Negation of the statement - ' There exists at least $4$ vertices each of degree at most $5$ ' ' There exists at least $4$ vertices, each of degree at most $5$ ' .
What is the negation of this statement??
My answers : Given any set of vertices, at most three of them has degree at most $5$
Given any set of vertices $S$ , any subset of $S$ with cardinality $4$ , contains a vertex of degree $\geq6$.
I think both of them are equivalent and both are negation of the statement.  Am I correct? Please correct me if I am wrong ?
 A: These are both correct but the second one involves invoking a somewhat nontrivial fact: if every subset of size $4$ fails the condition, then every subset of size greater than or equal to $4$ fails the condition. The truth value is equivalent but I wouldn't say that they are identical statements. 
A: The second statement can really be thought of as:  
'Given any $4$ vertices, at least one will have a degree $\geq 6$'
And yes, the two are then easily seen to be equivalent.
A: I find symbolic manipulation helpful for negating complex statements. I’d write this as 
$\neg\big(\exists \{v_1\dots v_n|n\geq4\}$ s.t. $\forall i\leq n$, deg$(v_i)\leq5\big)$
and we can negate by reversing symbols:
$\forall \{v_1\dots v_n|n\geq4\}, \exists i\leq n$ s.t. deg$(v_i)>5$.
I’d render the English version as “every collection of at least 4 vertices has a vertex of degree more than 5.”

What I'm doing here is using the following rule from symbolic logic:
Let P(x) be any logical statement which depends on x, and x takes on values from S. Then
$$\neg\big(\forall x\in S, P(x)\big) \iff \exists x\in S \text{ s.t. } \neg P(x)$$
and
$$\neg\big(\exists x\in S \text{ s.t. } P(x)\big) \iff \forall x\in S, \neg P(x).$$ 
This means we can negate chained together statements as follows (I'm abusing the notation here, but hopefully my meaning is clear).
$$\neg\big(\forall \sim, \exists\sim, \forall \sim P(x)\big)\iff \exists\sim, \forall \sim, \exists\sim \neg P(x)$$
