Boolos' Hardest Logic Puzzle Ever can be solved in no less than three admissible questions: axiomatic framework and rigorous proof

Abstract

A formal axiomatic mathematical framework for Boolos' Hardest Logic Puzzle Ever is presented and two theorems about its solvability are proved. By strictly following Boolos' instructions (in particular, the requirement that all gods are always obliged to answer), the novel concept of \textit{admissible questions} for the puzzle is introduced. It is then rigorously proved that Boolos' original puzzle can be solved, in an absolute deterministic way, in no less than three yes-no admissible questions. However, this does not mean that one could solve it in less than three admissible questions by just pure \emph{chance}. Hence, such probabilities are computed here as well.