#8 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8627)

The correct answer choice is (B)

As described earlier, when P is reduced, L cannot be reduced. When L is not reduced, M and R must be reduced, and hence answer choice (B) is correct.

[mgoss]

Hi,

Why wouldn't D also be correct? If P is selected, then L must not be. If L is not, then M and R must be (which is the correct answer). But following the logic chain, If R is selected (which it must be), then N must not be (due to rule 2). If N is not, then S must be. Therefore why would not answer choice D be correct as well.

I am confused because the double not arrows are as follows:

P L
N S
N R

Am I interpreting rule 2 incorrectly?

[Jeremy Press]

Hi,

Yes, you're right--you've incorrectly interpreted the rule governing the relationship between N and S. It's true that if N is selected, then S is not selected. But the logic of that rule allows for the possibility that neither N nor S is selected. In other words, just because N is NOT selected, doesn't mean S IS selected. S not being selected is the necessary condition of the rule, and a necessary condition can occur with or without its sufficient condition occurring. Thus S not being selected can occur when N is selected, but (and this is the key) it can also happen when N is not selected.

I hope this helps!

Jeremy

[mgoss]

Thanks, makes sense now.