YES(?,O(1))
* Step 1: TrivialSCCs WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B) -> f3(1,B)             True                     (1,1)
          1. f3(A,B) -> f3(1 + A,10 + -1*A) [-1 + A >= 0 && 10 >= A] (?,1)
          2. f3(A,B) -> f10(A,B)            [-1 + A >= 0 && A >= 11] (?,1)
        Signature:
          {(f0,2);(f10,2);(f3,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 2: UnsatPaths WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B) -> f3(1,B)             True                     (1,1)
          1. f3(A,B) -> f3(1 + A,10 + -1*A) [-1 + A >= 0 && 10 >= A] (?,1)
          2. f3(A,B) -> f10(A,B)            [-1 + A >= 0 && A >= 11] (1,1)
        Signature:
          {(f0,2);(f10,2);(f3,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,2)]
* Step 3: PolyRank WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B) -> f3(1,B)             True                     (1,1)
          1. f3(A,B) -> f3(1 + A,10 + -1*A) [-1 + A >= 0 && 10 >= A] (?,1)
          2. f3(A,B) -> f10(A,B)            [-1 + A >= 0 && A >= 11] (1,1)
        Signature:
          {(f0,2);(f10,2);(f3,2)}
        Flow Graph:
          [0->{1},1->{1,2},2->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
           p(f0) = 10        
          p(f10) = 11 + -1*x1
           p(f3) = 11 + -1*x1
        
        Following rules are strictly oriented:
        [-1 + A >= 0 && 10 >= A] ==>                    
                         f3(A,B)   = 11 + -1*A          
                                   > 10 + -1*A          
                                   = f3(1 + A,10 + -1*A)
        
        
        Following rules are weakly oriented:
                            True ==>          
                         f0(A,B)   = 10       
                                  >= 10       
                                   = f3(1,B)  
        
        [-1 + A >= 0 && A >= 11] ==>          
                         f3(A,B)   = 11 + -1*A
                                  >= 11 + -1*A
                                   = f10(A,B) 
        
        
* Step 4: KnowledgePropagation WORST_CASE(?,O(1))
    + Considered Problem:
        Rules:
          0. f0(A,B) -> f3(1,B)             True                     (1,1) 
          1. f3(A,B) -> f3(1 + A,10 + -1*A) [-1 + A >= 0 && 10 >= A] (10,1)
          2. f3(A,B) -> f10(A,B)            [-1 + A >= 0 && A >= 11] (1,1) 
        Signature:
          {(f0,2);(f10,2);(f3,2)}
        Flow Graph:
          [0->{1},1->{1,2},2->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(1))