Additive Inference Drill #2

[sim.LSAT]

Hi!

I have a question regarding question 2 of the Formal Logic Additive Inference Drill. I have taken a look at the online resource that goes through each question, but I still do not understand. Let me explain exactly what I don't understand....

So, I was able to correctly diagram the problem (yay!). I was also able to correctly make the inference W Y. However, I do not understand why the inference Z X cannot be made.

According to the diagram, we can take the some train from Z to Y. But, I thought we could also take the some train back across the arrow (using principle 7) to X. If not, is it because you would then be making an inference from 2 consecutive some arrows?

Is this also the reason why you cannot make the inference W :some:Z ? As this would require you to make an inference using 2 consecutive some arrows?

I hope that made sense...

Appreciate the help

[Dave Killoran]

Hi!

I have a question regarding question 2 of the Formal Logic Additive Inference Drill. I have taken a look at the online resource that goes through each question, but I still do not understand. Let me explain exactly what I don't understand....

So, I was able to correctly diagram the problem (yay!). I was also able to correctly make the inference W Y. However, I do not understand why the inference Z X cannot be made.

According to the diagram, we can take the some train from Z to Y. But, I thought we could also take the some train back across the arrow (using principle 7) to X. If not, is it because you would then be making an inference from 2 consecutive some arrows?

Is this also the reason why you cannot make the inference W :some:Z ? As this would require you to make an inference using 2 consecutive some arrows?

I hope that made sense...

Appreciate the help

Hi Sim,

So the scenario there is as follows:

• 2.
  All X's are Y's
  Some Y's are Z's
  Most X's are W's
  
  Initial diagram: W<---Most---X Y Z
  
  Answer key inferences:
  W Y

The question is, How would you make an inference such as W some Z? If you reuse the W Y inference, you get:


W Y Z

But two "Somes" in a row do not allow for an inference, so that doesn't work.

You could try this:

W<---Most---X Z

But a Most and a Some in a row also don't make an inference.

So, the problem is what you thought it might be: two Somes, or a Some and Most relationship are too weak to make a connection.

Let's also clarify something you said which relates to the above: "But, I thought we could also take the some train back across the arrow (using principle 7) to X." —Yes, you can do this!

"If not, is it because you would then be making an inference from 2 consecutive some arrows? " —Also correct, two Somes in a row yield nothing.

Please let me know if that helps. Thanks!

[Dreamsof180]

Hi, couple of years late, but why does X Z not count as an inference? I understand the W Y and that Z can't connect with W to make an inference, but it looks like you should be able to make an inference between X and Z.

[Robert Carroll]

Dreamsof180,

Dave's post above explains formally why that inference doesn't work. I'll also explain with examples.

For context, here is our set of initial conditions:

All X's are Y's
Some Y's are Z's
Most X's are W's

Let's use the following definitions:

X = law partner

Y = lawyer

Z = person working at his/her first job

W = person familiar to clients

We then have the following translations:

All law partners are lawyers.

Some lawyers are working at their first jobs.

Most law partners are familiar to clients.

The intended inference X Z would then be the following:

Some law partners are working at their first jobs.

...and that's not entailed by the statements we started with.

The reason the inference doesn't work is that they may be some Ys that aren't Xs - it's also possible that the Ys that are Zs don't include any Xs among them. In our translation, all law partners may be lawyers, but the lawyers at their first jobs don't include any law partners.

Robert Carroll