1 + A) [A >= 0 && 9 >= A]                (?,1)
          2. f3(A) -> f11(A)    [A >= 0 && A >= 10 && 0 >= 1 + B] (1,1)
          3. f3(A) -> f11(A)    [A >= 0 && A >= 10]               (1,1)
        Signature:
          {(f0,1);(f11,1);(f3,1)}
        Flow Graph:
          [0->{1},1->{1,2,3},2->{},3->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 10        
          p(f11) = 10 + -1*x1
           p(f3) = 10 + -1*x1
        
        Following rules are strictly oriented:
        [A >= 0 && 9 >= A] ==>          
                     f3(A)   = 10 + -1*A
                             > 9 + -1*A 
                             = f3(1 + A)
        
        
        Following rules are weakly oriented:
                                     True ==>          
                                    f0(A)   = 10       
                                           >= 10       
                                            = f3(0)    
        
        [A >= 0 && A >= 10 && 0 >= 1 + B] ==>          
                                    f3(A)   = 10 + -1*A
                                           >= 10 + -1*A
                                            = f11(A)   
        
                      [A >= 0 && A >= 10] ==>          
                                    f3(A)   = 10 + -1*A
                                           >= 10 + -1*A
                                            = f11(A)   
        
        
* Step 4: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A) -> f3(0)     True                              (1,1) 
          1. f3(A) -> f3(1 + A) [A >= 0 && 9 >= A]                (10,1)
          2. f3(A) -> f11(A)    [A >= 0 && A >= 10 && 0 >= 1 + B] (1,1) 
          3. f3(A) -> f11(A)    [A >= 0 && A >= 10]               (1,1) 
        Signature:
          {(f0,1);(f11,1);(f3,1)}
        Flow Graph:
          [0->{1},1->{1,2,3},2->{},3->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))