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.
Keywords
Cite
@article{arxiv.1804.05677,
title = {Boolos' Hardest Logic Puzzle Ever can be solved in no less than three admissible questions: axiomatic framework and rigorous proof},
author = {J. J. Colomina-Almiñana and P. R. Stinga},
journal= {arXiv preprint arXiv:1804.05677},
year = {2025}
}
Comments
13 pages, revised version, new title. The results were first presented as an Invited Colloquium in Philosophy of Logic at the American Philosophical Association Pacific Division Meeting in San Diego in March 2018. To appear in Logique et Analyse