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Related Discussions
Struggling with a valid argument form
CelerySoup
Alum Member
In the lesson Valid Argument Forms 4 - 9 of 9, a corollary to valid form 6 is introduced that reads—
A –> C
B –> C
∴ /A some /B
I understand how this form follows logically and how it relates to valid form 6, but it seems as though the /A some /B conclusion would NOT hold under the following scenario despite adhering to both initial premises.
Imagine you have A's.
A
A
A
All A's are C's.
AC
AC
AC
Imagine you have B's.
AC
AC
AC
B
B
B
All B's are C's (in this case, A and B do not intersect, without loss of generality).
AC
AC
AC
BC
BC
BC
Now, the inference should be "some non-A's are non-B's," however from the above scenario, all non A's ARE B's. Can somebody reconcile the above scenario with the valid argument form?
I've seen this example brought up in the lesson's comments section, however I have not seen it addressed directly.
Thanks!
Comments
@CelerySoup
I think I see two issues with your inference.
1) You wrote above all non-A's are B's. We don't know that in your world you can only be either A or B. What about things that are D? D is neither A nor B. We cant say "D", which is non-A, is a B.
2) In your above scenario you do have no intersection between A and B. But a "some" statement accounts for that possibility. "Some" includes anywhere from 1% - 100%. Therefore it includes "all".
- The problem with saying all non-A's are non-B's is that we don't usually know how many non-A's are non-B's. LSAT usually doesn't tell us, unlike your example above, how many A's and B's are not intersecting. So by saying "Some non-A's are non-B's" we are actually being precise in language and erring on the safe side. We are saying it could be some, most, or even all. But "some" is the word that allows us to encompass all three possibilities.
Let me know if that helped or answered your question.
Just briefly scanned this, but doesn't the argue form assume that the C's constitute a non-trivial partition of the domain? Of course, if the complement set of (the set of C's) is empty, then you could have the situation that you seem to be hitting upon. There, it seems that there needn't be one member of the domain which is a non-A and a non-B--but this is because every member of the universe is either an A or a B. However, I think that in such cases you assume that /C is non-empty, and therefore that there is at least one entity of the domain that is not in the set of C's. Since both the set A and the set B are subsets of the set C (because A-->C and B-->C), then you know that any such entity which is not in set C is also an entity which is a non-A and a non-B. Since "Non-A some non-B" only requires that there be one entity which satisfies the condition, the inference follows.
But you are right in a way. The inference appears to be invalid if you don't make the assumption that set C is a _proper_subset of the universe of discourse (i.e., the domain).
Thank you for your input, @Sami and @acsimon. Definitely helped me wrap my head around the apparent contradiction.
In particular, @acsimon, I think your mention of assuming that the C's constitute a non-trivial partition of the domain is spot-on. My knowledge of set theory is rudimentary, but from my understanding, it seems that this type of assumption (that C, B, A, or anything else for that matter, is not a universal set) is not unique to this argument form, but all arguments. Failing to make this assumption results in a contradiction.
Glad it helped. You shouldn't confuse the domain of discourse with the notion of a universal set though. It might be that the domain of discourse only contains ordinary objects and not sets, sets of sets, and so on. So, there is no need to worry about set theoretical paradoxs here, it seems.