this post was submitted on 04 May 2025
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Your example is incorrect.
The first two do not make the third.
You can have:
To fix this, reverse the first statement.
Any portion of d that intersects with p (some p is d) must also be c (since all p is in c). Hence some c, but not all c, is in the portion of p that intersects with d (some c is d).
Oops. I fucked up lol. I changed it with your edit :p
Mental note: don't do syllogisms at 1am.