Post
by soj » Sun Apr 10, 2011 9:09 am
According to Rule 4, exactly two of LMR must be reduced. The question says R is not reduced. Thus, L & M are reduced.
According to the contrapositive of Rule 3, because L is reduced, P is not reduced.
We now know two areas that are not reduced (P & R). There is exactly one more area that is not reduced, and that area must be N or S. Why? Because according to Rule 2, N and S cannot both be reduced.
Since R, P, and N/S are not reduced, G, L, M, W, and S/N are reduced. Thus, (A) is correct. Notice that GLMWN does not violate any rules, so you can be confident in eliminating (D).
Another quick way of eliminating (D) is by realizing right off the bat that N and S cannot both be reduced, and therefore, if (D) were correct, then (B) must also be correct. Since there can never be two correct answers, (D) is incorrect. Notice you cannot eliminate (B) for the same reason, however. To do so would be a Mistaken Reversal of the conditional statement above. Rule 2 alone does not prevent you from reducing neither N nor S. If this paragraph doesn't make sense to you, don't worry about it. Checking for ACs that imply each other is a useful but not necessary method in eliminating ACs.
