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bio website tangentstorm.com
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visits member for 2 years, 2 months
seen Aug 15 at 0:30

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Apr
5
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.
Apr
5
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.
Apr
5
comment Mapping a set of sets to a partitioning.
Corrected. Thanks. en.wikipedia.org/wiki/Boolean_algebra_(structure) looks promising.
Apr
5
revised Mapping a set of sets to a partitioning.
corrected the subscripts
Apr
4
asked Mapping a set of sets to a partitioning.
Feb
25
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.
Feb
24
awarded  Editor
Feb
24
revised The deep structure of logical formulas
fixed mismatched parens, removed extraneous word.
Feb
24
answered The deep structure of logical formulas
Feb
24
awarded  Supporter
Feb
24
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.
Feb
24
answered Proving Gabbay rule for Modal Logic
Jul
1
awarded  Yearling
Jul
27
awarded  Nice Answer
Jul
1
awarded  Teacher
Jul
1
answered Have there been efforts to introduce non Greek or Latin alphabets into mathematics?
Jul
1
awarded  Autobiographer