YES(?,O(n^1))
* Step 1: UnsatPaths WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B) -> div(A,B)        True                   (1,1)
          1. div(A,B)   -> div(A,-1*A + B) [B >= 1 + A && A >= 1] (?,1)
          2. div(A,B)   -> end(A,B)        [A >= B]               (?,1)
          3. div(A,B)   -> end(A,B)        [0 >= A]               (?,1)
        Signature:
          {(div,2);(end,2);(start,2)}
        Flow Graph:
          [0->{1,2,3},1->{1,2,3},2->{},3->{}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(1,3)]
* Step 2: TrivialSCCs WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B) -> div(A,B)        True                   (1,1)
          1. div(A,B)   -> div(A,-1*A + B) [B >= 1 + A && A >= 1] (?,1)
          2. div(A,B)   -> end(A,B)        [A >= B]               (?,1)
          3. div(A,B)   -> end(A,B)        [0 >= A]               (?,1)
        Signature:
          {(div,2);(end,2);(start,2)}
        Flow Graph:
          [0->{1,2,3},1->{1,2},2->{},3->{}]
        
    + Applied Processor:
        TrivialSCCs
    + Details:
        All trivial SCCs of the transition graph admit timebound 1.
* Step 3: PolyRank WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B) -> div(A,B)        True                   (1,1)
          1. div(A,B)   -> div(A,-1*A + B) [B >= 1 + A && A >= 1] (?,1)
          2. div(A,B)   -> end(A,B)        [A >= B]               (1,1)
          3. div(A,B)   -> end(A,B)        [0 >= A]               (1,1)
        Signature:
          {(div,2);(end,2);(start,2)}
        Flow Graph:
          [0->{1,2,3},1->{1,2},2->{},3->{}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
            p(div) = x2
            p(end) = x2
          p(start) = x2
        
        Following rules are strictly oriented:
        [B >= 1 + A && A >= 1] ==>                
                      div(A,B)   = B              
                                 > -1*A + B       
                                 = div(A,-1*A + B)
        
        
        Following rules are weakly oriented:
                True ==>         
          start(A,B)   = B       
                      >= B       
                       = div(A,B)
        
            [A >= B] ==>         
            div(A,B)   = B       
                      >= B       
                       = end(A,B)
        
            [0 >= A] ==>         
            div(A,B)   = B       
                      >= B       
                       = end(A,B)
        
        
* Step 4: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. start(A,B) -> div(A,B)        True                   (1,1)
          1. div(A,B)   -> div(A,-1*A + B) [B >= 1 + A && A >= 1] (B,1)
          2. div(A,B)   -> end(A,B)        [A >= B]               (1,1)
          3. div(A,B)   -> end(A,B)        [0 >= A]               (1,1)
        Signature:
          {(div,2);(end,2);(start,2)}
        Flow Graph:
          [0->{1,2,3},1->{1,2},2->{},3->{}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(n^1))