



 
Yes; you can model asking a man a question with correct answer A by the operation A^x where ^ is "exclusive nor" and x is the man's identity (1 if he tells the truth, 0 otherwise.) Asking the other man is modeled by A^(not x). The goal is to find an expression in this "algebra" that does depend on A and does not depend on x. So your solution reduces to A^(not x)^x = A^False = not A. A simpler solution: "If I asked you which door leads to heaven, which door would you point to?" A^x^x = A^True = A; take that door! Even reduced like this to Boolean algebra there is an element of trial and error in coming up with solutions. If you are interested in this kind of thing, I recommend the books of Raymond Smullyan. Good Hunting!  


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