I'm having trouble representing question # 3 in a propositional logic formula, from these lecture notes on propositional logic and propositional resolution:
3) Formalization and Proof. There are three suspects for a murder: Adams, Brown, and Clark. Adams says "I didn't do it. The victim was old acquaintance of Brown's. But Clark hated him." Brown states "I didn't do it. I didn't know the guy. Besides I was out of town all the week." Clark says "I didn't do it. I saw both Adams and Brown downtown with the victim that day; one of them must have done it." Assume that the two innocent men are telling the truth, but that the guilty man might not be. Write out the facts as sentences in Propositional Logic, and use propositional resolution to solve the crime.
So far I have this:
\begin{array}{l}
A & : \text{Adam did it} \\
B & : \text{Brown did it} \\
C & : \text{Clark did it} \\
S_1 = & (\lnot A \Rightarrow (B \lor C) ) & : \text{Not A. But the victim was old acquaintance of B's. } \\
&& : \text{But C hated him.} \\
S_2 = & (\lnot B) & : \text{Not B. B was out of town all the week.} \\
&& : \text{The fact that B was ot of town gives us no info.} \\
S_3 = & (\lnot C \Rightarrow (A \lor B) ) & : \text{Not C. C saw both A and B downtown with the} \\
&& : \text{ victim that day; one of them must have done it.} \\
S = & S_1 \lor S_2 \lor S_3
\end{array}
I feel like I should be deriving better interpretations. For example, should I interpret $S_1$ as:
$$ S_1' = (\lnot A \rightarrow (\lnot B \land C)) $$
Because B was victim's "acquaintance" (didn't kill him) and C "hated him" (so killed him)?
Best Answer
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.