First of all, sorry for my bad english. I'm not very comfortable writting about Boolean algebra in english but I'll try my best.
So my teacher came up with this problem. Express the following function using only AND + NOT gates, then, NAND only.
The function: $$ f = {a'}{b'}+{a}{b}+{b'}{c} $$
After converting f to its cannonical form I got the truth table so I can check if I made some mistakes when doing the circuit.
But my question is, do I have to apply the same process to express it as AND+NOT and NAND? They both use the same logic.
My solutions were:
NAND: $$ f'' = [({a'}{b'})'*({a}{b})'*({b'}{c})']' $$
AND + NOT: $$ f' = ({a'}{b'})'*({a}{b})'*({b'}{c})' $$
And here are the circuits:
If I understand the root of your question, you are wondering why AND + NOT and NAND are different in circuits but look more similar in boolean algebra.
You already mentioned that they are logically equivalent. The difference is in how you encode it. Logic gates have an operation for NAND.
You can emphasise the difference if you introduce a function or symbol that performs NAND in boolean algebra.
$$f = (a'\uparrow b')\uparrow(a\uparrow b)\uparrow(b'\uparrow c)$$
Which still means the same as
$$f = [(a'b')'\land(ab)'\land(b'c)']'$$
So, yes, whatever method you use to convert to AND + NOT can be used to convert to NAND.