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Hello all. This is from the trees game in prep test B. When applying the rules for embedded conditionals from the core curriculum, I come up with the following:
/Y → (L↔/O) = /Y and L ↔/O This could also be read as: /Y and O ↔/L
Contrapositive: O↔/L or Y Is this correct? It doesn't seem to make sense in the context of the game.
However, in the explanation of the game, the contrapositive is treated as a forever together biconditional
(L ↔ O) → Y or (/L↔/O) → Y
What am I missing here?
Admin Note: https://classic.7sage.com/lsat_explanations/lsat-b-section-2-game-2/
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Here is a visualization attempt, hope this is helpful: https://imgur.com/bnf7gta
/Y -> (O ↔/L)
If Y is out, then either O is in or L is in, but not both.
How do we falsify "either O is in or L is in, but not both?" Either O and L are both in, or O and L are both out. Formally:
(O & L) -> Y
(/O and /L) -> Y
There are two different options here to negate the necessary condition in the brackets, and to derive Y from there.
On '/Y → (L↔/O) = /Y and L ↔/O.' This seems false to me. (/Y and L) does necessitate /O, but I don't see why /O should also necessitate (/Y and L). /O could also work if you have e.g. (Y and L).
This is Prep Test B, Feb 1999, Game 2.
It's not about a specific question, rather, how the contrapositive is reached from the following: /Y → (L↔/O)
Please specify PT prep test #/ Date, and question # please.:)