A proof of $(\forall x P(x)) \to A) \Rightarrow \exists x (P(x) \to A)$ I recently asked this question. In that question I presented a hand-waving proof as part of the question. There was some confusion as to the validity of my hand-waving proof. So I wanted to make it more precise. The difficulty that I had is that it's been 25 years since I've written formal logic proofs. I've forgotten some of the rules (Like how to handle introducing and eliminating quantifiers) and rule names. So as you'll see below I just invented names of rules in the proof below.
A few questions.


*

*Assuming I used standard rule names (rather than made up ones), is the proof correct. Or do a few steps need to be "tightened" up. What is a good reference to "standard" rule names typically used on this site. 

*I never know how to typeset these sorts of proofs. Like in steps 3-5 of Case 2, I'd expect those to be indented to show that $c$ is only valid in the context of the existential. (Update: they are now indented using \quad) (I suppose this should be a separate question on meta or a LaTex site?)


A proof of $(\forall x P(x)) \to A) \Rightarrow \exists x (P(x) \to A)$
I'll prove by case analysis on $ \forall x P(x) $.
Case 1: $ \forall x P(x) $
$$\begin{align}
  & (\forall x P(x)) \to A && \text{Hypothesis} & \tag{1} \\
  & \forall x P(x) && \text{Case Hypothesis} & \tag{2} \\
  & A && \text{Implication 1, 2} & \tag{3} \\
  & \neg P(c) \vee A && \text{Or Introduction} & \tag{4} \\
  & P(c) \to A && \text{Definition of Implies, 4} & \tag{5}\\
  & \exists x (P(x) \to A) && \text{Existential Intro, 5} & \tag{6}\\
\end{align} $$
Case 2: $ \neg \forall x P(x) $
$$\begin{align}
  & \neg \forall x P(x) && \text{Case Hypothesis} & \tag{1} \\
  & \exists x \neg P(x) && \text{A Known Identity, 1} & \tag{2} \\
  & \quad \neg P(c)           && \text{Assumption, ref 2} & \tag{3} \\
  & \quad \neg P(c) \vee A    && \text{Or Introduction, 3} & \tag{4} \\
  & \quad P(c) \to A          && \text{Definition of Implies, 4} & \tag{5} \\
  & \quad \exists x (P(x) \to A) && \text{Existential Intro, 5} & \tag{6} \\  
  & \exists x (P(x) \to A) && \text{Existential Elim, 2,6} & \tag{7} \\
\end{align} $$
UPDATE: A commenter pointed out that the statement is false in the case that the universe is empty, so let's assume a non-empty universe.
UPDATE: An answer addressed my funny introduction of $\neg P(c)$ in case 2 step 3. I tried to make things more precise by using indentation and referenced EE as demonstrated here: http://softoption.us/content/node/277
 A: Addressing question 1 specifically:
I am not an expert in the subject but I took a couple of logic subjects at uni and never really saw a standard naming convention for the deduction rules in first order logic. You are likely to find different authors with different conventions. What is most important is you follow the rules correctly and you have in this case.
A: This depends on which rules you're allowed to use. If you're allowed to use those rules that you're using it seems OK - and all rules that's been used seem alowable too.
The comment of a non-empty universe is inherent in one or a few of the rules and since it's quite reasonable to require non-empty universe this fact doesn't ban these rules from being alowable.
Depending on how you view it's when using existential introduction (or perhaps when you introduce $c$ via or introduction). Either it's to conclude $\exists x \phi(x)$ from $\forall x \phi(x)$ that assumes non-empty or it's the assumption that $\phi(c)$ implies that $c$ refers to something existing.
Note that the second case might not need to rely on rules that imply non-empty universe since the case premise is false there. That's why you for example from the premise can conclude $\exists x\neg P(x)$
A: I'm a little bit puzzled by your proof in the case 2, i.e. under the assumption $\lnot \forall x \, P(x)$. As your proof should be valid classically but not intuitionistically (since it uses the law of excluded middle), you cannot infer from $\exists x \, \lnot P(x)$ that $P(c)$ (your step 3), the best you can classically do is apply the Herbrand's theorem. Indeed, in natural deduction (see Prawitz's Natural Deduction) the elimination rule for $\exists$ is NOT of the form 
\begin{align*}
\frac{\exists x A}{A[c/x]}
\end{align*}
So, I think that you have to reformulate the case 2 as follows. Assume $\lnot \forall x \, P(x)$; this means that $\exists x \, \lnot P(x)$. Let us suppose that $x$ is such that $\lnot P(x)$: then $P(x) \to A$  since any statement can be proven from a contradiction (ex falso quodlibet), in particular $A$. Hence, for any $x$ such that $\lnot P(x)$ we have proved that $\exists x \, (P(x) \to A)$. Therefore, under the assumption $\lnot \forall x \, P(x)$, we have proved that $\exists x \, (P(x) \to A)$.
I formalized your whole proof of $\forall x \, P(x) \to A \vdash \exists x \, (P(x) \to A)$ (following my argument for the case 2) in Prawitz's natural deduction (I suppose you are referring to natural deduction when you say "how to handle introducing and eliminating quantifiers"):

where $\pi$ is a derivation in classical natural deduction (but it is not intuitionsitically valid) of the instance $\forall x \, P(x) \lor \lnot \forall x \, P(x)$ of the law of excluded middle, and $\pi_0$ is a derivation in classical natural deduction (but it is not intuitionsitically valid) of $\exists x \, \lnot P(x)$ under the assumption $\lnot \forall x \, P(x)$. I can detail $\pi$ and $\pi_0$ but I think it is out of scope of this discussion.
