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General questions relating to LSAT Logical Reasoning.
 Johnclem
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#22513
Hi ,
This is from the bible explanation on formal logic.
And my question is why can't we infer G some I?


Question 3. Item: G --> H some-->I
This diagram appears quite similar to the diagram in #1, but it is slightly different, and that difference is significant enough to result in no inference being drawn. In this case, we start at I, since I is at one end of the chain and is involved in a some relationship. From I we can ride over to H. Once we arrive at H, is there a “track” leading away? No, there is only an incoming track from G, and thus we cannot travel further, and no inference can be drawn.

Thanx
John
 Nikki Siclunov
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#22518
Hi John,

You appear to have answered your own question... am I understanding your post correctly? Either that, or you've copied and pasted our own explanation to this problem. If you're having an issue with a particular aspect of it, I'd be happy to help you sort it out.

Let us know either way.

Thanks!
 Johnclem
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#22523
Hi ,
This is from the bible explanation on formal logic. Sorry I guess my question wasn't clear . I hope it is now .

My question is .. I get why we can't start from I and infer " I some G " (because of the direction of the arrow). BUT what I'm asking is why can't we Infer G--> I ? ( G is in an open relationship so why can't we start at G and follow the arrows to I"?

G --> H some-->I

thanks
John
 Nikki Siclunov
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#22527
Hi John,

Now your question makes sense. Here's a mnemonic trick called the Some Train that I suspect will help you out:

In a diagram such as A :some: B :arrow: C, we start at station A because A is the open variable in the :some: relationship, and the "some" ( :some: ) relationship is weaker than the all ( :arrow: ) relationship. Always start your Some Train from the weakest link onward, disregarding the temptation to only read left-to-right. From A, we can ride over to station B. Once at station B, we need a track “away” from B going to another station. Since the tracks are arrows, we need either an all arrow ( :arrow: ), a double arrow ( :dbl: ), or a double-not arrow ( :dblline: ). In this case we have the all arrow, and thus we can ride over to C. Thus, we can infer that A :some: C.

However, in your example (G :arrow: H :some: I), we need to start at station I and ride over to station H, because the weakest link is the some ( :some: ) relationship between H and I. Once at station H, we need a track “away” from H going to another station, but there is no track away, only an incoming track. Thus, we are stopped and there is no inference that can be made in this example.

The reason why we cannot start with the G :arrow: H statement deserves further discussion. Let me give you a similar example:
All dogs are animals.
Some animals are poisonous.

D :arrow: A :some: P
Does it follow that some dogs are poisonous? Of course not. The group of animals can be a very large group, and only a handful of its members can be poisonous. That subgroup need not overlap with the dogs, even if all of them are animals. We can represent this with a Venn diagram, but this is usually overkill for the purposes of the LSAT.

Hope this helps a bit! Let me know.

Thanks!
 Johnclem
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#22530
Sorry Im still lost I may have even confused myself further.

Okay so using the example from the bible again :
G-->H <---some ---->I

My issue is this :
1- we can't make G ->I because like I think you said it's not true that they overlap. So I kind of get that .
2 - if we infer I <some->G Then were combining two some chains . Which we can't ! And If some statements are read in both directions ..well then were saying G<---some--->I ! Well I though we had established that we couldn't assume there's an overlap.

Super lost ! And really am having some sort of a stroke.

John
 Nikki Siclunov
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#22535
Hey John,

Alright, so we have:

G :arrow: H :some: I

You need to read this chain starting from the I, because it's the open term in the weakest link of your chain (the "some" link). So, some I's are H's. Fine. But just because something is an H doesn't necessarily mean it has to be G: the arrow doesn't point away from H. So, there is no inference that can be drawn from this set of rules.

To answer your question, you're not combining two "some" chains. The mere fact that a "some" relationship can be read both ways doesn't mean you're combining two different "some" relationships. You aren't. The reason why we can't make an inference here has absolutely nothing to do with the rule prohibiting an inference using two "some" relationships. Here's an example of that:

A :some: B :some: C

In this scenario, as in your example, no inference can be drawn. The reason, however, is different: you cannot combine two "some" relationships and produce a valid inference. Just because some B's are A's and other B's are C's doesn't mean that some A's are C's. The A and C groups may not overlap.

Once again, I'd like to warn you against investing too much time in Formal Logic. At most one question on your test is likely going to involve this type of reasoning.

Thanks!

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