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Consider $\neg(X\wedge Y)\vdash\neg X \vee \neg Y$. The converse is valid since I can show that that inference is valid by using rules permitted. The above inference is valid classically but not intuitionistically. But, then, How can I show that certain inference is classically but not intuitionistically valid.

In classical case, to show that certain inference is not valid, it suffices to show that there exists a model in which the premises are all true but the consequent is false.

But in my textbook, it seems at least to me that there's no content on semantics of intuitionistic logic. How can I do?

Darae-Uri
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  • As you suspect, you need to establish a semantics for intuitionistic logic. There are several of these, although they may not be included in your text. See https://plato.stanford.edu/entries/logic-intuitionistic/ . Another option is to show that if the inference was valid, it would allow you to derive some other formula which is known not to be intuitionistically valid. However, that requires you to already know some other formula which is not intuitionistically valid, so in the end you need a semantics to establish that at least one formula is not intuitionistically valid. – Carl Mummert Dec 06 '16 at 14:35

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Typically, in the intuistionistic case, you would also do this by semantics. However, the semantics for intuistionistic logic is different than that for classical logic. One very common semantics for intuistionistic logics is that of Kripke frames.

A kripke frame which shows that $\neg (X \land Y) \not\vdash \neg X \lor \neg Y$ intuistionistically is the following: it consists of worlds $w,w_0,w_1$ with $w \leq w_X$ and $w \leq w_Y$, and $X$ holds at the world $w_X$ but $Y$ does not, and vice versa for $w_Y$. Then, at $w$, $\neg (X \land Y)$ holds, since in no "future world" $X \land Y$ holds, but $\neg X$ and $\neg Y$ both do not hold, since there is a "future world" where they do hold. Therefore, at $w$, $\neg X \lor \neg Y$ does not hold either.

Mees de Vries
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