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P(si )(1 P(si )) = 2 2
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When gentle persuasion proves unsuccessful, it is time to take up the cudgel. Having identified multiple infringers, the first question that must be addressed is which infringer(s) to sue first. The recommendation is often made, more often by well-intentioned laypersons (here a euphemism for idiots; known to lawyers as officious intermeddlers ) than by experienced attorneys, to first tackle the biggest and strongest infringer. The theory here is that, after successfully defeating such an infringer, all others will fall into line, taking licenses without further fuss. Do not listen to such people. Such advice is tantamount to butting one s head against a wall in preference to opening a door and walking through. Choose the weakest opponent, not the strongest one; choose the one most likely to settle, not the one most likely (and able) to resist. While there is invariably great reluctance on the part of most business executives to be the first to take a license under a patent, or patent portfolio, there is much less reluctance to be the second. As the number
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as mentioned above, the other two cases, i.e., parents(Ai+1 ) Wi Zi and parents(Ai+1 ) Yi Zi , are analogous. Since, as indicated above, X X|Uk | , Y Y|Uk | and Z = Z|Uk | , we can finally derive X Y | Z by (at most) two applications of the decomposition axiom to the statement X|Uk | W|Uk | Y|Uk | | Z|Uk | , thus completing the proof of the first part of the theorem. The second part of the theorem is much easier to prove than the first. Since we already know from the first part of the theorem that the local and the global Markov property are equivalent (recall that the semi-graphoid axioms are a subset of the graphoid axioms), it suffices to show that the pairwise and the local Markov property are equivalent. That the local Markov property implies the pairwise can be seen from the fact that the parents of a node are a subset of its non-descendants and thus we get any pairwise conditional independence statement for a node by applying the weak union axiom (cf. the observations made in the paragraph following Definition 4.1.19 on page 109). To show the other direction, we start from an arbitrary pairwise conditional independence statement A B | nondescs(A) {B}. Then we apply the intersection axiom, drawing on other pairwise conditional independence statements involving A, in order to shift attributes out of the separating set. Eventually only the parents of A remain and thus we have the desired local conditional independence statement.
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Conductivity of Solutions of Symmetrical Strong Electrolytes at Moderate to High Concentrations
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TIP: If the soon-to-be-married couple is going to have a prenuptial agreement to address the concerns of the parents making the gift of money for a down payment, be certain that both members of the couple have their own lawyers (never use one lawyer) review the agreement. Prenuptial agreements can have important legal (and emotional) ramif ications and should not be taken lightly.
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EJB & JSP: Java On The Edge, Unlimited Edition ISBN: 0764548026 by Lou Marco Your Guide to Cutting-Edge J2EE Programming Techniques.
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