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A --> B
-B --> -A
Is the following possible (could be true)?
-B --> -A --s-- B
So basically, is it a possibility (could be true) for the necessary condition to sometimes be the logical opposite of the sufficient condition?
Sorry if this seems stupid or even not pertinent to the lsat, but I'd really like to know.
Or maybe a better to ask this(?):
If -B --> -A, then is -B --> -A AND B necessarily false?
I hope I'm making sense
Edit:
I guess "sometimes" would not fit because of the 100% possibility sometimes implies. So I guess -B --> -A --x--> B could be true. ( --x--> meaning not necessarily, aka 0-99 % possibility)
Comments
Assuming A and B aren't identical sets(it is not the case that all As are Bs and all Bs are As) but A--->B then -B--->-A<--s-->B has to be valid
A logical opposite is defined as: for all things in set A, the logical opposite is everything not in set A
A--->B implies set B is larger than set A since set B contains all of set A and things not included in set A
Anything not included in set A is the logical opposite of set A and since set B has things not in set A, set B contains some things that are the logical opposite of set A.
Therefore A--->B<--S-->-A and
-B--->-A<--s-->B
For the second part of your comment, -B--->-A AND B, can never be the case. The simple answer is -B can never imply B since they're logical opposites. I'm assuming you made the infrence that because -B--->-A and -A<--s-->B, -B--->-B. But you can't make an inference that jumps from all to some. Hope that helps
My take is that -B --> -A --x--> B is MBF, but lol I've been wrong before. Would like to see what others say.
I tried to draw it out with circles and the only possibility I can see of it being true is if B means slightly different things? [-B --> -A --x--> Bx].
@BlueRiceCake ooooh that's gud
Thanks! I like to think in terms of circles too when dealing with conditional reasoning