Why can negation pass through multiple quantifiers? [Chartrand P52-53, Velleman P65] I'm mindful of the Quantifier Negation Laws and Negating a statement that ... several quantifiers.

  
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*$\neg \; \exists \; P(x) \equiv \forall \; x \; \neg \; P(x) $  
  
*$ \neg \; \forall \; x \; P(x) \equiv  \exists \; \neg \; P(x) $
  

How and why can negation permeate/pervade through (What's the proper term?) the quantifiers?
In other words, how does negation transform $(♦) \to (♣)$? To wit, how does negation effect $(♣)$?
For example, I negate the definition of uniform continuity :
$$\color{#FF4F00}{\neg}(\; \forall \; e > 0 \: \exists \; d > 0 \: \forall \; c \in S \: \forall \; x \in S \: {\LARGE{[}} \; |x - c|< d \implies |f(x) - f(c)| < e {\LARGE{]}} \;) \tag{♦}$$
$$\iff \exists \; e > 0 \; \color{#FF4F00}{\neg}( \; \exists \; d > 0 \: \forall \; c \in S \: \forall \; x \in S \: {\LARGE{[}} \; |x - c|< d \implies |f(x) - f(c)| < e {\LARGE{]}} \;) $$
$$\iff \exists \; e > 0 \; \exists \; d > 0 \: \color{#FF4F00}{\neg}(\; \forall \; c \in S \: \forall \; x \in S \: {\LARGE{[}} \; |x - c|< d \implies |f(x) - f(c)| < e {\LARGE{]}} \;) $$
After the negation suffuses the first four quantifiers and converts each, $(♦)$ becomes:
$$\exists \; e > 0 \: \forall \; d > 0 \: \exists \; c \in S \: \exists \; x \in S \:\: \color{#FF4F00}{\neg}{\LARGE{[}} \; |x - c|< d \implies |f(x) - f(c)| < e {\LARGE{]}} \tag{♣}$$
By virtue of Can $P \implies Q$ be represented by $P \vee \lnot Q $?, 
$$\begin{align}
\color{#FF4F00}{\neg}{\LARGE{[}} \; |x - c|< d \implies |f(x) - f(c)| < e {\LARGE{]}} 
&\equiv \color{#FF4F00}{\neg}{\LARGE{[}} \; \neg( \; |x - c|< d  \; ) \: \vee \:|f(x) - f(c)| < e {\LARGE{]}} \\
&\equiv \color{#FF4F00}{\neg}\neg( \; |x - c|< d  \; ) \: \wedge \: \color{#FF4F00}{\neg}( \;|f(x) - f(c)| < e \; ) \\
&\equiv  \; |x - c|< d  \;  \: \wedge \: |f(x) - f(c)| \ge e  
\end{align}$$
 A: I don't quite understand your question, but I'd like to note that
1) $\forall e > 0 : P(e)$ is a shorthand for $\forall e : (e > 0 \implies P(e))$
2) $\exists d > 0 : P(d)$ is a shorthand for $\exists d : (d > 0 \land P(d))$
Therefore $\forall e>0 \; \exists d > 0 : P(e, d)$ is a shorthand for
$\displaystyle \forall e : (e>0 \implies (\exists d: d>0 \land P(e,d)))$
Negating gives
$\begin{align}
\neg \forall e : [e>0 \implies (\exists d: d>0 \land P(e,d))] &\equiv \exists e: \neg \: [e>0 \implies (\exists d: d>0 \land P(e,d))] \\
&\equiv \exists e : \neg \: [\neg(e > 0) \lor (\exists d: d>0 \land P(e,d))] \\
&\equiv \exists e : e > 0 \land \neg \: (\exists d: d>0 \land P(e,d)) \\
&\equiv \exists e : e > 0 \land \forall d: \neg [d>0 \land P(e,d)] \\
&\equiv \exists e : e > 0 \land \forall d: [\neg (d>0) \lor \neg P(e,d)] \\
&\equiv \exists e : e > 0 \land \forall d: [d>0 \implies \neg P(e,d)]
\end{align}$
where the latter, using shorthand, looks like $\exists e > 0 \; \forall d > 0 \; \neg P(e, d)$
