Are these arguments valid?

Comments

• extramedium Alum Member

  April 2017 edited April 2017 419 karma

  Gonna try and help here. Makes sense for the most part.

  1.)
  Most beers are in cans. Most cans contain beer.
  B most C
  C most B

  In my understanding, there are no valid inferences to be drawn here. As far as the numbers are concerned, that sounds accurate. But from my experience, the LSAT typically doesn't go into this level of intricacy with regards to quantity.

  2.)
  Most beers are in cans. Most cans contain beer. Most beers contain alcohol.
  B most C
  C most B
  B most A

  The only inference you'd be able draw from this would be C some A. Linking up two most statements in chain doesn't provide you with an inference. Someone correct me if I'm wrong, but in this situation, an inference from two most statements can only be drawn with C some A rather than B some A, since C and A both link back up to B.

  Sorry if I wasn't much help here. I've been trying to clarify my understanding on these types of questions recently as well.

  Maybe someone else can chime in here and solidify this, but A most B most C doesn't yield any inferences I don't think, partly because Most cannot be reversed like Some can in these situations. The double arrows can't applied here for this reason. This probably goes outside your question here, but if it were A most B-->C then you would be able to say A most C or A<---some--->C since Some includes Most.

  I have a list of all the valid/invalid inferences with all/some/most statements. It's pretty concise and I refer back to it at least once a week. Message me if you'd like me to send it over.
• Blake160 Alum Member

  April 2017 97 karma

  I'm sorry, I miswrote #2. I have edited the original post.

  Also, thanks for taking the time to write such a thoughtful response. I will respond when I'm off work!
• extramedium Alum Member

  April 2017 edited April 2017 419 karma

  @Blake160 said:
  I'm sorry, I miswrote #2. I have edited the original post.

  Also, thanks for taking the time to write such a thoughtful response. I will respond when I'm off work!

  No problem. I enjoy it. Helps me sort out some of my misconceptions.

  With #2 though, the first sentence is pretty much irrelevant to drawing the inference C some A.

  That inference only can be drawn from the last two sentences on their own.
  B most C
  B most A

  Therefore:
  C some A

  You can't link up two most statements in a chain and draw an inference from that. You would need to have a Most or Some statement chained to an All statement to make an inference, and the Most/Some statement will always need to come first. I think some of the confusion might come from our understanding (me personally) of not being able to reverse Most statements like you can with Some. Actually came across an issue with that today.

  Also, some is usually represented as A<--some-->B, whereas most is A--most-->B if that makes it clearer.
• AllezAllez21 Member Inactive Sage Inactive ⭐

  April 2017 1917 karma

  I agree with @extramedium. Good analysis.
• Blake160 Alum Member

  April 2017 97 karma

  @extramedium

  Thanks for your input. I'm trying to learn most/some statements and I was going over JY's valid and invalid argument forms and trying to draw them out as buckets. Then I got thinking about some strange ideas.

  I understand that A ----Most---> B ---Most---->C cannot yield any inference about A and C. I tried to reinforce this in a diagram to aid my understanding:

  http://i.imgur.com/4NWkLbC.png

  Then I thought about adding the additional premise, B --Most--->A:

  http://i.imgur.com/6tEtILd.png

  I really couldn't think of any way to rearrange, resize, or overlap the buckets in a way that doesn't yield A some C.

  Can you think of a situation where most A's are B's, most B's are A's, most B's are C's, but no A's are C's?

  Maybe I'm overthinking this!
• AllezAllez21 Member Inactive Sage Inactive ⭐

  April 2017 edited April 2017 1917 karma

  I think it goes back to what @extramedium already said. You can draw the inference about AsomeC because of BmostA and BmostC. That's where you get your intersectional statement. It doesn't particularly matter that AmostB.
• Blake160 Alum Member

  April 2017 97 karma

  @AllezAllez21 You're right. Thank you!
• Blake160 Alum Member

  April 2017 97 karma

  @extramedium Thank you for your help! It would be much appreciated if you could send me the list of valid/invalid arguments you mentioned in your first post.
• extramedium Alum Member

  April 2017 419 karma

  @Blake160 said:
  @extramedium Thank you for your help! It would be much appreciated if you could send me the list of valid/invalid arguments you mentioned in your first post.

  No problem. Sorry for the lengthy responses. I'll send over that list shortly.