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LRB P459 Drill 3

Jerrymakehabit
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Can someone please help me with my question below?

From E :dblline: F :arrow: G :arrow: H, I can get E :dblline: F :most: G. If it is E then it is not F, most of F are G. Can I say E :dblline: G?

Thanks
Jerry
Brook Miscoski
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Jerry, I am not sure what edition you are using, so I cannot be sure that my answer helps you. If it does not, please provide edition information so that someone can follow up.

The short answer to your question is that the conclusion does not follow in the logic that you mapped out. Think of it this way. If you have F, then you don't have E, and you do have G. However, what do you know if you don't have F? You don't know anything at all, since the failure of a sufficient condition doesn't allow you to move forward. Thus, you cannot conclude that E and G are dissociated. But that leaves the question of what you're trying to interpret.

I think you are trying to diagram a LR method of reasoning question--argument part-- that features a psychologist.You do not need to diagram the stimulus to determine that the requested information is a conclusion.
Jerrymakehabit
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Brook Miscoski wrote:Jerry, I am not sure what edition you are using, so I cannot be sure that my answer helps you. If it does not, please provide edition information so that someone can follow up.

The short answer to your question is that the conclusion does not follow in the logic that you mapped out. Think of it this way. If you have F, then you don't have E, and you do have G. However, what do you know if you don't have F? You don't know anything at all, since the failure of a sufficient condition doesn't allow you to move forward. Thus, you cannot conclude that E and G are dissociated. But that leaves the question of what you're trying to interpret.

I think you are trying to diagram a LR method of reasoning question--argument part-- that features a psychologist.You do not need to diagram the stimulus to determine that the requested information is a conclusion.



Hi Brook,

I am using the 2019th Edition LGB.
From E :dblline: F :arrow: G :arrow: H, I can get E :dblline: F :some: G, Not E :some: G
Similarly, From E :dblline: F :arrow: G :arrow: H, I can get E :dblline: F :most: G, and then E :dblline: G. (If it is E, then it is not F. Most of F are G. Then, if it is E, then it is not G). Is this logic correct?
Based on your explanation, Not E :some: G would not be correct either. Can you please explain more?
Can you please tell which "LR method of reasoning question" you are referring to?

Thanks
Jerry
Robert Carroll
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Jerry,

It appears as if all the information you originally have (we'll get to inferences later) is as follows:

No E are F, so: E :dblline: F

All F are G, so: F :arrow: G

All G are H, so: G :arrow: H

From this, you cannot infer E :dblline: G. All F are G, but there could be additional G's that are not F. Those G's, because they aren't F, are allowed to be E. So E and G could overlap (though they certainly don't have to).

Not E :some: G can be inferred from the information you gave as long as you know, for instance, that there is at least one thing that is F. That thing is also G, via the conditional. But, being F, that thing can't be E, via the Double-Not Arrow. So there is something that is F, and G, but not E. So something out there is G and Not E at the same time, which is what your inference says.

Robert Carroll
Jerrymakehabit
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Robert Carroll wrote:Jerry,

It appears as if all the information you originally have (we'll get to inferences later) is as follows:

No E are F, so: E :dblline: F

All F are G, so: F :arrow: G

All G are H, so: G :arrow: H

From this, you cannot infer E :dblline: G. All F are G, but there could be additional G's that are not F. Those G's, because they aren't F, are allowed to be E. So E and G could overlap (though they certainly don't have to).

Not E :some: G can be inferred from the information you gave as long as you know, for instance, that there is at least one thing that is F. That thing is also G, via the conditional. But, being F, that thing can't be E, via the Double-Not Arrow. So there is something that is F, and G, but not E. So something out there is G and Not E at the same time, which is what your inference says.

Robert Carroll


Hi Robert,

I surely understand your explain that "From this, you cannot infer E :dblline: G. All F are G, but there could be additional G's that are not F. Those G's, because they aren't F, are allowed to be E. So E and G could overlap (though they certainly don't have to)."

But can you please point out where is wrong with my understanding below?
From E :dblline: F :arrow: G :arrow: H, I can get E :dblline: F :most: G, and then E :dblline: G. (If it is E, then it is not F. Most of F are G. Then, if it is E, it is not G).

Thanks
Jerry
Robert Carroll
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Jerry,

Even if most F are G, that doesn't mean there aren't additional G's that aren't F. As a concrete example:

"Most questions on the LSAT involve inferential reasoning. No question on the LSAT requires extensive math. So no question that requires extensive math involves inferential reasoning."

This has the same pattern as your example:

E = requires extensive math

F = question on the LSAT

G = inferential reasoning

I get E :dblline: F because "No question on the LSAT requires extensive math."

I get F :most: G because "Most questions on the LSAT involve inferential reasoning."

Yet I don't get E :dblline: G because it's not true that "No question requiring extensive math involves inferential reasoning."

Simply put, the "most" statement in your example is on the wrong side to combine with the Double-Not Arrow. There can be G things that are NOT F things, and those G things may or may not be E.

Robert Carroll
Jerrymakehabit
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Robert Carroll wrote:Jerry,

Even if most F are G, that doesn't mean there aren't additional G's that aren't F. As a concrete example:

"Most questions on the LSAT involve inferential reasoning. No question on the LSAT requires extensive math. So no question that requires extensive math involves inferential reasoning."

This has the same pattern as your example:

E = requires extensive math

F = question on the LSAT

G = inferential reasoning

I get E :dblline: F because "No question on the LSAT requires extensive math."

I get F :most: G because "Most questions on the LSAT involve inferential reasoning."

Yet I don't get E :dblline: G because it's not true that "No question requiring extensive math involves inferential reasoning."

Simply put, the "most" statement in your example is on the wrong side to combine with the Double-Not Arrow. There can be G things that are NOT F things, and those G things may or may not be E.

Robert Carroll


Thank you Robert!
I think the key thing I need to remember here is that from F :most: G, even if most F are G, there can be G things that are NOT F things. Got it!