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Formal Logic Additive Inference Drill Question

PostPosted: Wed Jul 12, 2017 10:19 pm
by PolskiCanuck
Hi,

I was working on the formal logic additive inference drills found on page 334 of the 2013 edition of the Logical Reasoning Bible. While doing the questions I ended up "discovering" some inferences that were not in the answer key and I would like some clarification.

1)
Some A's are B's
No B's are C's
All C's are D's

Initial diagram: A :some: B :dblline: C :arrow: D

The answer keys lists the inferences as:
A :some: notC
D :some: notB

My question is: Why isn't A :some: D a valid additive inference as well? Can we not recycle A :some: notC :arrow: D and ride the some-train to my answer?

Also, for question 2:

2)
All X's are Y's
Some Y's are Z's
Most X's are W's

Initial diagram: W<----(most)---X :arrow: Y :some: Z

Answer key inferences:
W :some: Y

Why isn't X :some: Z an inference?

Similarly, for question 6:

6)

Some N's are O's
No O's are P's
No P's are Q's
All Q's are R's

Initial diagram: N :some: O :dblline: P :dblline: Q :arrow: R

The answer key lists the following additive inferences:
N :some: notP
R :some: notP

Why isn't N :some: notQ an additive inference? Could we not recycle N :some: notP, ride the some-train, and end up with my answer?

My initial thought is perhaps I'm making a mistake when working from the middle of the problems instead of the edges? Perhaps I am also making an error regarding negatives?

Thanks in advance for your help!

Re: Formal Logic Additive Inference Drill Question

PostPosted: Thu Jul 13, 2017 7:04 pm
by Dave Killoran
Hi PolskiCanuck,

Thank you for your post! I actually wrote an extensive, additional set of explanation for the problems in this drill, as mentioned in the answer key. Here is the link to the full online supplement for our Bible Series: http://www.powerscore.com/lsatbibles.

For the specific explanations to this drill, you will want to go to the Logical Reasoning Bible, and click on "LSAT Logical Reasoning Bible Supplements" Once you click on that, everything is broken down by chapter. Here is the specific page you are trying to find: http://www.powerscore.com/lrbible/conte ... ndedAK.pdf

In short, your analysis of where you are going wrong is basically correct—good job! In each item, this is what happens:

#1: You can't jump across the negative is the short answer. At best you get A :some: C, but there's no way to connect to that C without using another "some" statement, and two "some" statements yield no inference. Basically, there's not a C :arrow: D statement to connect it is another way to see it. you get:

A :some: C
C :arrow: D

If you take the CP of the second statement, you have a "some" with an arrow leading in, which results in no inference.


#2: The direction of the arrow from X to Y is going the "wrong" way to make an inference. You again have a "some" with an arrow leading in, which results in no inference.


#6: Same issue, the inference between N and P is actual N :some: P. That negative on P totally changes the relationship and doesn't link directly to Q with a double-not arrow, but this instead:

N :some: P
Q :arrow: P

And again we have a "some" with an arrow leading in, which results in no inference. The nice thing is that with this identified, all your problems are resolved :-D

By the way, another way of seeing this particular one is that two double-not arrows in a row yield no inference.


Please let us know if you have any further questions!

Re: Formal Logic Additive Inference Drill Question

PostPosted: Fri Jul 14, 2017 12:17 am
by PolskiCanuck
Thank you very much for your answer and for the resource :) They were very helpful and I finally understand where I was going wrong.

Also, in case anyone else is having similar issues with not understanding why A :some: B :arrow: C yields an additive inference but A :some: B :larrow: C does not, page 321 in the 2013 edition explains it well (in newer editions, the section which first introduces the concept of the Some Train):

Basically, "with A :some: B :larrow: C, the B group could be so large that even though every C is a B, and some A's are B's, the groups of A's and C's do not overlap, and thus no inference between A and C is present."

That helped me finally understand it.

Thanks again Dave!

Re: Formal Logic Additive Inference Drill Question

PostPosted: Fri Jul 14, 2017 9:25 am
by Dave Killoran
Great, glad I could help!

