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If I have a rule of the form $\phi_{0\dots n}/\psi$, can I show that it is admissible, i.e. that if all premises are true then the conclusion is also true by showing that $\models\bigwedge_{k=0}^{n}\phi_k\rightarrow\psi$ provided that the concerned logic fulfills the deduction theorem?

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In general, no.

You can do such if you have a completeness meta-theorem and invoke it.

Say you work with a system which just has a conditional introduction rule and a conditional elimination rule:

{(cond $\alpha$ $\beta$), $\alpha$} $\vdash$ $\beta$

You might show that model-theoretically (semantically/using truth tables) the following holds:

|= (cond (cond (cond $\alpha$ $\beta$) $\alpha$) $\alpha$)

But, you can't admit the rule of inference

(cond (cond $\alpha$ $\beta$) $\alpha$) $\vdash$ $\alpha$ for the above system.

If you did then

$\vdash$ (cond (cond (cond $\alpha$ $\beta$) $\alpha$) $\alpha$)

which doesn't work for the above system, and you could find a model which satisfies the conditional introduce rule, the conditional elimination rule, but doesn't satisfy the last formula.

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    So for proving a rule being admissible, I would simply show that $\psi$ has to evaluate to 1 if all premises are set to do so?2017-01-05
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    @zzuussee Not unless you have the completeness theorem. If you have the rule of detachment, also known as *modus ponens*, if you show that $\vdash$(cond $\alpha$ $\beta$), then the rule of inference $\alpha$ $\vdash$ $\beta$ can get admitted.2017-01-06