Setup and Rule Diagrams

[Jon Denning]

Thanks for the reply!

Yes! You've got it now. It was the absence of one or the other (or both) that triggered the conditional.

Thanks!

[LSAT2018]

For your next question concerning rule #4 (a trickier rule, in my opinion), we once again have a two-part statement: if either L or O is not in the park, then F and S are both in. The "not L" or "not O" sufficient conditions would allow us to represent this with two diagrams:

(1) L F and S ; contra: F or S L

(2) O F and S ; contra: F or S O

Do you see how these would also cover the condition where both L and O are not in? In that case it would still be clear that F and S were in the park. Just remember that multiple conditions with "or" or with "and" require that you change between the two as you take contrapositives. So "or" becomes "and" and the terms get negated, and vice versa.

Jon

I would be very grateful for a clarification on this! How does this rule apply to the case where both L and O are not in? So 'not the case that the park contains both laurels and oaks' means Not (L and O) which means Not L or Not O? Wouldn't this be the only possibility then?

Can I know how the language used in the fourth rule 'not the case that the park contains both laurels and oaks' differs from that in the third rule 'either laurels or oaks, but not both' ?

Either L or O (Possibilities include L, O, or both L and O)
Either L or O, but not both (Possibilities include L or O)
Not Both L and O (Possibilities include L or O, or none)

[LSAT2018]

With that in mind, let's consider the contrapositives of both statements I gave showing the two separate ideas in rule #3:

(1) Y L or O ; contrapositive: L and O Y

(2) Y L or O ; contrapositive: L and O Y

Essentially (1) tells us that if both L and O are missing, Y must be in the park (Y's absence would force one of L or O in). And (2) tells us that if both L and O are in the park, Y must be as well (Y's absence would mean they cannot both be there).

Jon

I also have a question on the first part. When creating a chain I notices that you used the second one Y L or O with the contrapositive: L and O Y. But wouldn't the conditional statement if yews are not in the park, then either laurels or oaks, but not both are in the park. not be complete because you are only taking into account the but not both part, and the Y L or O ; contrapositive: L and O Y is not used for the conditional chain. Seems like using an incomplete part of the statement. Any thoughts?


Thank you so much

[Adam Tyson]

To answer your first question, LSAT2018, if both L and O are out then the the sufficient condition has been met to require that F and S be included. If one or the other OR BOTH of L and O are out, then F and S are in. "Or" is an inclusive, rather than exclusive, term on the LSAT, meaning "at least one." So if the rule is "if L and O aren't both in, then F and S must be in," then either or both being out will meet that sufficient condition. I'm not sure what you mean about being the only possibility, but I will go through several possibilities below.

For the second question, I think you are overlooking that Jon diagrammed both implications of Y being out. If Y is out, then you must have exactly one of L or O in. If they are both out (his first diagram in your last question), or if they are both in (his second diagram there) then Y must be in.

So, here are some possibilities for L and O:

1. Both are in. In this case, Y is also in, and using the first rule, M is out. We know nothing about F or S or P, but no more than two of them could be in, both because of the 5 tree maximum and because F and P cannot be in together.

2. Both are out. In this case, Y is again in, M is out, and now per the last rule F and S must be in. F in knocks P out, and we have a complete solution of YFS in and LOMP out.

3. L is in and O is out. We don't know anything about Y in this case, because the necessary condition for the third rule has been satisfied, telling us only that the sufficient condition COULD occur. The fourth rule is triggered, forcing F and S in, and P out per the second rule. Since we don't know about Y, we also don't know about M. Either could be in, or both could be out.

4. O is in and L is out, and we get the identical results as #3 except for that swap. F and S are in, P is out, M and Y are unknown.

I hope that clarifies things for you. A rule making something sufficient for "one or the other but not both" leads to a contrapositive of "if both or neither" as described by Dave earlier in this thread (he used "if not exactly one"), and for many folks it could be easier to treat the contrapositive as two rules, as Jon did.

