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 ChicaRosa
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#28009
When I was reading the table and their explanations about the relationships between none, some, most, and all I got confused with a few explanations.

1.How come none cannot be diagrammed with another none? If I were to diagram it would it look like this A :dblline: B :dblline: C or would there be two single arrows with the necessary condition crossed out like this: A :arrow: B :arrow: C?

2. The relationship between diagramming two "Alls" what are some situations that would allow an inference or not? Would A :arrow: B :arrow: C allow an inference? Would A :arrow: B < :longline: C not allow an inference?

3. What would be a circumstance where you cannot get an inference from diagramming two "Mosts"?

Thank You! Your help is appreciated :-D !
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 Dave Killoran
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#28028
Hi ChicaRosa,

Thanks for the questions. Let's get right to them:

1. Two "none"s don't add up to anything. If I dislike Joe, and you dislike Joe, does that tell me anything about whether you and I dislike each other? No :-D In the case of your diagram, you've mis-daigrammed it. The diagrams are:

A :arrow: B (you had this right)
B :arrow: C (this part you had wrong)

Thus, the arrow statements can't be connected, resulting in no inference.


2. A :arrow: B :arrow: C allows for A :arrow: C. This is a classic inference chain you see in the book multiple times. It also appear sin the LGB repeatedly.

A :arrow: B < :longline: C does not allow for an inference. What if there is 1 A, 1 C, and 1,000,000 Bs? Do the A and C then have to overlap? No, and hence there is no inference there.

A < :longline: B :arrow: C allows for the inference "A some C." Use the Some Train there to make that inference most easily.


3. A :most: B :most: C is one instance where you can't get an inference. I talk about this in the book in some detail, but without additional numerical information (which they've never done on the LSAT), you can't draw an inference. Another would be where the two "most" arrows point at the same variable, as in: A :most: B <-most-- C. No inference can be drawn there either!


Please let me know if that helps. Thanks!
 ChicaRosa
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  • Joined: Aug 23, 2016
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#28029
Dave Killoran wrote:Hi ChicaRosa,

Thanks for the questions. Let's get right to them:

1. Two "none"s don't add up to anything. If I dislike Joe, and you dislike Joe, does that tell me anything about whether you and I dislike each other? No :-D In the case of your diagram, you've mis-daigrammed it. The diagrams are:

A :arrow: B (you had this right)
B :arrow: C (this part you had wrong)

Thus, the arrow statements can't be connected, resulting in no inference.
Hi Mr. Killoran,

Your explanations were very helpful for questions 2 and 3 :-D With question 1 I was somewhat confused with the graphing and I want to make sure I understand this. Your example, where you mention that if you dislike Joe and I dislike Joe wouldn't it be diagrammed as A :arrow: B < :longline: C? From your example I got that the only thing we have in common is that we both dislike Joe but we can't make an inference if we dislike each other. And how would it be B :arrow: C?

Thank you!
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 Jonathan Evans
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#28030
Hi, Chica,

I hope you don't mind my jumping in briefly. Dave was not diagramming your mutual dislike of Joe. He was modifying your second diagram with reference to two "none" statements, that is:

A :arrow: ~B i.e. For the set of all A, any A is not a B (If A, then not B)

B :arrow: ~C i.e. For the set of all B, any B is not a C (If B, then not C)

You originally had:

~B :arrow: ~C i.e. For the set of all B, there is not a B that is not a C (If C, then B)

From your original diagram (a none/all relationship), you could infer that A :dblline: C.
 ChicaRosa
  • Posts: 111
  • Joined: Aug 23, 2016
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#28033
Jonathan Evans wrote:Hi, Chica,

I hope you don't mind my jumping in briefly. Dave was not diagramming your mutual dislike of Joe. He was modifying your second diagram with reference to two "none" statements, that is:

A :arrow: ~B i.e. For the set of all A, any A is not a B (If A, then not B)

B :arrow: ~C i.e. For the set of all B, any B is not a C (If B, then not C)

You originally had:

~B :arrow: ~C i.e. For the set of all B, there is not a B that is not a C (If C, then B)

From your original diagram (a none/all relationship), you could infer that A :dblline: C.
Hello Mr.Evans,

I don't mind at all :-D Your explanation helped clarify my misunderstanding with Mr.Killoran's explanation. Thank you very much!

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