#7 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=8548)

The correct answer choice is (E)

If H is in the forest, then G cannot be in the forest. If G cannot be in the forest, then W cannot be in the forest. At this point it has been established that H and M are in the forest and G and W are not in the forest. The only unaddressed birds are S and J, and at least one of them, possibly both, must be in the forest. Answer choices (A) and (B) are therefore incorrect because it is possible that both S and J can be in the forest. Answer choice (C) is incorrect because, due to the final rule, the forest always contains at least S or J. Answer choice (D) is incorrect because it is possible there is only one other kind of bird in the forest (S or J). Answer choice (E) is thus correct because at most S and J can be in the forest in addition to M and H.

[ascott64]

On page 251 of the most recent Bible, Question #7 says that answer choice E is correct..."because the only birds unaddressed are S and J, and at least one of them, possibly both, must be in the forest."

I'm utterly confused by this because one of the rules specifically states that if J is not in the forest, then S is. The contra positive of this rule is diagrammed and shows that if S is not in the forest, then J is. So how can it be that they are both possibly in the forest as a correct answer choice? This seems like a direct violation of the rules and so I just need some clarity of what I must have missed that will allow this rule to be dismissed.

Second, in the rule diagramming why is are only S and J shown to have contrapositives and none of the other rules, such as H and G? (It says if H is in, then G is out, but there is only a diagram showing H and G can't be together (the double arrow with a slash in the middle). What am I missing here? This is really frustrating me. Help! Anybody, somebody!

[Steve Stein]

When it comes to conditional reasoning, if the arrows of a diagram go in only one direction, you cannot go "against the grain." In other words, if the rule is
A B, then what can we tell from B's presence? nothing. If A is there, B will be there, but the arrow only goes in one direction.

On the same note, if J is not there, then S is:

NOT J S

So, if S is there, what can we tell about J? Nothing.

NOT S J

If J is there, what can we tell about S? Nothing.

So, it is indeed possibly that S and J can show up together (they can be present together, but they just can be absent together).


As for the double not arrow, that really spells out two sides of a conditional relationship. So, if C D, then we know that C and D don't get along. In other words, we know that if C is there, D won't be. And if D is there, C won't be:
C NOT D and D NOT C

Let me know whether that clears things up at all--thanks!

~Steve

[Laurah916]

Hello,

I have a question from the December 2000 test on LG number 7. It's on page 323 in the PowerScore LG Bible. Why would the answer be E? I do not understand how S and J can both be in the forest if rule number 4 says, "If Jay's are not in the forest, then shrikes are." Does that not mean you have to choose S or J? Why can you choose both if it goes against the rule? Any insight would be great! Thank you!

[Dave Killoran]

Note: for readers of the 2016 LGB, this game appears on page 359.

Hi Laura,

Thanks for the question! While you are correct that at least one of J and S must be in the forest, that rule also allows for both J and S to be in the forest. So, let's take a look at what can occur in this question.

The fourth rule in this game states that, "If jays are not in the forest, then shrikes are." The placement of the "not" here is critical, because it's not where it usually is. In this case, it's on the sufficient condition, not the necessary condition. Thus, the diagram is:


J S


As detailed on page 268 and 269 in your edition of the book (side note: for readers of the 2016 edition this discussion appears on pages 294-295), this presentation—where the sufficient condition is negative but not the necessary—is dangerous. Many students take this rule to mean that J and S can't be in the forest together, but the meaning is actually the polar opposite. The meaning of the rule as given comes down to: when one of J or S is absent, the other must be there. Thus, at least one of J or S is always selected, possibly both. Thus, under the rule there are three possible outcomes:

• 1. S not in forest, J in forest
  2. J not in forest, S in forest
  3. Both J and S in forest

That last possibility occurs because if either J or S is selected, that does not violate any condition in the J S rule.


So, in question #7, when we get down to M and H are in, and that G and W are out, only J and S are left, and you could have either one or both. Hence, answer choice (E) is correct, and the "at most" includes the possibility of four birds.

Here's the good news: now that you've seen this "negative sufficient condition" rule presentation, you won't fall for it again. And that's one of the great values of preparing—you get a chance to see what they can do and how you should best handle it. This rule type has been appearing with greater frequency recently, and so it's important to know how tricky it can be.

