Wrong hint 1 and solution?

Either my logic is completely failing me, or the solution presented is just wrong.

First of all, $\neg (A\Rightarrow B)\equiv A\wedge \neg B$, which is not the same as $\neg (A\wedge B)\equiv A\Rightarrow \neg B$. But this was claimed in Hint 1.

I think the solution should be: There exists a < b and there exists a function f, such that f is continuous and (last logical statement, B, is false), i.e. there exists y between f(a) and f(b) such that for all c in [a,b] it holds that $f(c)\not =y$.

I agree with your comment on Hint 1 and the solution. Can we just edit it our way?

Moreover, hint 2, in the setting of an exam hint, is just inappropriate too. It's not as if we're learning negation for the first time and need analogies. A proper hint to this question should be helping students to:

first, identify the premise (A) and conclusion (B) statements, second, identify the order of existential and universal quantifiers within both A and B and what sub-statements they quantify

Then for the solution, we do: thirdly, apply negation judiciously to obtain the negation statement with A and not B, then finally, rephrase the statement for linguistic elegance.

) I think I was quite tired when I wrote this! I don't have time to rewrite something now, if someone else would like to, please have a stab at a new solution!

I changed the hint and the solution, please let me know what you guys think about it.

much better, in fact, QG.