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A ---> B --SOME--> C; therefore A ----> C, is that a valid argument?

RaceTo180RaceTo180 Monthly Member

A ---> B --SOME--> C; therefore A ----> C, is that a valid argument?
I know A ---Some--> C is an invalid argument, but I am unsure about the not some situation.


  • Umar1100Umar1100 Alum Member
    6 karma

    It is not, as illustrated below by an example that conforms to those principles:
    This conforms to A --> B some C, yet it is not valid to say A --> C since it may not be the case that the Bs that are As also have Cs.

  • phosita_phoeatahphosita_phoeatah Yearly Member
    238 karma

    Poster above is correct.

    I think a slightly more intuitive way to think about this usage of Venn diagrams (which is just another way of representing a set, which is also what the poster above is doing).

    Also, A some C is a "weaker" relationship statement tying together A and C ("weaker" than A --> C). Former requires mere intersection; latter requires A to be completely within C (or else be equal to C).

    If the weaker relationship statement is invalid, it stands that the stronger statements (e.g. A -m-> C and A --> C) are also invalid. In this case, if there is not an instance of intersection, how can the entirety of A be within C.

  • KevinLuminateLSATKevinLuminateLSAT Alum Member
    edited December 2021 835 karma


    If you understand that Some As are C would be an invalid conclusion, how could All As are C be valid? In other words, if we can't be sure that there's even one A that's a C, why would we be sure that all As are C?

    I think your question reflects an understanding based almost purely on memorization of formulas, but not quite backed up yet by conceptual understanding. (I made some content on "inferences from quantified statements" that may be helpful to you on YouTube - let me know if you can't find it.)

  • poooumbaapoooumbaa Monthly Member
    26 karma

    It's easier to learn the 3 or 4 valid inferences that can be made with regards to some/most relationships. For every other relationship with some/most you can't draw inferences, so put them out of your mind.

    For LSAT purposes you don't need to know anything beyond memorizing those inferences, but if you're like me you'll want to understand the reasoning. When I'm not sure whether I can draw a particular inference I try to create an instance where that inference is not necessarily true. It only takes one counterexample to prove an inference is invalid.

    Try to think about the elements in terms of potential set sizes of indivisible elements. In this case let's assign the following set sizes: A contains 1, B contains 100, and C contains 1. So with the relationship A-->B some C, we have one [A] that is also a [B] where 99 [B] have no overlap with[A]. We also have one [B] that is also a [C] where 99 have [B] have no overlap with [C]. Who's so say that the one [B] assigned to [C] happened to be the [AB] element?

    So can you see the problem with saying all of the one [A] that we have is necessarily a [C]? We just don't know enough about the probabilities of assignment, so A does not necessarily imply C. We can even go a level deeper and say the elements are randomly assigned amongst the sets. Then we can say there is a 1% chance that all A-->C, so 99% of the time [A] will not imply [C]. We still can't draw the inference that you mentioned, as 99% of the time it would be incorrect.

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