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In FOL, consider the following statement in English. If

\begin{array} \\ \Gamma \vdash p_1 \\ \dots \\ \Gamma \vdash p_n \\ \hline \Gamma \vdash p \end{array}

then I guessed that $$ p_1, \dots, p_n \vdash p $$

But then I realized that if $\Gamma \vdash p$ holds and doesn't depend on $\Gamma \vdash p_1, \dots, \Gamma \vdash p_n$, then $p_1, \dots, p_n \vdash p$ doesn't necessarily hold. Is it a counterexample to the above statement?

I was wondering what change could be made to the statement to make it correct?

What about adding this condition to the statement:

None of $\Gamma \vdash p_1, \dots, \Gamma \vdash p_n$ could be omitted for $\Gamma \vdash p$ to hold.

? But I saw Exercise 6 on p130 in Enderton's A Mathematical Introduction to Logic:

  1. (a) Show that if $\vdash \alpha \to \beta$ , then $\vdash \forall x \alpha \to \forall x \beta$.

(b) Show that it is not in general true that $\alpha \to \beta \models \forall x \alpha \to \forall x \beta$.

So why is (b)?

Or can we change the conclusion? For example, how about this change:

If

\begin{array} \\ \Gamma \vdash p_1 \\ \dots \\ \Gamma \vdash p_n \\ \hline \Gamma \vdash p \end{array}

then

$$ \vdash ((\Gamma \to p_1) \And \dots \And (\Gamma \to p_n)) \to (\Gamma \to p) $$ ?

Tim
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  • I don't think you can eliminate $\Gamma$ so easily. For intuition, think of $\Gamma$ as a collection of axioms like ZFC. – Karl Oct 12 '23 at 04:24
  • Also, what exactly is the meaning of your proposed additional condition? Are there implicit quantifiers in the meta-statements about $\Gamma,p_i,p$? If so, where do they go? – Karl Oct 12 '23 at 04:27
  • @Karl that is my question of what changes to be made so that the statement is correct? Also see what change can be made to exercise 6(b), so that it holds? – Tim Oct 12 '23 at 04:45
  • In general the first form above is the correct one; consider e.g. the simple case of so-called $(\land \ \text {Introduction})$: \begin{array} \ \Gamma \vdash p_1 \ \Gamma \vdash p_2 \ \hline \Gamma \vdash (p_1 \land p_2) \end{array}. It is the correct expression for a rule of inference, and reads: "if we have a derivation of $p_1$ from the set $\Gamma$ of assumptions [$\Gamma \vdash p_1$] and we have a derivation..., then we can derive $(p_1 \land p_2)$ from the set $\Gamma$ [$\Gamma \vdash (p_1 \land p_2)$]. – Mauro ALLEGRANZA Oct 12 '23 at 12:45
  • The "if..., then..." construction of the rule is expressed with the standard symbolic form of inference rules, using the horizontal bar. – Mauro ALLEGRANZA Oct 12 '23 at 12:47
  • For Ex.6a, apply Gen.Th (no free $x$ in assumptions) followed by Ax.3. The same approach does not work for 6b. See the discussion regarding the proviso applying to "generalization" in your previous post. – Mauro ALLEGRANZA Oct 12 '23 at 12:53

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