Say that an $n$-place Boolean function $G$ is realized by a wff $\alpha$ whose atomic sentence symbols are $A_1,...,A_n$ if $G(X_1,...,X_n)$ is equal to the truth value of $\alpha$ when $A_1,...,A_n$ are given the truth values $X_1,...,X_n$ respectively.
The key to answering your question is the following result.
Theorem. Every $n$-place Boolean function $G$ is realized by some $\alpha$.
Proof. If $G$ is constantly false then $A_1 \wedge \neg A_1$ realizes $G$. Otherwise, suppose there are $k$ points at which $G$ is true: $$X_1 = (X_{11},...,X_{1n}),...,X_k=(X_{k1},...,X_{kn}).$$
Now let $\beta_{ij}$ be $A_j$ or $\neg A_j$ according as $X_{ij}$ is true or false. Let $\gamma_i = \beta_{i1} \wedge ... \wedge \beta_{in}$ and let $\alpha = \gamma_1 \vee ... \vee \gamma_k$. Now check that $\alpha$ realizes $G$ (I leave this to you as an instructive exercise). QED
Remark. Notice that $\alpha$ in the foregoing proof is in DNF. From this observation and the theorem just proved, we see that $\{\vee, \wedge, \neg \}$ is adequate.
Finally, we have
Theorem. The set $\{\vee, \neg \}$ is adequate.
Proof. Let $\alpha$ be a wff using only connectives in $\{\vee, \wedge, \neg \}$. We can show by induction on $\alpha$ that there is a tautologically equivalent wff $\alpha '$ using only connectives in $\{\vee, \neg \}$. And from this it follows that $\{ \vee, \neg \}$ is adequate, since if $\alpha$ realizes $G$ and $\alpha \equiv \alpha '$, then $\alpha '$ realizes $G$ too.
The base case in which $\alpha$ is an atomic sentence symbol is trivial (let $\alpha '$ be $\alpha$).
For the inductive step, first suppose $\alpha$ is $\neg \beta$. Then let $\alpha '$ be $\neg \beta '$.
Second, suppose $\alpha$ is $\beta \wedge \gamma$. Then let $\alpha '$ be $\neg(\neg \beta ' \vee \neg \gamma ')$. It's easy to check that $\alpha \equiv \alpha '$. QED
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
Consider just two propositional variables, say $p$ and $q$, and let's see what truth-functions of these we can express using $\neg$ and $G$. Using just $\neg$, we have $p,q,\neg p,\neg q$. Now let's apply $G$ to any triple of these, say $G(x,y,z)$. If two of $x,y,z$ are the same, the $G$ just produces that same one of $p,q,\neg p,\neg q$ as its output. So the only way to get anything new would be if $x,y,z$ are distinct elements of $\{p,q,\neg p,\neg q\}$. But then, since only one of $p,q,\neg p,\neg q$ is missing from $x,y,z$, this triple would contain either both $p$ and $\neg p$ or both $q$ and $\neg q$. But if some two of $x,y,z$ are each other's negations, then $G(x,y,z)$ agrees with the other one of $x,y,z$, so we still get nothing new. Conclusion: The only truth fnctions of $p$ and $q$ that can be expressed using $\neg$ and $G$ are $p,q,\neg p,\neg q$.