PTB.S2.G2 - Embedded conditionals and taking the contrapositive

Bomber Boi

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://7sage.com/lsat_explanations/lsat-b-section-2-game-2/

IISAC1007Aa

Please specify PT prep test #/ Date, and question # please.:)

Bomber Boi

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)

Clemens_

/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).

Clemens_

Here is a visualization attempt, hope this is helpful: https://imgur.com/bnf7gta