Identifying traitors Logic is one of the facets of math that is more 'fun', but this one is beyond me. Consider using logic statements and/or truth tables. 'Or' here is inclusive, that is 'A or B' means 'A or B or both, but not neither'.
King Warren suspects the Earls of Akaroa, Bairnsdale, Claremont, Darlinghurst, Erina and Frankston of plotting a conspiracy against him. He questions them in private and they each tell him:
Akaroa: If Darlinghurst and Erina are both loyal, then Frankston and either Bairnsdale or Claremont are traitors.
Bairnsdale: If either Claremont or Erina are traitors, then either Darlinghurst is a traitor and Claremont isn’t, or Erina is and Akaroa isn’t.
Claremont: Anyone here could be a traitor except for me.
Darlinghurst: If Frankston is a traitor and Bairnsdale’s involvement in the conspiracy implicates Erina’s involvement, then Claremont is a traitor.
Erina: If Akaroa is a traitor and the fact Bairnsdale is not a traitor means that Claremont is, then Akaroa is not a traitor.
Frankston: If Claremont is a traitor so is Akaroa.
Each traitor will give false information but each loyal Earl will give true statements. At least how many traitors are there, and who are they?
 A: Introduce boolean variables $A,B,C,D,E,F$, where $A$, for instance, is true if Akaroa is not a traitor and false otherwise, and analogously for the rest of the variables. Then we have five equations
$$\begin{align}
A&=(D\wedge E)\to (\neg F \wedge (\neg B \vee \neg C))\\
B&=(\neg C\vee \neg E)\to (\neg D\wedge C\vee E\wedge\neg A)\\
C&=C\\
D&=(\neg F\wedge(\neg B\to\neg E))\to\neg C\\
E&=(\neg A\wedge(B\to\neg C))\to  A\\
F&=\neg C\to\neg A.
\end{align}$$
We could simply try all $2^6=64$ combinations of truth values to $A,\ldots,F$ and see which solutions there are. It can be done by hand or easily by computer. There might, by chance, be a short elegant algebraic solution, but I don't want to look for it right now. This problem is an instance of 3SAT, and I don't want to spend superpolynomial time solving this problem (assuming P$\neq$NP).
Added: I input this into Mathematica, and if I didn't make any mistakes, there are 26 solutions. One of them is that everyone is a traitor, and the largest number of loyal earls is 5, which happens twice. The solutions are, sorted by the number of traitors:
$$\begin{matrix}
A&B&C&D&E&F\\
0& 0& 0& 0& 0& 0\\
0& 0& 0& 0& 0& 1\\
0& 0& 0& 1& 0& 0\\
0& 0& 1& 0& 0& 0\\
1& 0& 0& 0& 0& 0\\
0& 0& 0& 1& 0& 1\\
0& 0& 1& 0& 0& 1\\
0& 1& 1& 0& 0& 0\\
1& 0& 0& 0& 1& 0\\
1& 0& 0& 1& 0& 0\\
1& 0& 1& 0& 0& 0\\
0& 0& 1& 1& 0& 1\\
0& 1& 1& 0& 0& 1\\
0& 1& 1& 0& 1& 0\\
1& 0& 0& 1& 1& 0\\
1& 0& 1& 0& 0& 1\\
1& 0& 1& 0& 1& 0\\
1& 1& 1& 0& 0& 0\\
0& 1& 1& 0& 1& 1\\
1& 0& 1& 0& 1& 1\\
1& 0& 1& 1& 0& 1\\
1& 0& 1& 1& 1& 0\\
1& 1& 1& 0& 0& 1\\
1& 1& 1& 0& 1& 0\\
0& 1& 1& 1& 1& 1\\
1& 1& 1& 0& 1& 1
\end{matrix}$$
If one would happen to know the answer beforehand, or guess it, then it is easy to verify by hand: just check that $A=0,B=C=D=E=F=1$ is a solution and that $A=B=C=D=E=F=1$ is not a solution. Thus the answer to the puzzle is: the least number of traitors is one, and then it is either Akaroa or Darlinghurst.