Just as an aside for other readers, I current;y do NOT recommend obtaining editions of any of the Bibles that are older than 2016: http://blog.powerscore.com/lsat/the-201 ... ns-part-ii

I make changes every single year to each book, often adding material and expanding and refining explanations, many of which address the most common question (a good example is the question above, which because it came from an older edition did not indicate that I had added a lengthy, additional answer key online).

Current edition recommendations are:

LGB: 2016, 2017
LRB: 2016, 2017
RCB: 2017

Major changes are planned for the 2018 editions of the LGB and LRB, by the way. And the remake of the RCB will continue as well.

Thanks!

Re: Formal Logic Additive Inference Drill Question

PostPosted: Wed Jun 20, 2018 8:31 am
by Mikelo
Dave Killoran wrote:Hi PolskiCanuck,

Thank you for your post! I actually wrote an extensive, additional set of explanation for the problems in this drill, as mentioned in the answer key. Here is the link to the full online supplement for our Bible Series: http://www.powerscore.com/lsatbibles.

For the specific explanations to this drill, you will want to go to the Logical Reasoning Bible, and click on "LSAT Logical Reasoning Bible Supplements" Once you click on that, everything is broken down by chapter. Here is the specific page you are trying to find: http://www.powerscore.com/lrbible/conte ... ndedAK.pdf

In short, your analysis of where you are going wrong is basically correct—good job! In each item, this is what happens:

#1: You can't jump across the negative is the short answer. At best you get A :some: C, but there's no way to connect to that C without using another "some" statement, and two "some" statements yield no inference. Basically, there's not a C :arrow: D statement to connect it is another way to see it. you get:

A :some: C
C :arrow: D

If you take the CP of the second statement, you have a "some" with an arrow leading in, which results in no inference.


#2: The direction of the arrow from X to Y is going the "wrong" way to make an inference. You again have a "some" with an arrow leading in, which results in no inference.


#6: Same issue, the inference between N and P is actual N :some: P. That negative on P totally changes the relationship and doesn't link directly to Q with a double-not arrow, but this instead:

N :some: P
Q :arrow: P

And again we have a "some" with an arrow leading in, which results in no inference. The nice thing is that with this identified, all your problems are resolved :-D

By the way, another way of seeing this particular one is that two double-not arrows in a row yield no inference.


Please let us know if you have any further questions!


Hi, David. Thanks for the explanation. But, the fact that X some Z cannot be an inference is still strange to me. Here is what I'm thinking. I agree that if we start from Z, we cannot go from Y to X because the arrow is pointing inward. However, I think if we consider closed variable X, there's an arrow leading to Y and another double arrow leading to Z. So, my question is, why can't I start the journey from X to Z? The arrows are pointing to the right directions for inference to be made. Since I have also read that an inference doesn't have to be "makeable from both sides in order to be valid". What am I missing, please? Thank you.

~
M

Re: Formal Logic Additive Inference Drill Question

PostPosted: Wed Jun 20, 2018 4:25 pm
by Jonathan Evans
Hi, Mikelo!

Welcome to our forums! Good question.

It is true that with a biconditional/equivalence double arrow ( :dbl: ) you can make an inference such as the one you suggest. For example:

    M :arrow: N :dbl: P

In the above situation, since we know that every M is also an N and that N and P are equivalent (N's are P's and P's are N's), we may correctly infer:

  • M :arrow: P
  • P :some: M

However, the problem is that in the drill in the book, we do not have a biconditional arrow but instead a "some" arrow. The some arrow is where things break down here. What do we know and what can we infer?

    X :arrow: Y :some: Z

We know that all X's are Y's. This means every X is also a Y. However, this does not mean every Y is an X. This would be a Mistaken Reversalâ„¢. What can we infer about Y with respect to X? Since every X is a Y, we may infer that some Y's are X's.

    Y :some: X

Now add the above statement to the other "some" statement:

    Z :some: Y :some: X

Notice that while it is true that some Y's are Z's and also that some Y's are X's, these need not be the same Y's! The Y's that are Z's could be different from the Y's that are X's. This is why we cannot infer *X :some: Z

I hope this helps!