Understanding the conversion of the existential quantifier to the universal quantifier 
I am trying to learn set theory on my own .. and the book introduced these suppositions I understand all of them except for the last one that converts the existential quantifier into the universal quantifier .. I understand what universal, existential quantifiers and negation mean .. so I understand that the conversion of one equals some form of negation of the other .. what I don’t understand though is the negation in front of the formula on the far right .. is it necessary ? why is it there ? What does it mean ? What does the whole sentence mean in this exact form ?
Thanks in advance
 A: To see what's going on here let's work with a concrete example. To do that we need to assign meaning with an interpretation, $\mathscr{I}$.
$\begin{array}{lrl}\mathscr{I}:&&\\
&\text{Domain:}&\text{{Alison, Bob, Craig}}\\
&\text{F:}&\ldots~\text{is a logician}\\
&\text{a:}&\text{Alison}\\
&\text{b:}&\text{Bob}\\
&\text{c:}&\text{Craig}\\
\end{array}$
Next, we need to know what it means for the formulas $\exists x[Fx]$ and $\forall x[Fx]$ to be true.
$\exists x[Fx]$ is true if, and only if, at least one member of the domain is an $F$. In other words, the formula is true if, and only if, Alison or Bob or Craig is a logician, in the inclusive sense of "or". Symbolically, we could represent that as $(Fa\lor Fb)\lor Fc$.
$\forall x[Fx]$ is true if, and only if, all members of the domain are an $F$. In other words, the formula is true if, and only if, Alison and Bob and Craig are logicians. Symbolically, that becomes $(Fa\land Fb)\land Fc$.
We can think of $\exists x[\dots]$ as disjunction, $\lor$, and $\forall x[\dots]$ as conjunction, $\land$. With that in mind, it's quite intuitive why $\exists x[Fx]=\lnot(\forall x[\lnot Fx])$.
Let's say $\exists x [Fx]$ is true. That means that someone is a logician. Consider the formula $\forall x[\lnot Fx]$ - can it be true at the same time? If someone is a logician, can we also claim that no one is a logician? No, that would be a contradiction. Hence, $\lnot\forall x[\lnot Fx]$ - it is not the case that no one is a logician. Symbolically, given $(Fa\lor Fb)\lor Fc$ is true $(\lnot Fa\land \lnot Fb)\land\lnot Fc$ is a contradiction, hence, $\lnot((\lnot Fa\land \lnot Fb)\land\lnot Fc)$, which is simply Double Negation and De Morgan's - $(P\lor Q)=\lnot\lnot(P\lor Q)=\lnot(\lnot P\land \lnot Q)=(\lnot\lnot P\lor\lnot\lnot Q).$
A: Reading the sentences $\quad\lnot\forall v\:\,\lnot\phi\quad$ and $\quad\exists v\:\, \phi\quad$ literally:

*

*It is false that every $v$ fails to satisfy $\phi.$


*There is some $v$ that satisfies $\phi.$
Does this clarify their logical equivalence?
