The assignment of technicians to computers is fully determined if which one of the following is true?

Hi, D is the correct answer. However, the diagram drawn in the Real Time is false. Matt Shiners place S in F; but S cannot be in F when/if X is not in G per the contrapositive to the first rule. I am concerned that I might be missing something fundamental, otherwise I am correct that Matt Shiners incorrectly inferred S going into F when/if both T and W are assigned to G per answer-choice D. To be clear, here is my analysis; and please correct me if I am wrong: Per answer choice D, T and W examine G; thereby leaving no more for anyone else, including X, because the setup of the game states that "each computer will be examined by exactly two technicians." So X can no longer examine G. Rule one states that if S examines F, X examines G. Accordingly, the contrapositive to rule one is: if X does not examine G, then S does not examine F. And since, per answer-choice D, G is at full capacity of technicians for being examined by T and W; it follows that X does not examine G, which triggers S not examining F. Accordingly, the only place for S to go is in H. So far, we have T and W examining G, and S examining H. Per rule three, R and T must be together. R and T cannot go in G or H because there is no room for the both of them; therefore, R and T must go in F. In retrospect, looking at the diagram which would look like this, F: R and T G: T and W H: S and ? We determine that T is the one doubling up, and X has not examined any computers. The setup states that "each technician will be assigned to at least one technician." So X must examine at least one computer, it has examined none, and the only space for it to go is in H. Consequently, the assignment is now fully determined with X coupling with S, both of whom examining H: F: R and T G: T and W H: S and X Would an expert please tell me if I missed something here? Thank you Mazen