Prop. 3 of Sec. 5.6 of Fulton's "Algebraic Curves" is:
Let $C$ be an irreducible cubic, $C'$, $C''$ cubics. Suppose $C' ⋅ C = ∑_{i=1}^9 P_i$, where the $P_i$ are simple (not necessarily distinct) points on $C$, and suppose $C'' ⋅ C = ∑_{i=1}^8 P_i + Q$. Then $Q = P_9$.
Proof. Let $L$ be a line through $P_9$ that doesn't pass through $Q$; $L ⋅ C = P_9 + R + S$. Then $LC'' ⋅ C = C' ⋅ C + Q + R + S$, so there is a line $L'$ such that $L' ⋅ C = Q + R + S$. But then $L' = L$ and so $P_9 = Q$.
It seems to me that Fulton is implicitly assuming that this is a proof by contradiction with $P_9 ≠ Q$. But then it seems to me that the proof doesn't actually need to be structured that way, i.e. it can be proved directly. Here's my attempt:
Let $L$ be a line through $P_9$ such that $L ⋅ C = P_9 + R + S$. Then $LC'' ⋅ C = C' ⋅ C + Q + R + S$, so there is a line $L'$ such that $L' ⋅ C = Q + R + S$. But then $L' = L$ and so $P_9 = Q$.
As far as I can tell, nothing relies on $Q$ not equaling $R$ or $S$. Am I missing anything?
(From this answer, Fulton's proof has a small gap. A version with the fix would be:
Let $L$ be a line through $P_9$ such that $L ⋅ C = P_9 + R + S$ with both $R$ and $S$ being simple points (this is possible by Prob. 5.14). Then $LC'' ⋅ C = C' ⋅ C + Q + R + S$, so there is a line $L'$ such that $L' ⋅ C = Q + R + S$. If $R ≠ S$, then two points uniquely determine a line, so $L' = L$. Otherwise, if $R = S$, the unique line meeting $R = S$ with multiplicity two is the tangent line to the cubic at $R = S$, so $L' = L$. In either case, $P_9 = Q$.
)