Please let me know if that helps. Thanks!

[pinsyuanwu]

. The only unaddressed birds are S and J, and at least one of them, possibly both, must be in the forest.

According to Rule #4 If Jays are not in the forest, S are in
How could it possible that both J and S are both in the forest.
Please kindly assist

[Dave Killoran]

Hi Pin,

Thanks for the question! You've misunderstood how this rule works, so please take a look at this article, which explains how rules with negative sufficient conditions work: https://blog.powerscore.com/lsat/the-mo ... rule-lsat/

This is a rule construction we talk about frequently, and it's in our books and courses—it's something you definitely want to be solid on before the LSAT!

Thanks!

[bfromang]

Hello,
I am not understanding how both J and S can be in the forest. I understand this:
If H is in the forest, then G cannot be in the forest. If G cannot be in the forest, then W cannot be in the forest. At this point it has been established that H and M are in the forest and G and W are not in the forest. The only unaddressed birds are S and J, and at least one of them, ( DON'T GET THIS PART:
possibly both, must be in the forest. )

[Stephanie Oswalt]

Hello,
I am not understanding how both J and S can be in the forest. I understand this:
If H is in the forest, then G cannot be in the forest. If G cannot be in the forest, then W cannot be in the forest. At this point it has been established that H and M are in the forest and G and W are not in the forest. The only unaddressed birds are S and J, and at least one of them, ( DON'T GET THIS PART:
possibly both, must be in the forest. )

Hi bfromang!

Thanks for the post! I have moved your post to the thread discussing this topic. Please review the above discussion and let us know if this helps, or if you still have any additional questions.

Thanks!

[rench.co]

Note: for readers of the 2016 LGB, this game appears on page 359.

Hi Laura,

Thanks for the question! While you are correct that at least one of J and S must be in the forest, that rule also allows for both J and S to be in the forest. So, let's take a look at what can occur in this question.

The fourth rule in this game states that, "If jays are not in the forest, then shrikes are." The placement of the "not" here is critical, because it's not where it usually is. In this case, it's on the sufficient condition, not the necessary condition. Thus, the diagram is:


J S


As detailed on page 268 and 269 in your edition of the book (side note: for readers of the 2016 edition this discussion appears on pages 294-295), this presentation—where the sufficient condition is negative but not the necessary—is dangerous. Many students take this rule to mean that J and S can't be in the forest together, but the meaning is actually the polar opposite. The meaning of the rule as given comes down to: when one of J or S is absent, the other must be there. Thus, at least one of J or S is always selected, possibly both. Thus, under the rule there are three possible outcomes:

• 1. S not in forest, J in forest
  2. J not in forest, S in forest
  3. Both J and S in forest

That last possibility occurs because if either J or S is selected, that does not violate any condition in the J S rule.


So, in question #7, when we get down to M and H are in, and that G and W are out, only J and S are left, and you could have either one or both. Hence, answer choice (E) is correct, and the "at most" includes the possibility of four birds.

Here's the good news: now that you've seen this "negative sufficient condition" rule presentation, you won't fall for it again. And that's one of the great values of preparing—you get a chance to see what they can do and how you should best handle it. This rule type has been appearing with greater frequency recently, and so it's important to know how tricky it can be.

Please let me know if that helps. Thanks!

I have the 2020 edition of the Powerscore books. At my local library I also have the 2022 edition available if needed for any information that has been updated in the bookset. I finished reading all of the books, but I have to finish a few rechallenge game sets in the Powerscore Logic Games Bible.

Dave states "on page 268 and 269 in your edition of the book (side note: for readers of the 2016 edition this discussion appears on pages 294-295)." In the 2020 or 2022 edition of the book which pages can I find this information on? I've spotted a weakness with my understanding of a negative in the sufficient condition so now I'm trying to learn this again to engrain it in my brain better and take more notes of this section of the book. I need a refresher on when the sufficient is negative vs. when the necessary is negative.....that being said, when the necessary is negative (say the situation is, if jays are in the forest, then shrikes are not instead of the rule "If jays are not in the forest, then shrikes are"). This would mean that if there is one then there is not the other. Is this correct?