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 Apr5 comment Number of 5 letter words with at least two consecutive letters same First, you're counting some words twice here. For example the word AABBB would get counted in both of the first two lines. Second, your calculations aren't considering permutations. For example, the sequences { AAA, B, C } can be arranged in 6 different ways, but you're multiplying by 3. Apr5 comment Mapping a set of sets to a partitioning. I see. These are the semantics I want, I'm just lacking the mathematical vocabulary. (I come from a programming background and am writing a little logic system to help me learn.) I simply mean that none of the resulting $2^n$ sets overlap. Being able to show that some of them are empty is kind of the goal. Apr5 comment Mapping a set of sets to a partitioning. Corrected. Thanks. en.wikipedia.org/wiki/Boolean_algebra_(structure) looks promising. Apr5 revised Mapping a set of sets to a partitioning. corrected the subscripts Apr4 asked Mapping a set of sets to a partitioning. Feb25 comment Proving Gabbay rule for Modal Logic Perhaps I was wrong about $(\Box p \wedge p)$: I don't really understand the nature of the proof you're trying to write. I don't know what "the $\leftarrow$ part" means or "the other part". All I know is that the only possible way for the statement $(\Box p\rightarrow p) \vee A / A$ to be valid is if $A$ is a theorem and $(\Box p\rightarrow p)$ is an antitheorem. I then showed (by the first truth table) that this is only possible in one situation (world x), and that a frame containing world x cannot be reflexive because $(\Box p\rightarrow p)$ is a necessary condition for reflexivity. Feb24 awarded Editor Feb24 revised The deep structure of logical formulas fixed mismatched parens, removed extraneous word. Feb24 answered The deep structure of logical formulas Feb24 awarded Supporter Feb24 comment Proving Gabbay rule for Modal Logic Well, it's implied by the first truth table. For example, if $A = p$, then world x contains a contradiction, because $p$ is true and $A$ is false, so that line would have to be removed from the list (along with all the other lines where $A$ and $p$ don't match). The only way you get all 8 possible combinations of 3 boolean variables is to make the variables completely independent. Feb24 answered Proving Gabbay rule for Modal Logic Jul1 awarded Yearling Jul27 awarded Nice Answer Jul1 awarded Teacher Jul1 answered Have there been efforts to introduce non Greek or Latin alphabets into mathematics? Jul1 awarded Autobiographer