LSAT Discussion Forum

Hello! I'm studying Formal Logic now, and had a question about making inherent inferences from somes and mosts.

The Logic Bible says that A -(M)-> B -(M)-> C does not yield an inference, but would it yield A <-(S)-> B <-(S)-> C because going backwards against an arrow yields the inference some?

Some clarity on the relationship between somes and mosts inferences would be helpful. Thank you so much in advance!

Hi, me again! One more question. I wanted to clarify if contrapositives work the same with formal logic, ie., in the following example:

One of the answers in the workbook was D <-(S)-> B negated. I wrote B <-(S)-> D negated. Are these the same?

Thank you kindly and in advance!!

Sorry, one more formal logic question (I'm having a hard time with this!)

In this example: W <-(M)- X --> Y <-(S)-> Z, the answer says that the only valid additive inference is W <-(S)-> Y. I understand how we arrived here, but can someone explain why X <-(S)-> Z is not also valid?

Last question, I swear! In this diagram, S <-- T <-(S)-> U --> V, how is this inference made: S <-(S)-> V? I'm confused because I thought that some arrows are not enough to make the leap, so how do we get to V from S <-(S)-> T?

hyperfang9000 wrote: ↑Mon Apr 28, 2025 4:22 pm Hello! I'm studying Formal Logic now, and had a question about making inherent inferences from somes and mosts.

The Logic Bible says that A -(M)-> B -(M)-> C does not yield an inference, but would it yield A <-(S)-> B <-(S)-> C because going backwards against an arrow yields the inference some?

Some clarity on the relationship between somes and mosts inferences would be helpful. Thank you so much in advance!

Yes, it would, but note that doesn't allow you to make an inference between A and C either, and that A/C connection is what the test makers would be looking at. In a problem using a relationship like this, wrong answers would typically focus on someone mistakenly thinking A and C can be connected.

Also, the inherent inferences that "go backward" aren't what we are looking for when we talk inferences, typically. We want the additive inferences that connect variables that are not directly connected (like A and C here).

Thanks!

hyperfang9000 wrote: ↑Mon Apr 28, 2025 5:11 pm Hi, me again! One more question. I wanted to clarify if contrapositives work the same with formal logic, ie., in the following example:

One of the answers in the workbook was D <-(S)-> B negated. I wrote B <-(S)-> D negated. Are these the same?

Thank you kindly and in advance!!

They are NOT the same. Contrapositives with Some and Most statements don't really exist because the contrapositive depends on the absoluteness of the necessary condition.

The way to see why the above doesn't work is to use a simpler statement and compare: Is Some Dogs are not White the same as Some White things are not Dogs? No, those aren't identical.

Thanks!

hyperfang9000 wrote: ↑Mon Apr 28, 2025 5:16 pm Sorry, one more formal logic question (I'm having a hard time with this!)

In this example: W <-(M)- X --> Y <-(S)-> Z, the answer says that the only valid additive inference is W <-(S)-> Y. I understand how we arrived here, but can someone explain why X <-(S)-> Z is not also valid?

The most direct answer is that the arrow from X to Y is going the wrong direction: X --> Y <-(S)-> Z. If you start at Z, you ride over to Y on the Some Train. From Y is there a "track" leading away? No, so no inference can be drawn.

If it were X <-- Y <-(S)-> Z, then there would be the inference there. The Some Train would wok, and would yield Z some X.

Thanks!

hyperfang9000 wrote: ↑Mon Apr 28, 2025 5:27 pm Last question, I swear! In this diagram, S <-- T <-(S)-> U --> V, how is this inference made: S <-(S)-> V? I'm confused because I thought that some arrows are not enough to make the leap, so how do we get to V from S <-(S)-> T?

There's a note in the Answer Key which should help: "(Made by recycling either one of the two previous inferences into the original diagram. This form has appeared in several LSAT questions and this final inference has always been the correct answer. See the book website for an expanded explanation.)" the book site explanation goes into much greater detail in explaining those. It is also explained in lesser detail in the LRB Formal Logic Chapter discussion of this problem at: viewtopic.php?f=1363&t=25763. the key is to re-use either of the sub inferences and recompile it. It's a valid inference!

Thanks!

Thanks!

Thank you so much for these thorough and thoughtful responses, Dave! Really really appreciate it. I have one more general question; when the LSAT Bibles discuss The 11 Principles of Making Formal Logic Inferences, #8 "Watch for the relevant negativity," has a note that the double-not arrow will always appear just before the last station. I'd like to understand the logic behind this general rule -- isn't it possible for a double-not arrow to appear in the middle stations or just generally a station that isn't second to last?

As in this example from the Formal Logic drills N <-(S)-> O <-|-> P<-|-> Q --> R

Could someone clarify this general rule? Thank you so much!!!

Hi hyperfang,

Double-not arrows can appear in the middle of a formal logic chain, as in the example from the drill that you cite:

N <-(S)-> O <-|-> P<-|-> Q --> R

However, no inferences can be made beyond "first" double not arrow in the chain.

For example, starting on the left, we can make an inference from this part of the chain:

N <-(S)-> O <-|-> P

The inference is:

N <-(S)-> Not P

However, we cannot make any inferences beyond "Not P" from this direction. For example, there are no inferences to be made between N and Q or R.

Now going from right to left, we can also make an inference from this part of the chain:

P<-|-> Q --> R

Using the inherent inference that some Rs are Qs, we can then infer:

R <-(S)-> Not P

Again, however, we cannot make any inferences beyond "Not P" from this direction either.

To see why, it may be helpful to diagram the double-not arrows as single, one way arrows. You will see that both O and Q have arrows pointing to "Not P," so there is nowhere to "go" from "Not P," (i.e no outbound arrow leaving the "Not P" station). "Not P" is the final destination so to speak.