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May 12, 2013 at 2:14 comment added Joel David Hamkins Yes, you seem to have some need for greater interaction of the $R_n$'s with each other. But that specific proposal is refuted by Sam Robert's example. I believe that some kind of bisimilarity type relation will be required to make the conclusion go through.
May 11, 2013 at 14:13 comment added Wolfgang Schwarz Thanks! Your response suggests that the claim might be true if each $R_n$ extends $R_{n-1}$ in the sense that whenever $(x_1,\ldots,x_n)R_n(y_1,\ldots,y_n)$, then $(x_1,\ldots,x_{n-1})R_{n-1}(y_1,\ldots,y_{n-1})$. (That would actually be enough for my application.) Does that seem right?
May 11, 2013 at 14:13 vote accept Wolfgang Schwarz
May 11, 2013 at 11:11 history edited Joel David Hamkins CC BY-SA 3.0
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May 11, 2013 at 11:05 history answered Joel David Hamkins CC BY-SA 3.0