Formal Logic Problem Set #4

[Lawyered]

In the Formal Logic Problem Set:

Q.4 Can we diagram it as CH </> Tra <- Pte -/-> Hom?

I realized later that I put the conclusion itself into the problem but initially it made sense since that would paint the full picture.

While the book went into a long explanation, much of it beyond me, I figured the two not-not signs in the problem are dealt with similar to the way #3 was and hence CH = Hom would work. It was one of the answer choices and it did...

I do want to run my Logic by y'all though.

Also, in #3 is it okay to diagram the thing from left to right rather than 2 different thingies? Ex.

OV <- GS <-M- SS -M-> H ??

Thanks!

[Jon Denning]

Hey Lawyered - thanks for the questions! I'll give your diagrams a look below and let you know what I think, but before I do I just did a quick search for "chordates" from our main Forum page to see if this question had any other discussions about it and found the following:

lsat/viewtopic.php?t=3579&p=9458

lsat/viewtopic.php?t=8421&p=21568

So be sure to both give those a look, and use the search feature first on future questions—it's amazing how much content has been explained, meaning answers are often already in place to help you out (saves you the time of typing out questions and waiting on a reply)!

For this one I'd make sure to keep the premises and conclusion separate, since you're really trying to evaluate the nature and validity of arguments when authors make them. To do that you need to clearly see what's given as fact, and how the author attempts to use that information to arrive at a conclusion. So let's do premises first, then see if the conclusion follows (and if not, why not?).

You've set the premises up perfectly!

Ch Tra Pte (note I just used the little arrow icons over to the right of this text window)

You've got the conclusion correct as well!

Pte Hom

So far, so good. We just need to see how to get to that new term in the conclusion, Hom, and show that it's incompatible with Pte from our premises. And if all Hom are a part of Ch, Hom Ch, then sure enough Pte and Hom can't touch!

Nicely done!

As for #3, you can connect it all together, sure:

Happy <--most-- Serious Grad Overworked (no left-facing most arrow, so I typed it out)

I happened to flip the diagram you made, but it comes to the same thing

Good work!

[akanshalsat]

hello!

I know that for something like this: A B C the inference is: A [NOT] C (you're accounting for the negativity which is transferring over to C

but for something like this:

A [NOT] B C

why is the inference:

A C? why doesn't B's "not" get transferred over to C? I'm referring to practice problem 2 on page 423 in the LRB (Some train diagramming)

[Adam Tyson]

I'm not seeing the same diagram that you are in that drill, akanshalsat. In both my 2016 and 2018 copies, #2 in that drill doesn't have a Most, but a Some. It looks like this:

D E F

So, some Ds are not Es, and anything that is not an E is an F. Therefore, at least some Ds are Fs.

The negative doesn't carry over because F is positive, not negative. It might help to assign values to these symbols to make it clear:

D = Duke fan

E = Evil

F = Funny

This diagram means that at least some Duke fans are not evil. Let's say that of the millions of Duke fans out there, 3 of them are not evil. Next, everyone who is not evil is funny. We can infer that all three of those Duke fans that are not evil are, therefore, funny. Can we infer that any Duke fans are NOT funny? No, because there is nothing sufficient her for "not funny."

It works in the double-not-arrow situation because, in your example, B is sufficient for C, so if any As are Bs, they must not be Cs. It's not so much about "transferring over the negative" but about following the sufficient conditions to their necessary conditions, and in this problem the necessary condition is a positive and not a negative.

Just follow the conditional arrows to their necessary conditions, and you'll get it!

[sim.LSAT]

Hi!

I have taken a look at the links that relate to the Formal Logic Problem set, but I can't seem to find a a response that answers my question....

For Q4 in the problem set I understand why using a Mechanistic Approach is the best option. So it leaves us with answer options A B and D.However, I don't understand why it states "the first relationship that comes in mind H T." How was it determined that this relationship is to be assumed if the conclusion is to follow logically? In other words, why is it that H T is the relationship that we need in order for the conclusion to follow?

[KelseyWoods]

Hi sim.LSAT!

To see this, it might help to breakdown the relationships a bit more and use contrapositive pairs instead of Double-Not Arrows:

P: C T (C T, T C)
P: P T
C: P H (P H, H P)

You could combine those premises in this way:

P: P T C

So now it's a bit easier to see that to get from P to not H (P H), you just need to get from either T to not H (T H) or get from not C to not H (C H).

Answer choice (B) gives us H C, which is just the contrapositive of not C to not H (C H).

So if we add that to the premises we get:

P T C H

This gets us from P to not H (P H), which is our conclusion.

Hope this helps!

Best,
Kelsey