[amcrust]

For the third rule, you are correct in recognizing that it essentially produces two separate ideas:

(1) Not Y, then either L or O: Y L or O

(2) Not Y, then NOT L or NOT O: Y L or O (this is
the same as saying they cannot both be in the park)

I feel fairly sure I understand this, but this is the first time I've seen someone take a compound condition and negate it like this. I'm sure it's covered in the Bibles and would love to locate that if someone can point me in the right direction.
I did want to check, though -- is it because the rule states "but not both"? So "if not Y, then L or O" but since it's just one of those variables, it would be effectively the same to write "not Y, then not L or not O"?

Lastly, how do I know when to do that sort of negation? Is it just when it's a "but not both" scenario?

Thank you!
Anna

[Adam Tyson]

It is indeed the "but not both" that turns this into a more complex conditional, Anna. Y being out requires that EXACTLY ONE of L or O be in, so if they are both in, or if they are both out, then Y must be in. You could also view that as "if Y is out, then either L is in and O is out or else O is in and L is out."

Because, you know, conditional reasoning isn't challenging enough without them throwing these kind of curveballs at us, am I right?

Good job, you've got this!

(Oh, and it might be the first time you have seen it, but it probably won't be the last time, because LSAC has done a LOT of these "but not both" conditional rules over the years. Keep your eyes peeled for them!

[lsatstudent5555]

Is this a good game to use the 'identify the templates' method?

I've done this game perfectly (twice) without it, but I think it took me too long (it took me 37 mins to do the entire LG section the second time around)

I went back and looked at the rules to see if I could identify templates, and what I came up with is (in | out):

L O Y F/P S/ | M F/P /S (italics denote that they don't have to be in - it makes sense to me but I understand if it's confusing to others)

F S L O Y | M P

F S L/O M/Y | M/Y P L/O

F S Y | M P L O

Did I miss anything? Or is there a reason why this wouldn't be a good approach to this game? Thanks!

[Robert Carroll]

l,

I think the number of templates is too large to be particularly worthwhile in this game. For instance, your third template doesn't seem to capture all the options: you have one of M/Y in, the other out, but can't both be out? The first template you list includes the second template as an option, so that second template is superfluous, so you'd either want to list out all the options for the first template, or just leave it what it is. Being left as it is means the first template is pretty loose; being filled in with more options increases the number of templates, including the ones that would have to come from fully listing the options for the third.

A good number of the questions are locally anyway, meaning you'd end up doing diagrams for them, so even a person not doing templates should be able to answer them fairly well.

Robert Carroll

[thecooperperspective]

LSAC Lawhub Superprep I PrepTest B February 1999
GAME 2 Trees in Park

I am not sure about my inference(s). Usually, if I diagram the rules and inferences correctly, I do well.

Question(s):
10. MBT: If firs are not in the park, then which one of the following must be true?

Process of Elimination (POE): Answer: A) F  M

I am sure I missed some inferences. I do not know how to arrive at this answer other than by POE?

Trees: FLMOPSY, Max 5/7

RULES:
R1: M + Y (BOTH COULD BE OUT)
R2: F + P (BOTH COULD BE OUT)
R3: Y  L or O (NOT BOTH), Contrapositive: L and O  Y
R4: L or O  F and S, Contrapositive: F or S  L and O

INFERENCE(S):
I1: (R1, R3): L and O + M

[Adam Tyson]

The best way to answer Question 10, thecooperperspective, is to create a conditional chain based on the local restriction of F being out (if you hadn't already done so as part of your initial diagram). Here's how that would work:

If F is out, that triggers the contrapositive of the last rule. This means that L and O must both be in.

Having both of them in triggers the contrapositive of the third rule. This means that Y must be in.

This triggers the contrapositive of the first rule, which means that M is out.

P and S are randoms. Either or both could be in or out. Thus, we have this solution:

In: LOY

Out: FM

Random: PS

Thus, answer A must be true. No Process of Elimination required, it's all dictated by the conditional rules!

B is wrong because S is random and could be in or out.

C is wrong because Y must be in.

D is wrong because P is random, like S.

E is wrong because S is random.

Create those conditional chains, and inferences like this will come faster and easier, and you won't need to test answer choices!