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 Johnclem
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#22529
Hi there ,
I am having trouble understanding the the bible explanations for the chapter on formal logic. Please look at the inferences below to see where I am suffering.

Here is an example :
Some A's are B's
No B's are C's
All C's are D's


A <---some---> B :dblline: C---->D

Inferences :
1- A --> not c I get this inference !
2- D <---some---> not B How does this inference make sense ? We're going in the opposite of the arrow and combining two some statements! This was not clear at all in the explanations. Especially when before the drill we were given a share saying we could never combine two "some " statements.


Thanks
John
 Adam Tyson
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#22538
John,

I have to ask, but is that first inference (A->C) in the book? Because that is not correct - you cannot get there from the statements given. The second one, however, is good. You don't have to combine two Some statements to get there, but rather a None (the one about B and C) and an All (the one about C and D). Since all Cs are Ds and no Cs are Bs, then at least one D must not be a B (any D that is a C cannot be a B, and there must be at least one D that is a C).

Try making it real - let's make up some things to substitute for those letters, and it should become more clear:

A = Animals
B = Buffalo (Buffaloes? Buffalos? Buffali?)
C = Chickens
D = Delicious

So, some Animals are Buffaloes.
No Buffaloes are Chickens
All Chickens are Delicious

Does the inference that all Animals are not Chickens work? Nope - just because some Animals are Buffaloes doesn't mean they all are, and so some of the Animals that are NOT Buffaloes COULD be Chickens.

Can we infer that some Delicious things are not Buffaloes? You bet - at a minimum, the Chickens are Delicious but they are not Buffaloes.

Now I want Buffalo wings. I'm out! Hope that helped.
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 Stephanie Oswalt
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#44407
We recently received the following question from a student. An instructor will respond below. Thanks!
Hi,

My name is Giorgia and I have recently bought the three PowerScore bibles.

I have a question regarding the Logical Reasoning section and it would be great if you could help me solve the following doubt.

1. On p. 458 of the 2018 LR edition, the solution to question no. 1 states that there are two inferences - and I agree with the solution.
However, I was wondering whether we could also infer that A <—|—> D. Would it make sense? If not, why? Thanks.

I look forward to hearing from you, and thank you in advance for your time.

Kind regards,
Giorgia
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 Jonathan Evans
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#44411
Hi, Georgia,

Thanks for the great question!

We cannot infer that A :dblline: D because all we know is A :some: C. It is possible, but not necessary, that A :some: C, and therefore those As would also be Ds (because C :arrow: D).

But let's ask the next logical question: could we infer that A :some: D? Seems like it makes sense. After all, A :some: C and C :arrow: D, thus couldn't we conclude A :some: D?

No actually we cannot.

D is a necessary condition for C, D is not a sufficient condition for C. In other words, it is not true that D :arrow: C. This is a Mistaken Reversal™. Thus it could be possible that ALL As are Ds even though A :some: C.

This might be hard to follow. Let's look at this through Venn diagrams.

Image

Let's see whether our conditions check out.
  • C :arrow: D. Yes, this is true. The entire C circle is contained within the D circle.
  • A :some: C. Yes, this is true as well. There is only partial overlap between the A circle and the C circle.
  • A :some: D. No, this is NOT true. In this illustration, the entire A circle is contained within the D circle. Therefore, we cannot infer that A :some: D.
Is it possible that A :some: D? Yes, this is possible. The bottom line is that we do not have a necessary relationship between A and D. You cannot make a connection between A and D.

I hope this helps!
 graymo
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#74661
Hi!

I just finished reading/working through Chapter 13: Formal Logic in the Logical Reasoning Bible. I'm relieved that I'm understanding "most" (51-100 :-D ) of the concepts in the chapter. I did, however, eventually encounter some difficulty when it came to listing the additive inferences derived from complex Formal Logic statements.

Diagramming for me is not the issue -- I feel very confident in my efficiency/accuracy of diagramming the statements (I usually always get these right). I understand the Logic Ladder and how we can make inherent inferences "climbing down" the rungs. I also understand the concepts of recycling and checking the closed variables. What I am unsure of, however, is when to use and not use these tools. For example, when is it necessary to infer a some relationship from an all relationship, and when is it redundant? When is recycling advantageous, and when is it a waste of time/an unnecessary step?

For some of the drills listed on page 465 of the 2020 Logical Reasoning Bible, I found myself listing inferences that were not matched in the example key. I found all the inferences listed as correct, but I always had a couple extras. Perhaps these inferences are inherent and thus are not listed (as per the instructions of the drill), but I clearly seem to be unable to differentiate the two -- which inferences are relevant to the statements and which are not.

Let's take drill #1 for example, which reads:

Some As are Bs.
No Bs are Cs.
All Cs are Ds.

I'm able to quickly diagram this problem:

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

I'll talk you through my thought process and perhaps you can tell me where I'm going awry. The first thing I do is look at the ends of the chain (Principle 1). I see the some relationship at one end between A and B, and automatically my brain thinks Some Train. I ride the Some Train and establish the relationship between A and C:

[A :some: B :dblline: C] :arrow: D
[A :some: C] :arrow: D
INF. #1: A :some: C

And there's my first additive inference (connecting A and C through B, then removing B)! Yay, all is good so far (I hope).

So now that I've established an inference from A, B and C, I turn to the other end of the statement to see if I can infer anything from the B, C and D relationship. Well, Principle 3 says to not start with a variable involved in a double-not arrow, so I can't infer anything starting with B (moving toward D). D, on the other hand, has an arrow pointing toward it, but if I take that relationship (the relationship between C and D) and climb down the Logic Ladder, I can infer that...

