Is there only one negation of the statement? I am very much confused with the negation of the following statement. 

A sequence of real numbers is divergent $\implies$ either it is unbounded or there must exist at least one pair subsequences of the original sequence such that their limits are not equal.

Actually this problem was given to one of my friend who tried to prove this by assuming the negation of this statement. So his negation was,

If sequence of real numbers $\{x_n\}$ is bounded and there exists no (or exactly one, which is discarded due to meaning nonsense) subsequence of it which converges to two different limits then the sequence is convergent.

But yesterday when I told this to one of my professors, he replied that it seems to him that there must be two negations, not one. But he was in hurry and didn't tell me what the other negation is. Instead he told me to figure it out myself. But I can't. 
Any help will be appreciated. If the negation is shown explicitly using logical connectives it will be the best.  
 A: The statement :

(a sequence) "either it is unbounded or there must exist at least one pair subsequences of the original sequence such that their limits are not equal"

has the "logical form" :

$Unbound(s) \lor \exists x \exists y (x \ne y \land Q(x,y,s))$

where the subformula $Q(x,y,s)$ abbreviates : "$x$ and $y$ are subsequences of the original sequence $s$ such that their limits are not equal". 
Thus, its negation must be :

$\lnot Unbound(s) \land \lnot \exists x \exists y (x \ne y \land Q(x,y,s))$

i.e.

$\lnot Unbound(s) \land \forall x \forall y \lnot (x \ne y \land Q(x,y,s))$

i.e.

$\lnot Unbound(s) \land \forall x \forall y (x = y \lor \lnot Q(x,y,s))$.

Thus ("unwinding" the subformula $Q(x,y,s)$), we can say that :


$\lnot Diverg(s)$ iff $Bound(s) \land \forall x \forall y (x = y \lor \lnot Sub(x,s) \lor \lnot Sub(y,s) \lor \lnot (Lim(x) \ne Lim(y)))$.


We can rewrite it as :

$Converg(s)$ iff $Bound(s) \land \forall x \forall y ((x \ne y \land Sub(x,s) \land Sub(y,s)) \to (Lim(x) = Lim(y)))$.


Now you have to check if it makes sense ...
