Most Inferences

[Mookit]

Hi everyone,

I just finished the Formal Logic chapter in the LRB (2009 version, so chapter or page numbers may not match newer versions.)

I need some help understanding this concept:

Question: What inferences can be drawn?

G ----> H --m--> I

Correct answer: there are no inferences that can be drawn (as per the "Most Training Mini-Diagramming Drill" in the middle of the Formal Logic chapter.)

What I do not understand is why you cannot say that since ALL G's are H's, and MOST H's are I's, then at least (some/most?) G's are I's. As far as I understand it, if you have a group, G G G G G, and all G's are also H's, so GH GH GH GH GH, and if most H's are I's, so HI HI HI, wouldn't you also have to AT LEAST be able to say that some G's are I's, if G's are equal to being H's, which in turn are mostly related to I's?

To clarify, I do understand how it works by starting from the Some/Most arrow, for example:

G <--s--> H ----> I

Correct answer: Some G's are I's (G <--s--> I), because if some G's are H's, and ALL H's are I's, then it follows that some G's are also I's.

I may have this all confused in my head, but from my perspective at the moment, I can't understand why I cannot go with the flow from the All arrow through the Most arrow to make an inference.

Thanks for any help!

[Dave Killoran]

Hi Mookit,

Thanks for the question! This particular non-inference is always a tough one to make sense of at first. So, let's take a look at it more closely:

• Question: What inferences can be drawn?
  
  G H I
  
  
  In the above relationship, the problem often starts with the G H relationship, because at a glance it can be easy to think that G and H are pretty much the same size or overlapping group. And they can be like that. The problem is that they do not have to be, and when that occurs no inference can be drawn. I'll use a numerical example to show how this works.
  
  Let's start by saying there are 100 Hs. And, via the "most" relationship with I, that means 51 of the Hs are also Is. Now, if there were 100 Gs (or 75 or 60 or any number of 50 or above), then it would indeed be the case that at least one of the Gs would be Is. However, what if it was the case that there was just a single G? What happens then—does that single G have to be in the group of Hs that are also Is? No, it doesn't, and that's why no inference can be drawn here.
  
  Generally, if G and H are about the same size in numbers, then an inference would be able to be drawn. But, the possibility exists that the number of Gs is far smaller than the number of Hs, and when that occurs, there does not have to be overlap between G and I, and hence no inference.

Please let me know if that helps. thanks!

[Mookit]

Thank you Dave, that really helps! I was always assuming that G is completely equal to H, including group size.

[Dave Killoran]

Great, glad I could help! Yes, that's a common error, and a quite understandable one. While every G is an H, not every H has to be a G, and that's where the problems start. Please let me know if you run into any other issues.

Thanks!