A :some: [B :dblline: C :arrow: D]

...can be reduced to (using the Inherent Inference rule):

A :some: [B :dblline: C :some: D]

Which now allows me to take the Some Train from D to B, and I find a second inference:

A :some: [B :dblline: C :some: D]
A :some: [B :some: D]
INF. #2: B :some: D

Now these two inferences are all that the answer key on the following page (page 466) lists. My thought process at this stage in inferring is that I haven't completely explored every avenue to see if I've found all the inferences I can find. What about recycling?

So I go ahead and throw inference #2 into the equation and see if I can infer anything.

A :some: [B :dblline: C :arrow: D]
A :some: [B :some: D]

Principle 9. Two consecutive somes do not yield an inference. Nothing I can do here.

Let's throw inference #1 back into the equation.

[A :some: B :dblline: C] :arrow: D
[A :some: C] :arrow: D

This looks, to me, like I could derive a third inference from the statement. I ride the Some Train (from A to D) and find a third inference:

A :some: C :arrow: D
A :some: D
INF. #3: A :some: D

And this is the inference that is not listed in the answer key. Is this inference inherent, not additive? Have I taken an unnecessary step? Is my thought process correct in that I should (in the case of three individual relationships) be looking at the two relationships on the either end, then recycle any inferences I find into the original statement to potentially find more inferences (that's a confusing sentence)?

In some of the following drills on page 465, I found myself questioning whether or not I needed to reduce all statements to some statements to see if there was a possible inference I could make. Should we always be reducing to some relationships to potentially infer? Are those additive inferences? I think I'm confusing myself even more by typing this... but I hope my questions are somewhat clear?

Thank you!
Last edited by graymo on Mon Apr 06, 2020 9:25 am, edited 2 times in total.
 Jeremy Press
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#74683
Hi graymo,

The reason your final inference is not showing up in the chapter is that it's actually not a valid inference. You're correct to say that Some A's are not C's. But that inference cannot validly be connected to the rule stating the relationship between C and D, because that rule is C :arrow: D. We don't know whether (or how) "not C" connects to D.

It's likely the double-not arrow depiction that's tripping you up there. Just remember that when we go from C to D in your diagram, we're going from the positive (non-negated) C to D, not from the negative (negated) C to D. So, if you want to involve (positive) D in any inferences, they have to be inferences that would involve (positive) C. By the way, that's why your second inference involving "not B" works. We know some D's are C's (the positive form of C) and we also know that all C's are not B's. So we can validly say some D's are not B's.

I hope this helps!

Jeremy
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 Emma Beck
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#93861
Hello,

In question 1 (p.438, 2017 edition), we must make inferences from the following chain:
A :some: B :dblline: C :arrow: D

The answer choice details the following inferences:
A :some: ~C (because A :some: B :dblline: C)
D :some: ~B (because B :dblline: C :arrow: D)

However, the answer explanation does not say anything about A :some: D. I came to this inference by "recycling" A :some: ~C and adding the " :arrow: D" element to reach A :some: ~C :arrow: D => A :some: D. Is my reasoning correct? Are there three additive inferences here instead of just two?

I would be very grateful for someone's help!
Thank you,
Emma
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 Emma Beck
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#93863
Emma Beck wrote: Tue Feb 22, 2022 9:44 am I have the same doubts about the question 6: N :some: O :dblline: P :dblline: Q :arrow: R.

The answer explanation shows that N :some: ~P and R :some: ~P are the valid additive inferences here, which is part of what I got when doing the question. That said, I also got the following inferences:
I recycled N :some: ~P and added :dblline: Q to get N :some: ~Q ;
I recycled R :some: ~P and added O :dblline: to get R :some: ~O ;
and finally, I got R :some: N by trying to ride the Some Train all the way through the logic chain.

I'm probably getting this all wrong and overcomplicating this, but I'm quite confused right now. Thanks to anyone who can provide me with an explanation!
Emma
Edit: I looked over the rules and remembered that two negations (i.e., :dblline: and :dblline: ) do not yield any inferences. Apologies for the confusion, this question makes sense now, but I still have yet to figure out the inferences for Q1 outlined in the post above.
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 Stephanie Oswalt
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#93870
Emma Beck wrote: Tue Feb 22, 2022 9:17 am Hello,

In question 1 (p.438, 2017 edition), we must make inferences from the following chain:
A :some: B :dblline: C :arrow: D

The answer choice details the following inferences:
A :some: ~C (because A :some: B :dblline: C)
D :some: ~B (because B :dblline: C :arrow: D)

However, the answer explanation does not say anything about A :some: D. I came to this inference by "recycling" A :some: ~C and adding the " :arrow: D" element to reach A :some: ~C :arrow: D => A :some: D. Is my reasoning correct? Are there three additive inferences here instead of just two?

I would be very grateful for someone's help!
Thank you,
Emma
Hi Emma,

Thanks for the post! I have merged your question with a prior thread about this same topic. Please review the response from Jeremy Press above, and let us know if that helps, or if you still have further questions.

Thanks!
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 Emma Beck
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#93871
Hello,

Thank you so much for pointing me in the right direction!

So, in N :some: O :dblline: P :dblline: Q :arrow: R, we can't say N :some: R not just because inferences between :dblline: and :dblline: are impossible, but also because the P in N :some: ~P (the inference we make from N :some: O :dblline: P) is not the same as the P in P :dblline: Q :arrow: R (which was what I attempted to add onto N :some: ~P when trying to recycle N :some: ~P).

To use the PowerScore analogy of train rides, we can't ride the Some Train all the way through these stations because P and ~P are in fact different stations!

Thanks again for the help :)
Emma

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