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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!
Thank you for your input, @ and @. Definitely helped me wrap my head around the apparent contradiction.
In particular, @, 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.