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In my previous question, Mees de Vries talked about Kripke semantics and reading his links, I tried to show that $(X\wedge \neg Y)\rightarrow Z \vdash X\rightarrow (Y\vee Z)$ is valid using Kripke semantics.

Here is my answer, but I'm not sure whether it is right or not:

Since $(X\wedge \neg Y)\rightarrow Z$, for any world $u\ge w$, if $X\wedge \neg Y$ holds at $u$, then $Z$ holds at $u$. What I want to show is for any world $u \ge w$ if $X$ holds at $u$, then $Y\vee Z$ holds at $u$.

Let's call $u_X$ be the world where $u_X\ge u$ and $X$ holds at $u_X$. Then For any such $u_X\ge w$, either $Y$ or $\neg Y$ holds at $u_X$. If $\neg Y$ holds at $u_X$, then $(X\wedge \neg Y)$ holds at $u_X$. So, by the premise above, $Z$ holds at $u_X$, ans so does $Y\vee Z$. If $Y$ holds at $u_X$, then $ Y \vee Z$ holds at $u_X$. So, it is valid intuitionistically.

Is it right?

Community
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Darae-Uri
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  • Must be : What I want to show is for any world $u≥w$ if $X$ holds at $u$, then $Y∨Z$ holds at $u$.So, last paragraph must be : If $Y$ holds at $u_X$, then $Y∨Z$ holds at $u_X$. If $¬Y$ holds at $u_X$, then $(X∧¬Y)$ holds at $u_X$ and thus also $Z$ and $Y∨Z$. – Mauro ALLEGRANZA Dec 06 '16 at 15:54
  • @MauroALLEGRANZA There's some error in my question. I edited. Thanks :) – Darae-Uri Dec 06 '16 at 15:59

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There is an incorrect step where you say, "either $Y$ or $\neg Y$ holds at $u_X$." In Kripke semantics, you do not always have, for an arbitrary node $\alpha$ and sentence $\varphi$ that $\alpha\Vdash\varphi$ or $\alpha\Vdash\neg\varphi$. What you do have is that either $\alpha\Vdash\varphi$ or $\alpha\nVdash\varphi$, but $\alpha\nVdash\varphi$ does not imply $\alpha\Vdash\neg\varphi$.

Nathan
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