Daniil wrote an excellent post, but just to add to that a little bit:
As Daniil pointed out, you can't capture any truth-functions that non-trivially depend on more than $1$ variable, such as $P \land Q$, with only a $\neg$. So, let's restrict ourselves to functions defined over one variable, $P$, and see if maybe we can capture all those using a $\neg$?
Unfortunately, the answer is still no. Again, as Daniil already pointed out, we can't capture any tautology or contradiction. That is, we can't capture the truth-function that always returns true (i.e. the function $f$ such that $f(T)=f(F)=T$), nor can we capture the truth-function that always returns false (i.e. the function $f'$ such that $f'(T)=f'(F)=F$)
So in this post I just wanted to show you how you can prove that result using induction. In particular, let's prove the following:
Claim
For any expression $\phi$ built up from $P$ and $\neg$ alone, it will be true that if $v$ is the valuation that sets $P$ to true (i.e. $v(P)=T$), and $v'$ is the valuation that sets $P$ to false (i.e. $v'(P)=F$), then either $v(\phi)=T$ and $v'(\phi)=F$, or $v'(\phi)=T$ and $v(\phi)=F$ (in other words, $v(\phi)$ and $v'(\phi)$ will always opposite values, meaning that $\phi$ can not be a tautology or contradiction, for that would require that $\phi$ has the same value for any valuation)
Proof
We'll prove the claim by structural induction on the formation of $\phi$:
*Base: *
$\phi=P$. Then $v(\phi)=v(P)=T$, while $v'(\phi)=v'(P)=F$. Check!
Step:
If $\phi$ is not an atomic proposition, then there is only one possibility: $\phi$ is the negation of some other statement $\psi$, i.e. $\phi = \neg \psi$.
Now, by inductive hypothesis we can assume that $v(\psi)=T$ and $v'(\psi)=F$, or $v'(\psi)=T$ and $v(\psi)=F$
Well, if $v(\psi)=T$ and $v'(\psi)=F$, then $v(\phi)=v(\neg \psi)=F$ and $v'(\phi)=v'(\neg \psi) =T$. On the other hand, if $v(\psi)=F$ and $v'(\psi)=T$, then $v(\phi)=v(\neg \psi)=T$ and $v'(\phi)=v'(\neg \psi) =F$. So, we can conclude that $v(\phi)=T$ and $v'(\phi)=F$, or $v'(\phi)=T$ and $v(\phi)=F$, as desired.
Best Answer
I presume you are working with propositional logic. You are given that $\{\lor, \land, \neg\}$ is expressively complete. This means that any formula in propositional logic can be rewritten so that it only uses the connectives $\lor$, $\land$, and $\neg$.
Now, to show that $\{\land, \neg\}$ is also expressively complete, all you need to do is show that you can rewrite any formula that uses only $\lor$, $\land$, and $\neg$ so that it only uses $\land$ and $\neg$, or more to the point, you need to show that you can write $\lor$ using $\land$ and $\neg$.
This can of course be done using De Morgan's laws. One of De Morgan's laws says that $$\neg (A \lor B) \equiv \neg B \land \neg A.$$ Now, if we negate both sides of this equivalence, we get $$\neg \neg (A \lor B) \equiv (A \lor B) \equiv \neg (\neg B \land \neg A),$$ which tells you how to write $\lor$ in terms of $\neg$ and $\land$.