Proving an implication by proving its dual My textbook "Discrete and Combinatorial Mathematics, an Applied Introduction" by Ralph P. Grimaldi contains the following definition:

Let $s$ be a statement. If $s$ contains no logical connectives other than $\wedge$ and $\vee$, then the dual of $s$, denoted $s^d$, is the statement obtained from $s$ by replacing each occurrence of $\wedge$ and $\vee$ by $\vee$ and $\wedge$, respectively, and each occurrence of $T_0$ and $F_0$ by $F_0$ and $T_0$, respectively.

(I apologize for not knowing the formatting conventions here)
and the following theorem:

The Principle of Duality
  . Let $s$ and $t$ be statements that contain no logical connectives other than $\wedge$ and $\vee$. If $s \leftrightarrow t$, then $s^d \leftrightarrow t^d$.

Is anyone familiar with these? Can I use them to reason as follows:


*

*Start with a premise

*take its dual

*manipulate the dual

*arrive at the dual of our conclusion

*take its dual


This would allow me to do something like:
to prove that $\forall x\  P(x)\vee \forall\ Q(x) \rightarrow \forall x\ [P(x)\vee Q(x)]$
1. Ax,P(x) V Ax,Q(x)  premise
2. Ax,P(x) ^ Ax,Q(x)  the dual of (1)
3. Ax,P(x)            conjunctive simplification
4. P(c)               universal specification
5. Ax,Q(x)            conjunctive simplification from (2)
6. Q(c)               universal specification
7. P(c) ^ Q(c)        rule of conjunction from (4) and (6)
8. Ax[P(x) ^ Q(x)]    universal generalization
9. Ax[P(x) V Q(x)]    the dual of (8)

When I wrote this proof I didn't know that you could use universal specification to go from (1) to the dual of (7), so I needed a way to separate (1) into two parts. Taking the dual seemed to be a thing I can do. I worry though, because it seems like I could prove the converse this way, and the converse is false. 
So what's wrong? Why can't I use the concept of the dual of a statement in this way?
Thanks everyone!
a.
PS oh and I solved the question in another way, so this is just my curiosity.
PPS thank you for being patient with my ugly post, I'll try to learn your mark up.
 A: Your steps (1)-(2) and (8)-(9) don't work. The dual is defined in your quote only under the condition

If $s$ contains no logical connectives other than $\land$ and $\lor$, 

but here you also have quantifiers appearing in addition to $\land$ and $\lor$, and then the definition does not apply as written.
The principle behind the dual is that instead of the the formula $\phi(A,B,C,\ldots)$ where $A$, $B$, $C$ are propositional variables, you look at $\neg\phi(\neg A,\neg B,\neg C)$ and then use De Morgan's laws to push the negations down through the formula until they eliminate each other. On the way they change every $\land$ to $\lor$ and vice versa. You can then do some transformations and dualize again, giving you something equivalent to $\neg \neg \phi(\neg \neg A,\neg \neg B, \neg\neg C,\ldots)$, which of course is just $\phi(A,B,C,\ldots)$.
When there are quantifiers involved, you need to handle those using their appropriate De Morgan rules:
$$\neg(\forall x\,\phi) \leftrightarrow \exists x\,(\neg \phi) \qquad \qquad
\neg(\exists x\,\psi) \leftrightarrow \forall x\,(\neg \psi)$$
so when you dualize $(\forall x\,P(X)) \lor (\forall x\,Q(X))$ you should get $(\exists x\,\bar P(x)) \land (\exists x\,\bar Q(x))$, where $\bar P$ and $\bar Q$ are the duals of $P$ and $Q$.
