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# PT34.S4.G4 - biconditional

Alum Member
edited January 2020 71 karma

Basically, my issue is with the whole explanation from JY. But for discussion sake let's focus on Q23.

If you read the question paragraph or question stem without reading Q23, you will figure out that J and K are bidirectional i.e. cannot be in the same clinic together (proof below).

So my question is why is then Q23 can have A too as the right answer choice.

Proof
s and r are locations

Js -> Kr
~Kr -> ~Js
Ks -> Jr (because only s and r are there)

Therefore, Ks <-> Jr

• Yearly Sage
edited January 2020 6839 karma

Js -> Kr and Ks -> Jr does not form a biconditional statement. What you've done here is translate the same rule two different ways, by taking a contrapositive (which ends up with the same statement by definition), and then translating that contrapositive into positive terms (which still makes the same statement, since it's just a translation). At the end of the day, all you've done here is say the same thing three different ways.

You're missing the /Js -> /Kr (or Kr->Js, if you prefer) side of the puzzle.

• Alum Member
edited January 2020 71 karma

@Jonathan Wang, My bad. It is not a biconditional.

But, they cannot be in the same group right from my proof ?

https://www.manhattanprep.com/lsat/forums/diagram-t268.html#p482

• Yearly Sage
edited January 2020 6839 karma

Nope. Your proof starts with Js->Kr, which is given. Fine. Then you contrapose: ~Kr -> ~Js. Great. Then, you translate that contrapositive into Ks -> Jr. Still fine. But you haven't actually changed anything - you've taken a contrapositive (no change in meaning), and then translated the contrapositive (still no change in meaning). You still have a single directional conditional statement that triggers on either Js or ~Kr, and you have no idea what happens when you start with ~Js or Kr.

Put another way, the "always apart" statement Ks <-> Jr requires two components: Ks -> Jr and Jr -> Ks. You know that Ks -> Jr. Where is your Jr -> Ks component?

Or, take this scenario -

Souderton: ONP
Randsborough: JKL

First rule: Js -> Kr. Js isn't happening, so this rule is irrelevant and is discarded.
Second rule: Jr -> Os. J is in R, so O is in S. This rule is satisfied.
Third rule: Ls -> Nr and Pr. Ls isn't happening, so this rule is irrelevant and is discarded.
Fourth rule: Nr -> Or. Nr isn't happening, so this rule is irrelevant and is discarded.
Fifth rule: Pr -> Ks and Os. Pr isn't happening, so this rule is irrelevant and is discarded.

This example shows that J and K can be together in Randsborough with no violation of your rules. J and K cannot be in Souderton together, but that is very different than saying that they cannot ever be together.

One more way to look at it - let's say I told you: If I go to McDonald's, you must go to Burger King. Is it true then that we can't ever end up in the same store? What if I went to Burger King?