Additive Inference Drill

[zwglanz]

So I'm understanding all the concepts of this chapter, but in action i'm having trouble.

When I hit a <--/-->, does the some train stop there? For instances, #1 of the drill is Some As are Bs, No Bs are Cs, and All Cs are Ds. I diagram it all correctly, but I would've thought that you could conclude that some A->D. #2 All Xs are Ys, Some Ys are Zs, and Most Xs are Ws. I wanted to make X -> Y into a some and make it a <--->, which I thought would lead me to be able to conclude that some w->Z. But then you get to #3, No Es are Fs, All Fs are Gs, and All Gs are Hs. Why is it on this one that I can turn it all into somes and work my way back up to the /E? Can someone help me identify where i'm going wrong?

[Dave Killoran]

Hi Z,

You are actually running into multiple problems here, which is why this probably seems quite confusing. A few thoughts:

The doesn't stop the inferences in the Some Train. It just turns the relationship into a negative one. Compare the following two relationships:

• 1. A B C
  
  Inference: A C
  
  2. A B C
  
  Inference: A C

Same inference, one just has a negative on C.

Second, you should go back and review the Rules of Reversibility. They play a role in some of your confusion as far as I'm seeing.

Third, go use the book site code in the front cover of the 2023 versions, or see Chapter 1 for the book site in other versions. I wrote an expanded set of explanations for each of the items in this drill that takes the inferences step-by-step There are also some additional discussions of each drill in this forum.

Thanks!

[HarmonRabb]

Hi, I have some of the some confusions as OP. From the drill on page 463 (2024 edition) we come up with the diagram:

Code: Select all

 A <--s--> B <--|--> C --> D

I don't understand why we can't make the inference:

Code: Select all

A <---s--> D

We ride the some train from A to B
We ride the double not arrow from B to C
We ride the arrow from C to D

Where have I gone wrong?

[Dana D]

Hey Harmon,

Your last inference is where you're making the mistake - looking back at Dave's example, the last inference is A C - that's a negated C. The inference is that some (at least one) A does not meet the criteria of C, and we know that has to be true because some A B, and B C - which means absolutely no B are ever C.

If absolutely no B C, but we know at least one (some) A are B, then we can be sure there is at least one A C - or at least one A that does not fit the criteria for whatever C is, because it is a B.

In your example, we don't actually know that A D. We know some A are definitely not C, or A C, but what about the rest of A? We have no idea about the relationship between A and D, so we can't say for sure A D.

Hope that helps!

[HarmonRabb]

Thanks for your reply Dana, I really appreciate the level of detail!

The logic in your explanation makes sense to me but I'm still struggling with trying to match it to the rules in the chapter.'

I thought when a double not arrow is followed by an (all) arrow we can always make an inference. Is it a case of the combination of a some arrow follow by a double not arrow cannot produce any further inferences after the double not?

[HarmonRabb]

Thinking about this further, it is because the first inference gives us "Not C" and thus we cannot recycle the inference because the relationship is C--->D ?

[Jeff Wren]

Hi Harmon,

That's exactly right.

The Double-not Arrow can be tricky, especially when combining it in longer chains.

Remember that what it represents is two conditional statements in one.

Here B <-|-> C

Means

B -> Not C

And its contrapositive

C -> Not B

If you were to diagram this problem using:

A some B -> Not C

Then you'd more easily spot that "All Cs are D, diagrammed C -> D" doesn't connect to make an inference.

[HarmonRabb]

Thanks, I appreciate the detail