Is it possible to write this express with only nand and NOT?

Question

Find an equivalent expression using only $ nand $ and $ \lnot $ as well as grouping parenthesis. You may use $ A $, $ B $ and the operators any number of times.

(i) $ A \land B $

(ii) $ A \lor B $

(iii) $ A \Rightarrow B $

I was able to figure out (i). It is simply $ \lnot(A\,nand\,B) $

What about (ii) and (iii)? I cannot figure out how to flip the binaries in the middle of the truth table.

Answer 1

(i) $A\land B=\neg(A\uparrow B)$

(ii) $A\vee B= \neg(\neg A\land \neg B)=\neg(\neg(\neg A\uparrow \neg B))=\neg A\uparrow \neg B$

(iii) $A \Rightarrow B = \neg A \vee B= A\uparrow \neg B$

Answer 2

Hint: First try to express $\lnot A$. Then use that $\lnot (A \lor B) = (\lnot A) \land (\lnot B)$, and that $A \Rightarrow B = (\lnot A) \lor B$