Converting to CNF - Distribution I am still trying to get my head around what is valid CNF and how distribution works, I am not even sure if the examples I have tried to convert are correct for one, but I am also unsure of the rule for distribution and when I have looked at other posts here it seems so complicated and the actual rule does not seem to me to be obvious.
I think the correct process is:

Elimate Bi-Directionals
Eliminate Implications
Apply De Morgans Law
Apply Distributive Law

Clause One
A V B =&gt; C
-(A V B) V C -- Remove implication
(-A ^ -B) V C -- De Morgans Law
(-A V C) ^ (-B V C) -- Is this right? I am unsure of exactly how to distribute the or.

Clause Two
A ^ B =&gt; C
-(A ^ B) V C -- Eliminate Implication
(-A V -B) V C -- Is this now valid CNF or is there something more to do?

Clause Three
-A ^ -B =&gt; C
(A ^ B) V C -- Elimate Implication - Is this correct for a negated expression?
(A V C) ^ (B V C) -- Distribute OR - Is this correct?

I am not able to access the set of maths symbols for some reason so have used the notation in my course book.
I am currently taking a self taught course and have an exam in two weeks and I know there will conversion to CNF involved so I am trying to make sure I understand the steps now but I am still not entirely sure what is valid CNF.
The course material states it is a conjunction of disjunctions which I get, but then it seems to imply that: A V B V C V ..n is valid CNF?
By definition - for example Herbert B. Enderton - A Mathematical Introduction to Logic-Academic Press (2001), page 53 - simply saying CNF is conjuction of disjuntions or product of sums.

So, as we agreed earlier,
$$(I \lor M) \Rightarrow H = (\neg I \lor H) \land (\neg M \lor H )$$

For $(I \land M) \Rightarrow H$ we can get $\neg (I \land M) \lor H=\neg I \lor \neg M \lor H$ and this last is CNF

For third case for $(\neg I \land \neg M) \Rightarrow H$ answer is analogical $2$d case: $\neg(\neg I \land \neg M) \lor H = I \lor M \lor H$ and this also is valid CNF.


And answer for the last question is yes - $A_1 \lor A_2 \lor \cdots  \lor A_n$ is valid CNF if each $A_i$ is  either a sentence symbol or the negation of a sentence
symbol - it's allowed to have just one conjuct.
Let me add, that one symbol is as CNF, so DNF.