Q:

First Order Logic - exactly one predicate is true

I am working on a question for an assignment and I am to declare a clause in Conjunctive Normal Form that says exactly one of three predicates is true. Or given a bit of context - given three suspects and exactly one is guilty.
My first attempt to capture this in FOL (where A, B, and C are constants) Guilty$(A) \lor \text{Guilty}(B) \lor \text{Guilty}(C)$ but I do not think that this restricts to exactly one. 
Attempts to better capture as only one can be guilty is Guilty$(x) \land \lnot \text{Guilty}(y) \land \lnot \text{Guilty}(z)$ thought I do not think this is quite right either as it maybe wouldn't play nice when using resolution proofs?
I know I mentioned this is for homework but this an isolated clause - the actual question is to write an answer extraction proof which I am not asking about as I want to figure it out on my own but this clause is causing a bit of trouble with that.
Thanks in advance. 

A:

If you prefer a more systematic approach, the CNF of a statement can be found easily with the following 3 steps:

$1.$ Negate the statement.
$2.$ Construct the DNF of the negated statement.
$3.$ Negate the statement again (once it is in DNF), and apply DeMorgan's Laws.

An example below...

Given propositions $G(a),G(b),$ and $G(c)$, consider each possible scenario:
scenario $1$: $G(a)$ is true while $G(b),G(c)$ are false
scenario $2$: $G(b)$ is true while $G(a),G(c)$ are false
scenario $3$: $G(c)$ is true while $G(a),G(b)$ are false
Now we can translate this rather intuitively into DNF as follows:
$(G(a) \wedge \neg G(b) \wedge \neg G(c)) \vee (G(b) \wedge \neg G(a) \wedge \neg G(c)) \vee (G(c) \wedge \neg G(a) \wedge \neg G(b))$
But we have a problem here: we have our statement in DNF, but what we really want is the DNF of the negation. So, let's begin step $1$ above by negating the statement and then use DeMorgan's Law to find...
$\neg \bigg[ (G(a) \wedge \neg G(b) \wedge \neg G(c)) \vee (G(b) \wedge \neg G(a) \wedge \neg G(c)) \vee (G(c) \wedge \neg G(a) \wedge \neg G(b)) \bigg]$
$\Leftrightarrow \bigg[ \neg G(a) \vee G(b) \vee G(c) \bigg] \wedge \bigg[ \neg G(b) \vee G(a) \vee G(c) \bigg] \wedge \bigg[ \neg G(c) \vee G(a) \vee G(b) \bigg]$
We now have the negated statement in CNF. My suggestion is to convert this into DNF by applying the distribution law numerous times and then simplifying with other applicable equivalencies. Once you have done this, you have completed step $2$, and you may complete the final step - step $3$ above - rather straightforwardly.

