Thank you, Ominvium, for the right answer. Where I made a major mistake is not putting the clues (who said what) into sentences. I only considered each person's statement as a whole, and created sentences that merely captured whether each person was lying or not.
Here is the complete solution (version 1), described by Ominvium. Since the original question is from the Propositional Resolution (PR) section of my notes, the solution below is in that form.
(comments very much appreciated)
\begin{array}{lll}
A & = & \text{Adams did it} \\
B & = & \text{Brown did it} \\
C & = & \text{Clark did it} \\
p & = & \text{Brown knew victim} \\
q & = & \text{Brown was in town } \\
r & = & \text{Clark was in town} \\
t & = & \text{Adams was in town} \\
S_A & = & (\lnot A \land p) \\
S_B & = & (\lnot B \land \lnot p \land \lnot q) \\
S_C & = & (\lnot C \land q \land t) \\
\end{array}
Quick recap on Propositional Resolution (PR)
PR depends on clausal form :
\begin{array}{ll}
(p \lor q) = & \left\{p,q\right\} \\
(p \land q) = & \left\{p\right\} \\
& \left\{q\right\} \\
\end{array}
Key thing to note for this question is that following the PR procedure, if we get an empty clause {}, the sentences contain a contradiction. For example:
\begin{array}{ll}
(p \land \lnot p) = & \left\{ p \right\} \\
& \left\{ \lnot p \right\} \\
& \text{--------} \\
& \left\{\right\}
\end{array}
So in my approach, I took the three scenarios: 1) Adams is lying, 2) Brown is lying, or 3) Clark is lying, and worked through each to see which results in an empty clause.
We have 3 version that we need to check:
1) Adams is lying and the others are telling the truth:
\begin{array}{lll}
\lnot S_A & = \lnot ( \lnot A \land p ) & \qquad\qquad \\
& = ( A \lor \lnot p ) \\
\end{array}
\begin{array}{lll}
(S_B \land S_C ) \land \lnot S_A \\
1) \lnot B \\
2) \lnot p \\
3) \lnot q \\
4) \lnot C \\
5) q \\
6) t \\
7) \left\{A, \lnot p\right\} \\
\text{--------} \\
8) \left\{\right\}& \mbox{(3,5) Contradiction} \\
\end{array}
2) Brown is lying and the others are telling the truth:
\begin{array}{lll}
S_B & = \lnot ( \lnot B \land \lnot p \land \lnot q) & \qquad\qquad \\
& = \lnot ( B \lor p \lor q) & \\
\end{array}
\begin{array}{ll}
(S_A \land S_C ) \land \lnot S_B \\
1) \lnot A \\
2) p \\
3) C \\
4) q \\
5) t \\
6) \left\{B, p, q \right\} \\
\text{--------} \\
7) \mbox{No Contradiction} \\
\end{array}
3) Clark is lying and the others are telling the truth:
\begin{array}{lll}
\lnot S_C & = \lnot ( \lnot C \land q \land t ) & \qquad\qquad \\
& = ( C \lor \lnot q \lor \lnot t ) \\
\end{array}
\begin{array}{lll}
(S_A \land S_B ) \land \lnot S_C \\
1) \lnot A \\
2) p \\
3) \lnot B \\
4) \lnot p \\
5) \lnot q \\
6) \left\{C, \lnot p, \lnot q \right\} \\
\text{--------} \\
7) \left\{\right\} & \mbox{(2,4) Contradiction} \\
\end{array}
Since (1) and (3) have contradictions, only (2) can be true.
$\therefore $ Brown is lying, and is the murderer.
Best Answer
Let $\varphi$ be some sentence in the language you just described. We prove by induction on the complexity of $\varphi$ that either $\Gamma \vdash \varphi$ or $\Gamma \vdash \neg \varphi$.
The base case is just atomic sentences. That is, $\varphi$ is of the form $S_i$ for some $i$ (or maybe $\top$ or $\bot$, depending on your definitions). So this case is covered by the assumption that either $S_i \in \Gamma$ or $\neg S_i \in \Gamma$, which translates into $\Gamma \vdash S_i$ or $\Gamma \vdash \neg S_i$.
If you want to try the rest of the induction for yourself, you may want to consider to stop reading here.
There are two induction steps: one where $\varphi$ is of the form $\neg \psi$ and one where $\varphi$ is of the form $\psi_1 \lor \psi_2$. Let us consider each case.
If $\varphi$ is of the form $\neg \psi$, then by induction hypothesis we have $\Gamma \vdash \psi$ or $\Gamma \vdash \neg \psi$. In the first case we thus have $\Gamma \vdash \neg \varphi$, and in the second case we find $\Gamma \vdash \varphi$. So again, either $\Gamma \vdash \varphi$ or $\Gamma \vdash \neg \varphi$.
If $\varphi$ is of the form $\psi_1 \lor \psi_2$, we can again use the induction hypothesis to find a two different cases. The first case is that $\Gamma \vdash \psi_1$ or $\Gamma \vdash \psi_2$ (or both), which gives us $\Gamma \vdash \varphi$. The other case is that $\Gamma \vdash \neg \psi_1$ and $\Gamma \vdash \neg \psi_2$, which means that $\Gamma \vdash \neg \varphi$. Once more concluding that indeed $\Gamma \vdash \varphi$ or $\Gamma \vdash \neg \varphi$.
This completes the induction, and so we can conclude that $\Gamma$ is complete.