Variable use restriction in a Hilbert-style predicate calculus system

Question

I am solving an exercise from A Primer for Logic and Proof (Hirst & Hirst) using the book's axiom system. One of the (meta)theorems available is leaving me unsure if I applied it correctly, as it is stated intuitively and not super formally.

Here are the relevant axioms/theorems I use in the proof, as stated in the book:

Axiom 4: $\forall x A(x) \Rightarrow A(t)$ $\quad$ provided $t$ is free for $x$ in $A(x)$

Rule C: If $\exists x A (x)$ is a previous line in a proof, we may write $A(\underline{c})$ as a line, provided the following two conditions hold:

1. $\underline{c}$ is a new constant symbol (That is $\underline{c}$ doesn't show up in any earlier lines of the proof, or in any proper axioms we ever plan to use.)
2. If some variable (say $y$) appears free in the formula $\exists x A(x)$, then Generalization is never applied to $y$ in the proof.

Adding $\exists$: If $\forall x A(x) \Rightarrow A(t)$ is a correct instance of Axiom 4, then from $A(t)$ we may deduce $\exists x A(x)$

There are other lemmas I use, but I omit them here for brevity, as they do not cause problems. The exercise and the proof are as follows:

$$\text{Prove: } (\forall x A(x) \wedge \exists x B(x)) \Rightarrow \exists x (A(x) \wedge B(x))$$

Proof.

$\textbf{1. }\forall x A(x) \wedge \exists x B(x) \quad \text{Given (setup for Deduction Theorem)}$

$\textbf{2. } \forall x A(x) \quad \text{Conjunction elimination (lemma), 1}$

$\textbf{3. } \exists x B(x) \quad \text{Conjunction elimination (lemma), 1}$

$\textbf{4. } B(\underline{c}) \quad \text{Rule C, 3}$

$\textbf{5. } \forall x A(x) \Rightarrow A(\underline{c}) \quad \text{Axiom 4}$

$\textbf{6. } A(\underline{c}) \quad \text{Modus Ponens, 2,5}$

$\textbf{7. } A(\underline{c}) \wedge B(\underline{c}) \quad \text{Conjunction introduction (lemma), 4,6}$

$\textbf{8. } \exists x (A(x) \wedge B(x)) \quad \text{Adding } \exists \text{, 7}$

$\textbf{9. } (\forall x A(x) \wedge \exists x B(x)) \Rightarrow \exists x (A(x) \wedge B(x)) \quad \text{Deduction Theorem, 1,8}$

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My main concern is this: Is it legal to use Axiom 4 in line 5 containing the constant $\underline{c}$? The restriction for Rule C states that $\underline{c}$ may not appear in any proper axioms we ever plan to use - but the term "proper axiom" is never defined in the book. There aren't many results for this online, but it seems to refer to a non-logical axiom. Does this make Axiom 4 exempt from the restriction, allowing the use of $\underline{c}$? I also use the axiom after the use of Rule C, hopefully abiding to the restriction that $\underline{c}$ may not appear before the rule, but should be okay to appear again after. I can not find these exact axioms/rules with such variable restrictions elsewhere, there seem to be other variances but not exactly the same.

Edit Deduction Theorem:

If there is a proof of $A \vdash B$ with no applications of generalization to any variables that occur free in $A$, then there is a proof of $\vdash A \Rightarrow B$

Answer 1

Line 5 is fine, but your questions show you haven't understood the spirit of Rule C.

Rule C says: if you know that a predicate is satisfied by some object, you are allowed to "choose" such an object and give it a name.

Condition 1 just says that that name must not be already in use. Otherwise we could write $A(\emptyset)$ or $A(\mathbb{N})$ on a line, which would make a statement about an existing constant instead of introducing a new constant.

There is no problem at all with using $\underline{c}$ after the application of Rule C. In fact, that is the whole point of it. You introduce a new constant symbol so that you can later refer to it. Directly after the introduction, the one and only thing you know about $\underline{c}$ is that $A(\underline{c})$ holds, and everything you later state about $\underline{c}$ will be deduced from that.