IN/OUT Grouping - Game

[Yit HanS103]

I just started in/out grouping chapter 9 LG (5TH EDITION) PG 351

the game:
"A horse breeder is deciding which of seven horses N, O, P, R, S,T,V- to sell at auction this year. The breeder will determine which horses to sell according to the following restrictions:

rules:
1. R <--/--> P ( if R is sold auction, P is not: does it mean, if one of these two is IN the other one must be OUT, however can they both be out or no?)

2. O<--/-->N (Either O or N is sold at auction, BUT NOT BOTH: does this mean, O or N one must be in and the other one out?)

3. -S---> -R contra: R-->S

4. -T---> V (T is not sold at auction, V is.- what I know is that, when the negative in on the sufficient, that means one of those two must be in when the other one is absent, however they both can be in. )


*When I have a negative in the necessary, I do this kind of diagram ei: R <---/--> P is this correct ?
* I'd like to know what are the big difference between rule 1 and 2: I feel like I get it but I'm still a little confuse. when I see the inferences that manhattan makes in pg 352.
the inferences that I made in this game was:

in | out
R/P (in and out)
O/N (in and out)
R &S (in)
T/V (in)

Though, i'm not sure wether R/P should be shown in the inference as "out" as well, for example, I'm sure that O/N must be in and out, because rule2 says one or the other one, but rule 1 R/P says only when this rule is trigger.

Also, I was confused at first because in the previous chapters (ordering) when the game have conditional statements, MH suggest to not add them in the diagram. I think that changes then when you don't have ordering games?

Going back to this particular game, explanation part (from book)

pg 353 explanation for question #1
I got the right answer B, however, the book in this pg has B: r, s, t,v
when it should be p,r,t,v (as stated in the original game pg 351) Please, tell me if this was a book mistake or am I missing something.

pg. 356 explanation for question #4 - I did not answer and the explanation has me even more confused
"if either R or S is sold, but not both, which of the following could be a complete and accurate list of horses that are NOT sold?"

from rule # 3 we know the if R-->S
so in the IN group must be S, placing the R in the OUT group.
When R is out, then P must be In

The book says D is the answer:
I don't see D as being the answer because it has O, P, R as not sold and R/P cannot be together, or does R/P not being together is only in the IN group?

Thank you very much!

[ohthatpatrick]

These are a lot of questions for one post! I probably can't hit all of them. If you have any questions about a specific game / question, you should go to the thread for that test / section / question and see if it's already been answered.

If you suspect any mistake in the books, check the Errata page first:
https://www.manhattanprep.com/lsat/errata/

pg. 353 is indeed a typo.

There is a big difference between those first two rules:
1. R <--/--> P ( if R is sold auction, P is not: does it mean, if one of these two is IN the other one must be OUT, however can they both be out or no?)

2. O<--/-->N (Either O or N is sold at auction, BUT NOT BOTH: does this mean, O or N one must be in and the other one out?)

For the 2nd rule, it's saying that one is IN and one is OUT, so you could put an O/N placeholder in the IN and a complementary N/O placeholder in the OUT.
For the 1st rule, it's saying they can't both be IN, but they could both be OUT.

(To check: just put them both out and then read the rule to see if it violates the rule. As soon as we read "If R is sold ..." we think "Well, R is not sold in this scenario, so clearly this rule does not apply". If the rule doesn't apply to a scenario then you can't possibly break the rule with that scenario)

So you would NOT write it as some mutually exclusive anti-rule. You would just write it as it's written:
R --> ~P
P --> ~R

Since they can't both be IN, at least one is OUT. If you want, you can put a P/R+ placeholder in the OUT column.
(the "+" always means "at least", so this is reminding you that "AT LEAST ONE of them is out")

This answers your question for Q4.
P and R can certainly both be OUT, so there's no trouble with an answer putting them there.

Again, consider the scenario (D) describes ...
SOLD: S, T, V, ..... NOT SOLD: O, P, R

If you check this scenario against the rules, the first rule says
"If R is sold ..."
Stop right there. R is NOT being sold, so this rule immediately disappears from relevance.

[VendelaG465]

I had a question in regards to a problem from the LG textbook pg. 309 #5. I was trying to translate " F cannot be selected unless H is also selected." I had read it as F cannot be selected if not H but still translated it as -H---> -F . How is the correct translation F---> H? wouldn't we have to wait until H is selected to determine if F is in/selected?

[ohthatpatrick]

Those are the same thing.

-H---> -F
is the same thing as
F---> H

If you're not already in the habit of always writing the contrapositive to any conditionals you see, you might want to start doing so.

In this case, it would make it easier for you to see that you symbolized the rule correctly.

[OliviaF872]

I also have a question about this game.

One of the rules says "If Sam is not sold at auction, neither is Renegade."

I diagramed this as:
RULE: -S --> -R (i.e. if S is out, R is out)
CONTRAPOSITIVE: R --> S (i.e. if R is in, S is in)

But then Question 4 asks "If either R or S is sold, but not both, which of the following could be a complete and accurate list of horses that are not sold?"

Does this question stem not break the "If S is not sold, neither is R" rule? Does this mean it would have to be S that is sold in order for R to not be sold? Because if R was sold, the contrapositive rule is triggered and S would also be sold?

Thank you!!

[Laura Damone]

Hi OliviaF872! Please forgive the delayed response here. We had a technical issue that caused this post to get buried!

You are correct that in order to satisfy the condition that either S or R, but not both, are selected, it would have to be S in and R out, because R in and S out would violate the R --> S rule. Well done!