Biconditions on in/out vs. grouping games

itsemmarobyn

Hey friends, quick clarification question for you #help.

In 17.01.G4, the rule "J is on the same team as K," translates as a forever together biconditional, where J ↔ K. Similarly, the second rule, "K is not on the same team as N," translates as a forever apart biconditional, K ↔ /N. In his explanation (see link below), it seemed like J.Y. was implying that this was because there were only two groups for the pieces to be sorted into.

Does this mean that a similar rule would not translate as a biconditional if there were multiple groups? If so, someone please explain this to me, because I don't understand why it couldn't be a biconditional in both game types.

17.01.G4 video explanation: https://7sage.com/lsat_explanations/lsat-17-section-1-game-4/

Jonathan Wang

K <-> /N breaks down into two statements: K -> /N and /N -> K. As with all conditional statements, executing these is not optional - when you trigger the sufficient, you are stuck with the necessary.

K -> /N is not a problem - when I see K, I won't see N. Fine. I will only see K once, so that category can't have N in it. That jives with my understanding of "always apart" no matter how many groups there are.

BUT, /N -> K says that any time you don't have N, you have K. In two groups, that's not a problem; the group without N is necessarily the one with K. But if there are three groups, you'll end up with one N and two groups without N, meaning you're going to have to place K in each of those groups according to the rule, which is obviously not correct. In four groups, it's even worse, and so forth.

In three groups, the rule should be K -> /N, because that indicates properly that in the one place where you see K, you won't see N, and the contrapositive says that in the one place that you see N, you won't see K. This is the proper manifestation of "always apart" in 3+ group scenarios.

Is it confusing that you have to swap from a biconditional in a two group scenario into a single conditional in 3+ groups? Sure, but that's just how it is. Two group setups have special characteristics that go away once you go to 3 or more groups. Rather than bang your head against a table trying to remember that, though, I'd suggest instead that there are probably easier and more intuitive ways for you to represent that concept than to immediately jump to formal conditional logic.

Hope this helps.

itsemmarobyn

@"Jonathan Wang" said:

BUT, /N -> K says that any time you don't have N, you have K. In two groups, that's not a problem; the group without N is necessarily the one with K. But if there are three groups, you'll end up with one N and two groups without N, meaning you're going to have to place K in each of those groups according to the rule, which is obviously not correct. In four groups, it's even worse, and so forth.

In three groups, the rule should be K -> /N, because that indicates properly that in the one place where you see K, you won't see N, and the contrapositive says that in the one place that you see N, you won't see K. This is the proper manifestation of "always apart" in 3+ group scenarios.
Hope this helps.

This was super helpful, thanks so much!