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I stumbled across this rule:

$$\frac{\Gamma\vdash q\rightarrow p(a) }{\Gamma \vdash (q \rightarrow \forall x. p(x))}$$

where $a$ also needs to be a fresh constant, so with that in mind you could re-write the rule as the following:

$$\frac{\Gamma\vdash q\rightarrow p(a)\hspace{1cm} a\text{ is fresh}}{\Gamma \vdash (q \rightarrow \forall x. p(x))}$$.

The calculus I'm working with is a standard sequent calculus, without the weakening rules, and for the quantifier, I also have the following rule, which is very similar:

$$\frac{\Gamma\vdash p(a)\hspace{1cm} a\text{ is fresh}}{\Gamma \vdash \forall x. p(x)}\text{$\forall$R}$$.

I tried several approaches to derive the first (or second) rule without any success, is it possible to do that?

Taroccoesbrocco
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1 Answers1

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The idea is that (using a top-down approach) you first have to decompose the formula $q \to p(a)$ using the rules $\to_L$ and cut, then you apply $\forall_R$ to $p(a)$, and finally, you recompose the formula $q \to \forall x. p(x)$ using $\to_R$.

Note that this is possible because you are assuming that $a$ is fresh, which means that it does not occur free in $\Gamma$ and $q$, otherwise the rule $\forall_R$ could not be applied.

$$ \dfrac { \lower{1.5ex}{\Gamma \vdash q \to p(a)} \qquad \dfrac{ \dfrac{}{q \vdash q}\text{ax} \qquad \dfrac{}{p(a) \vdash p(a)}\text{ax} } {q, \, q\to p(a)\vdash p(a)} \to_L } { \dfrac { \dfrac{\Gamma, q \vdash p(a)}{\Gamma, q \vdash \forall x.p(x)}\forall_R } {\Gamma \vdash q \to \forall x.p(x)} \to_R } \text{cut} $$

Taroccoesbrocco
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