Here is my thought
to prove the theorem we should get $\{\exists x_i (A \to B), A\} \vdash \exists x_i B $
then, I don't know how to process...
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kengkeng
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Informally, pick $x_i$ such thhat $A\to B$. Then using $A$ we conclude $B$. Hence for this $x_i$, we have $B$. – Hagen von Eitzen Jan 18 '15 at 17:52
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modus ponens... – vadim123 Jan 18 '15 at 17:52
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What are your inference rules? – Git Gud Jan 18 '15 at 17:55
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1As in your previous post, you have to specify the proof system you are using; with Natural Deduction it is quite easy, following Hagen's hint. – Mauro ALLEGRANZA Jan 18 '15 at 17:56
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you can use rules from this link@GitGud – kengkeng Jan 18 '15 at 17:57
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Think of an example. Assume: "there is a number $x_i$ such that: if the moon is made of green cheese, then $x_i$ is an odd perfect number" Deduce: "If the moon is made of green cheese, then there is an odd perfect number $x_i$". – GEdgar Jan 18 '15 at 17:57
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I understand the sentence, but I just have a hard time to formally prove it – kengkeng Jan 18 '15 at 17:58
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It is probably more useful for kengkeng to look at those rules and see which one applies, than to ask Git Gud to do it for him. – GEdgar Jan 18 '15 at 17:58
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I would recommend using $A \rightarrow B \equiv \lnot A \lor B$ as well as $(\exists x ~ A \lor B) \equiv \bigg((\exists x ~ A) \lor (\exists x~B) \bigg)$ – DanielV Jan 19 '15 at 04:35
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In the long answer to this post you can find a proof with Enderton's axioms. – Mauro ALLEGRANZA Jan 19 '15 at 07:12
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thank you all, I derived it. – kengkeng Jan 19 '15 at 08:08
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Use $\exists{x_i}(A\rightarrow{B})\equiv\neg \forall{x_i}\neg(A\rightarrow{B})\equiv\neg \forall{x_i}(A\land\neg{B})$ and the instantiate universal quantifier.
jimbo
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