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Concerning the proof of C34 (pp. 40-41): can the argument used to prove that $$(\exists x)(\forall y)R \implies (\forall y)(\exists x)R$$ is a theorem be applied to the (false) converse?

In detail (working in $\mathscr{T}_0$): for any relation $R$ and any letter $x$, we have the theorem $R \implies (\forall y)R$ by C27. Thus $$(\forall y)(\exists x)R \implies (\exists x)R \implies (\exists x)(\forall y)R$$ is a theorem by C30 and C31.

Randy Randerson
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  • C27 says: If $R$ is a theorem of a logical theory $\mathcal T$ in which the letter $x$ is not a constant, then $(\forall x)R$ is a theorem of $\mathcal T$. It does not say that $R\implies(\forall x)R$ is a theorem of $\mathcal T$ (supposing that $R$ is a relation of $\mathcal T$ and $x$ a letter which is not a constant). The assumption that $x$ is not a constant is crucial. (+1 for your nice question!) – Pierre-Yves Gaillard Jan 08 '16 at 05:57

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There is no error...

C27. says :

if $R$ is a theorem, then $(\forall x) \ R$ is a theorem.

In more "modern" notation :

if $\vdash R$, then $\vdash (\forall x) \ R$.

This is not the same as : $\vdash R \to (\forall x) \ R$, that is not valid.

Mauro ALLEGRANZA
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  • @RandyRanderson - because the Deduction Th (C14) must be "restricted" when interacting with $\forall$. If we apply it without restriction, from $R \vdash (\forall x) \ R$ we may conclude with : $⊢R→(∀x) R$, which is invalid. Think at the counterexample : $(x=0) \to (∀x) (x=0)$. – Mauro ALLEGRANZA Dec 21 '15 at 07:38