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So for this tricky SA question ...
I was looking for a statement that would say “D”
A and D ——> M
Not M ——-> Not A or Not D
Conclusion: Not M --> Not A
Would a way to conclude Not A is if we have D?
Would this be logically sound?
Can another right AC be just D? Why did they decide to go with A -> D?
Any explanation would be greatly appreciated.
Thanks!
Admin note: edited title and added link
https://classic.7sage.com/lsat_explanations/lsat-71-section-3-question-11/
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1 comments
Yeah, you're right that establishing D would result in /A. We can't have both of them, so if we establish that we do have D, we must kick A out. That would be sufficient, but it's not the only way. What they chose to do instead is to connect the two in a way that no matter which one you don't have, you don't have A. If you kick A out, your conclusion is good. What's the alternative? Kicking out D. So the correct AC gives us if not D then not A. Now, no matter which one we choose to boot, we're covered. If A is out then A is out. And if D is out, A is out.
It's a clever question, not quite as cookie cutter as we're used to. That's becoming something of a hallmark of the contemporary test though, so definitely worth studying.