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# PT34.S4.G4 - each of exactly six doctors

Alum Member
edited December 2016 161 karma
Hi,

I am beyond confused on one of the logic games
Each of exactly six doctors- Juarez, Kudrow, Longtree, Nance, Onawa, and Palermo is at exactly one of two clinics: Souderton or Randsborough. The following conditions must be satisfied:
Kudrow is at Randsborough if Juarez is at Souderton
Onawa is at Souderton if Juarez is at Randsborough
If Longtree is at Souderton then both Nance and Palerno are at Randsborough
If Nance is at Randsborough, then so is Onawa
If Palerno is at Randsborough, then both Kudrow and Onawa are at Souderton

So @"J.Y. Ping" solved it one way

L-> /N --> /O --> J --> /K --> P
--> /P

In order to get rid of the contradiction with the P's he negated /L and got
/N-->/O --> J--> /K--> P

However, I decided to find the contrapositive of the first statement and got

/P--> K--> /J --> O --> N --> /L
P ---> /L

But I have no idea how to remove the contradicting P's because L is already negated. I'm so confused regarding what to do next?
I feel like I'm missing a basic logic concept but I tried going back to the material and I still dont get it.
https://7sage.com/lsat_explanations/lsat-34-section-4-game-4/

• Alum Member
edited December 2016 2426 karma
I don't remember all the details on this game but from what you wrote, I think you would just put L out, since P in L is out, P out L is also out, then might as well just put L out, and when necessary satisfied, rule goes away, which would free up P, leaving the sequence chain of /P--> K--> /J --> O --> N. The key point is L cannot be in, for if L is in, run the contrapositive, we will get a contradiction of P again, which we are just drawing circles there.

I personally think it's easier/clearer the way J.Y. did it, but it's perfectly okay if you decided to run the contrapositive version, logically it's the same thing.
• Alum Member Sage 7Sage Tutor
10731 karma
Hey,

So lets focus on your contrapositive. We now know that L is always out because its negated. Also because there is no way to keep L "in", if you do you are going to have a contradiction with P. So L is always out.

So from our lessons if you remember when the necessary condition is satisfied, the rule that its attached to becomes irrelevant. So now that we have /L, anything that /L is attached to as a necessary condition just "falls way". Because of that, we only have the following left:
/P--> K--> /J --> O --> N

For this one: P ---> /L, we know that necessary is satisfied since we always have a /L so the rule becomes irrelevant. We can cross this out. P ---> /L

That's it.

The only chain that matters is this: /P--> K--> /J --> O --> N ; unless there are other ones in the game; I didn't go back to the game here and I am just working with the chains you gave me.

so that's it. you only have one chain to work with now, and depending on what's going on , P can be in or out based on this chain.

I hope this helped.
• Alum Member 🍌🍌
8684 karma
I think this is a very important game to be familiar with. The presence of the contradiction signals that LSAC expects us to recognize when it one present and what that means for our gameboard. I will be drilling this game several times form until June.