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General questions relating to the LSAT Logic Games.
 LustingFor!L
  • Posts: 80
  • Joined: Aug 27, 2016
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#38012
I just want to confirm the following rule would be diagrammed as a double arrow:

"If A is not interviewed then B must be interviewed, and if B is interviewed than A cannot be interviewed"

A :arrow: B
B :arrow: A

A :dbl: B

And that diagram means that A will always be with B? So, because this is an in and out grouping game they will never be in the same group?
 LustingFor!L
  • Posts: 80
  • Joined: Aug 27, 2016
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#38013
Also, this is from the 2017 Edition LSAT Logic Games Bible Workbook in the practice drills on page 12 of Chapter One.
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 Dave Killoran
PowerScore Staff
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#38014
Hi Lusting,

Think of it more as when ever B is around (B), A won't be there (A). So, in a two group game of any type, if B shows up in a group, A is out. and, via the contrapositive of the underlying rules, whenever a appears in a group, B won't be there. A and B literally will not appear in the same group together.

Thanks!

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