$(\forall x \in D) (P(x)) \Leftrightarrow (\forall x)(x\in D \implies P(x)) $. how are two the same? Assuming all the variables are naturals, why are those two equal?
I don't get how $\implies$ is introduced in the latter equation: 
 $(∀x)(x∈D \implies P(x))$
Thanks.
 A: Because that's what the $\forall x\in D. P(x)$ notation is defined to mean.
A: If you interpret both D and P as sets, both of them say precisely that D is a subset of P.
A: As Peter Smith notes, most commonly 
$$\forall x \in D.\ P(x) \quad \text{ is defined as }\quad\forall x.\ x \in D \implies P(x).$$
However, should this be unsatisfactory explanation for you, there is alternative argument below.

First
\begin{align}
\mathtt{true} \implies P(x) \quad \text{ is equivalent to } \quad P(x), \\
\mathtt{false} \implies P(x) \quad \text{ is equivalent to } \quad \mathtt{true}.
\end{align}
Hence,
\begin{align}
\forall x \in D.\ P(x) &\quad\text{ is equivalent to }\quad \forall x \in D.\ x \in D \implies P(x), \\
\forall x \notin D.\ \mathtt{true} &\quad\text{ is equivalent to }\quad \forall x \notin D.\ x \in D \implies P(x).
\end{align}
Then, if we join them with
$$P'(x) = \begin{cases}P(x) & \text{for }x \in D \\ \mathtt{true} & \text{for }x \notin D\end{cases}$$
we get
$$\forall x.\ P'(x) \quad\text{ is equivalent to }\quad \forall x.\ x \in D \implies P(x).$$
I hope this helps $\ddot\smile$
