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I like your answer, but I'm going to refuse to concede that it's correct, out of stubbornness. No wait, I can do better. For some values, $($(x)) is the equiva
by jes5199 14y ago
I like your answer, but I'm going to refuse to concede that it's correct, out of stubbornness.
No wait, I can do better. For some values, $($(x)) is the equivalent to $(x). So, for some values $(y) = y. So for the domain of y where $(y) = y, $() is the identity function, and now it's a monad.
- nbouscal 14y agoNow it's a monad how? What category is it a monad on, and in what way does it satisfy the monad laws? Are you really saying that sometimes $() is return and other times it's id, and presto magico it's a monad?
- jes5199 14y agoLook, if you want to have a discussion with a reasonable person who's going to try to follow your arguments and retort them under the exact same level of rigid discourse that you are using, maybe you should try someone other than me. But yes, presto magico. I am outwitting your narrow-minded mathematics.
- nbouscal 14y agoWell, as long as it's all in good fun :-)