Representing a strong syllogism by the product rule in probability - What is the condition for this fraction to be defined?

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In Jaynes' book Probability Theory: the Logic of Science (page $65$ in the pdf, i.e. page $35$ in the book), the author says that the strong syllogism

$$A\implies B$$ $$ A$$ $$\therefore B$$

corresponds to the product rule:

$p(B|AC) = \frac{p(AB|C)}{p(A|C)}$ where $C$ is defined as $C\equiv A\implies B$

My question is (since it hasn't been specified in the book) under which conditions do we have $p(A|C)>0$ (so that the fraction is defined)?

My guess is that, since $P(A|C) = \frac{p(AC)}{p(C)}$ and that $p(C)$ is finite, the condition for $p(A|C)$ to be non-zero is that $p(AC)$ is non-zero as well. This, in a discrete space would mean that $A\cap C \neq \emptyset$ .

Is my reasoning right ?

Thank you in advance.