Can you deduce an statement from the two statements "No B is an A" and "Some C are B"

Question

In my textbook it says that there is no statement you could deduce from the above two statements in the title. However in my opinion it would be possible to deduce "Some C are no A" from it.

$$ (\forall B \rightarrow \neg A) \wedge (\exists C \rightarrow B) \Vdash (\exists C \rightarrow \neg A) $$

Intuitively this would make sense for me and I also did a quick proof with tableau calculus. Now I am not sure if there is an error in my textbook or if I'm missing something. Either I wrongly converted the statements into boolean logic or the entailment is in fact not true.

Answer 1

In my textbook it says that there is no statement you could deduce from the above two statements in the title. However in my opinion it would be possible to deduce "Some C are no A" from it. $$ (\forall B \rightarrow \neg A) \wedge (\exists C \rightarrow B) \Vdash (\exists C \rightarrow \neg A) $$

"No B is an A" ought to be translated as $$¬∃x \;(Ax\land Bx)\tag1$$ or, equivalently, $$∀x \;(Bx\to¬Ax).$$

"Some C is B" ought to be translated as $$∃x \;(Bx\land Cx)\tag2.$$

"Some C is not A" ought to be translated as $$∃x \;(¬Ax\land Cx)\tag3.$$

Indeed, $(1)$ and $(2)$ together logically entails $(3).$

Answer 2

"In my textbook it says that there is no statement you could deduce from the above two statements in the title." Really? For a start you can deduce "No A is B" and "Some B are C". Then as @ryang notes, you are right that you can deduce "Some C are not A" [though NB his corrections of your badly misused quantifier symbolism]. You can also deduce "Some C are not both C and A', and "Some C are not both B and A", etc. etc.

So I wonder: does your textbook really make that silly claim you report?