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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.

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    $\begingroup$ "Some C are B" is $\exists x (Cx \land Bx)$ $\endgroup$ Jun 23 at 12:04
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    $\begingroup$ Having said that, your intuition is correct: we can conclude that $\exists x (Cx \land \lnot Ax)$ (that is: "Some C is not A". $\endgroup$ Jun 23 at 12:06
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    $\begingroup$ It's amazing that you can claim to have done a proof using the tableaux calculus when you do not even have syntactically valid premise and conclusion! Learn the basics properly. Since your textbook is so lousy, read "Language, Proof and Logic" instead. $\endgroup$
    – user21820
    Jun 23 at 12:52
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    $\begingroup$ The textbook's claim is horribly wrong. You can deduce any one of those two statements from those two statements. You can deduce the conjunction of those two statements. You can deduce the disjunction of any one of those two statements with any other statement. You can deduce any tautology from those two statements. There are infinitely many things you can deduce from those two statements. Including the one you indicate. $\endgroup$
    – Bram28
    Jun 23 at 15:33
  • $\begingroup$ If the issue has been resolved, do consider accepting and upvoting answers (clicking on the checkmark and up-arrows): this signals resolution, scores points, prevents the page from being bumped, and influences the site's search results, cleanup activities, and other behind-the-scenes processes. $\endgroup$
    – ryang
    Jun 25 at 15:48

2 Answers 2

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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 $(3).$

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"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?

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