I'll interpret the given statement as
[tex]\neg \bigg( \neg s \lor \big( \neg r \land s \big) \bigg)[/tex]
where [tex]\neg x[/tex] means "not x", [tex]\lor[/tex] means "or", and [tex]\land[/tex] means "and".
If r is false, then [tex]\neg r[/tex] is true.
s is given to be false, so [tex]\neg r\land s[/tex] (basically "true and false") is false.
If s is false, then [tex]\neg s[/tex] is true.
Then [tex]\neg s \lor \big(\neg r \land s\big)[/tex] (i.e. "true or false") is true.
Take the negation of that and you end up with a false statement.
If you intended "~r < s" to mean something like "not r is implied by s", so the original statement is actually
[tex]\neg\bigg(\neg s \lor \big(\neg r \impliedby s \big)\bigg)[/tex]
then [tex]s\implies \neg r[/tex] is true because s is false. Then [tex]\neg s \lor \big(\neg r \impliedby s\big)[/tex] is still true, so the statement still ends up being false.