What's the negation of “At least three of the sentence are false”? Following with the question I asked before
What's the negation of "One of the sentence is false"?
The negation of “At least three of the sentence are false” would be "any, one or two of the sentence is/are false"?
 A: Call $f$ the number of false sentences.
Your "At least $3$ of these sentences are false" is 
$$
f \ge 3
$$
The negation of this is clearly
$$ f < 3$$
Thus, "the number of false sentences is smaller than $3$", i.e., "there are $0$, $1$ or $2$ false sentences", i.e., "there are at most $2$ false sentences."
A: With f-o logic, the original statement is:

$\exists x \ \exists y \ \exists z \ [(x \ne y \land x \ne z \land y \ne z) \ \land \ (False(x) \land False(y) \land False(z))]$.

Thus, negeatin it:

$\forall x \ \forall y \ \forall z \ \lnot [(x \ne y \land x \ne z \land y \ne z) \ \land \ (False(x) \land False(y) \land False(z))]$,

i.e.

$\forall x \ \forall y \ \forall z \ [(x \ne y \land x \ne z \land y \ne z) \ \to \ \lnot (False(x) \land False(y) \land False(z))]$,

i.e.


$\forall x \ \forall y \ \forall z \ [(x \ne y \land x \ne z \land y \ne z) \ \to \ ((False(x) \land False(y)) \to \lnot False(z))]$.


The final formula matches with the "informal" negation of the original statement:

"there are at most two false sentences."

A: The negation of “At least three of the sentence are false” is "At most two of the sentences are false".
Assuming variables range over sentences, the "at least three" statement can be represented by 
$$
\exists x,y,z(False(x)\land False(y)\land False(z)\land x\ne y \land x\ne z \land y\ne z).
$$
Its negation is (equivalent to)
$$
\forall x,y,z(False(x)\land False(y)\land False(z)\to x=y \lor x=z \lor y=z).
$$
A: No! The negation of the sentence "At least one of the three sentences is false" is "All three of the sentences are true" or, equivalently "None of the sentences are false".
