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How to show in propositional logic, that $\mathrm{Cn}(\mathrm{Cn}(A)) = \mathrm{Cn}(A)$?

I thought of first showing $\mathrm{Cn}(\mathrm{Cn}(A)) \subseteq \mathrm{Cn}(A)$ and then $\mathrm{Cn}(\mathrm{Cn}(A)) \supseteq \mathrm{Cn}(A)$. I managed to show the second but don't exactly know how to show the first inclusion.

  • 5
    What does $\mathrm{Cn}$ mean?2012-05-02
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    What is $\operatorname{Cn}$ in your system?2012-05-02
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    Cn(A) are all the consequences of $A$. So all $a$ with $A \models a$.2012-05-02
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    Do you know that $A\models a$ iff $A\vdash a$?2012-05-02
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    Given $\phi$ such that $Cn(A)\models\phi$ you want $A\models\phi$. To show this you just need to show that for all models of $A$ satisfy $\phi$. Take an arbitrary model of $A$, let's call it $M$. Then by the definition of $Cn(A)$ we have that $M$ is also a model of $Cn(A)$. Since we assume that $Cn(A)\models\phi$ we get that $M$ satisfies $\phi$, which is what we wanted.2012-05-02

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