Inherent Inference Recognition Drill #5

[askuwheteau@protonmail.com]

Thank you Jeff...I appreciate your advice. I'll focus upon finishing the chapter.

[askuwheteau@protonmail.com]

Jeff,

I'm currently completing the Some Train diagramming mini-drill (LRB 2023 Edition, Chapter 13, pg. 435) and ran into a challenge. Below's my reasoning below:

Legend:

• ^s means a lowercase letter s is present over the conditional diagram which begins the some train.
  <> means double arrow
  <means single arrow
  > means single arrow
  <|> means double not arrow

Problem #4: J<K<^s>L yields the following inference according to the answer key: L<^s>J . However, my answer is J<^s>L.. Is my answer correct given that the some diagram goes in both directions?

Problem #7: S<|>T<^s>U yields the following inference according to the answer key: U<^s> S (negated...struckthrough). However, my answer is a negated S<^s>U . Is my answer correct given that the some diagram goes in both directions?

I ask because the textbook seems to give two different interpretations of how to form inference diagrams that go from right to left (pg. 433, paras. 1 & 2 juxtaposed against pg. 434, paras. 1-3)

Thanks,

Jonathan

[Jeff Wren]

Hi Jonathan,

The short answer to both questions is that your inferences were correct and were identical in meaning to what was written in the answer key.

As you correctly mentioned, "some" is reversible, so the statement "Some doctors are lawyers" is identical in meaning to the statement "Some lawyers are doctors." Both of these statements mean that there is at least one (i.e. some) person who is both a doctor and a lawyer.

[askuwheteau@protonmail.com]

Hi Jeff,

Thank you for explaining this. I appreciate it.

Best,

Jonathan

[askuwheteau@protonmail.com]

When doing the Most train drill, a seemingly similar situation arises as with my previous inquiry re the Some train drill.

Re the Most train drill:

#4: I have as my answer J<^m L whereas the answer key has: L > ^m J. In either case, the conditional statements are saying that Most L's are J's. Is my answer incorrect?

#7: I have as my answer Negated S < ^m U whereas the answer key has: U>^m S (negated letter S). In either case, the conditional statements are saying that Most U's are not S's. Is my answer incorrect?

[Jeff Wren]

Hi Jonathan,

The short answer to both questions is that your inferences were correct and are identical in meaning to what was written in the answer key.

The "direction" of your arrows (meaning whether they point up, down, left, right, diagonal, etc.) does not matter in diagramming. All that matters is what term is at the "beginning" of an arrow and what term is at the "end" of an arrow.

Having said that, we usually diagram left to right most of the time as that is often easier to follow since that is how we read English.

For example, if I were to diagram the statement "All As are Bs," I would diagram it

A -> B

Of course, I could diagram this statement:

B <- A

Only the physical direction (left to right) has changed, but in each diagram the arrow points from the "A" to the "B." These two diagrams are identical.

Note that these diagrams are not reversible. So diagramming the statement,

B -> A would be a Mistaken Reversal.

As for why we would ever diagram in a different direction than left to right, when you start linking multiple formal logic statements, you need to "branch out" in different directions so that you can read them all and make the appropriate inferences.

I realize that your questions were about "Most" statements, but they follow the same rules.

[askuwheteau@protonmail.com]

Got it! Thanks for clarifying. I went back and reread my textbook section on this concept and figured this must be what the textbook means when it says on pg. 429 that a.) Always combine common terms and b.) there is no traditional direction in logic.

[askuwheteau@protonmail.com]

Good afternoon,

I'm nearly completed with the Formal Logic chapter and am reading the 10th Principle out of the Total 11 for making Formal Logic inferences. The 10th principle has to do with analyzing compound statements. On page 445 (LRB, 2023 Edition), the following problem is presented: F<^s>A>LH<|>H

Three separate additive inferences are derived from the above problem:

• F<^s>LH
  F<^s>H (negated letter H)
  A<|>H

It is with the third and final additive inference that I am having trouble understanding. If A and LH are the closed variables and we go ahead and transform the A>LH into the Some Train, then wouldn't the inference look like this: A<^s>H (negated letter H)? This would result from the Some Train exercise transforming A>LH<|>H into A<^s>H (negated letter H).

Please assist...thanks.

[Adam Tyson]

We shouldn't be transforming the relationship between A and LH into a Some Train, askuwheteau@protonmail.com, because it's a pure conditional relationship. ALL A is LH, so it's not just some A that are not H, it's all of them. Every single A is an LH, no exception, and every LH is not an H, also no exceptions. Thus, no A could possibly be an H.

To illustrate:

Every human is a mammal, and no mammal is a fish.

Would you say that some humans are not fish? No, you would say that all humans are not fish. No human is a fish.

Of course, a conditional relationship like that does also carry with it an underlying "some" inference. If all humans are mammals, then some mammals are humans. And we can further infer that some things that are not fish are humans. But expressing those relationships in terms of "some" is too weak. When you can infer a conditional relationship, do so.