Predicate calculus-is there a way to prove that a false term can implicate anything and is therefore true? I'm currently trying to solve the following:
(∃x(¬A(x))) → [∀x    (A(x)) → B(z)   ]
using only the rules of predicate and propositional calculus. I've had a few stabs at the problem. My chief idea has to do with the A(x) statements. I understand that if there exists at least one false A(x), then A(x) is not all true. I understand that falsities can imply anything. How do I join these two ideas together with the proper formal logical notation?
For example, I first assume:
(∃x(¬A(x))) 
which means:
    [∀x    (A(x)) → B(z)   ]           (=> E)
        [∀x    (A(x))]              (assume)
        A(a)                        (∀E)
        F

So I have proved it's false. Now what?
I realise that there is a very good chance I am barking up the wrong tree but I cannot see any other way to prove this. B(z) is totally unrelated to A(z) and while existential elimination will let me arrive at a "q" statement, I do not see how I can get "q" from the other two terms. 
 A: You're being a bit inconsistent with parenthesis in the statement you're trying to prove, and maybe your notation is a bit unclear. But the idea is correct.
What you're basically relying on is the DeMorgan theorem for quantifiers, that is the fact that $\exists x(\neg\phi(x)) \leftrightarrow \neg\forall x(\phi(x))$. The proof may be a bit difficult to word in a natural way since we're doing a conditional proof twice and also in doing that we're throwing in an assumption that we're using in RAA. 
In addition it might seem a bit odd too that we assume something that directly contradicts what we're already have assumed, but that's perfectly allowed.
If we write it using symbols instead it may be a bit more "clear":
$$\begin{align}
\tag{1}\exists x(\neg A(x)), \forall x(A(x))&\vdash \exists x(\neg A(x))
&\text{ Assumption}\\
\tag{2}\exists x(\neg A(x)), \forall x(A(x))&\vdash \neg\forall x(A(x))
&\text{ Assumption}\\ \tag{3}\exists x(\neg A(x)), \forall x(A(x))&\vdash \forall x(A(x))
&\text{ DeMorgan(1)}\\
\tag{4}\exists x(\neg A(x)), \forall x(A(x))&\vdash B(z)
&\text{ RAA(2,3)}\\
\tag{5}\exists x(\neg A(x))&\vdash\forall x(A(x)) \rightarrow B(z)
&\text{ Cond(4)}\\
\tag{6}&\vdash \exists x(\neg A(x)) \rightarrow (\forall x(A(x)
 \rightarrow B(z))
&\text{ Cond(5)}
\end{align}$$
To be clear here, in formal logic we're not concerned with statements being true or false, we're concerned with statements being proved (possibly under a set of conditions to the left of $\vdash$ symbol). What RAA then is about is that if we have proven $\phi$ and $\neg\phi$ we can conclude $\psi$.
A: Here is a natural deduction proof (somehow informally stated). The details of the proof may vary according to how exactly your proof system is stated, but the following derivation should offer a good blueprint for adapting it to various proof systems.
The goal is to prove: $(\,\exists x\ \lnot A(x)\,)\ \to\ [ \,(\forall x\ A(x))\to B(z) \,]$.
1) Assume $\exists x\ \lnot A(x)$.
Now the goal is to prove: $(\forall x\ A(x))\to B(z)$ from assumption 1).
2) Assume  $\forall x\ A(x)$.
Now the goal is to prove: $B(z)$ from assumptions 1) and 2).
3) From 1), by instantiation of a existentially quantified formula, you get: $\lnot A(t)$, where $t$ is a term not appearing in the proof so far.
4) From 2), by instantiation of a universally quantified formula, you get $A(t)$.
5) From 4), by inserting a disjunction, you get $A(t)\lor B(z)$.
6) From 3) and 5), by "disjunctive syllogism", you get $B(z)$.
Now you discharge the assumption and you get the desired result, i.e.
7) By discharging assumption 2), you get: $(\forall x\ A(x))\to B(z)$.
8) By discharging assumption 1), you get: $(\,\exists x\ \lnot A(x)\,)\ \to\ [ \,(\forall x\ A(x))\to B(z) \,]$.
