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A duck walks into a bar. Animal control is promptly called and the duck is released into a near by park.

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2d
comment Morphisms in the category of natural transformations?
@OmarAntolín-Camarena: Ah, okay, I see. The case where the hom-set is actually a group is common enough, so maybe there would be a Yoneda lemma variant.
2d
comment Morphisms in the category of natural transformations?
I wrote $\mathrm{hom}(G,G)$ because it's the set of arrows on the one object $G$, which are exactly the group elements of the group $G$ - you take $y$ only out of $H$ (it wasn't in the answer, but indeed it makes sense, otherwise the multiplication $y·g(x)·y^{-1}$ can't be made sense of). The thing still looks so close to the Yoneda lemma expression - can any functor $g:G\to H$ be represented somehow?
2d
revised Weaker, but not thin
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2d
comment Morphisms in the category of natural transformations?
In the last two lines you're claiming $\mathrm{nat}(g,f)\cong \mathrm{hom}(G,G)$. Is this stronger than the Yoneda lemma $\mathrm{nat}(\mathrm{Hom}(G,-),f)\cong fG$? Maybe you use that there is only one object in the domain category and that all morphisms are invertible. Anyhow you seem to imply all functors are representable or something.
2d
revised Weaker, but not thin
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2d
asked Weaker, but not thin
Oct
27
awarded  Necromancer
Oct
26
comment How to prove $p \vee q \vdash \neg(\neg p \wedge \neg q)$?
@ItaloPessoa: fyi, I translated the proof to a program in the sense of Curry-Howard‌​.
Oct
26
revised How to prove $p \vee q \vdash \neg(\neg p \wedge \neg q)$?
added 321 characters in body
Oct
26
revised How to prove $p \vee q \vdash \neg(\neg p \wedge \neg q)$?
added 321 characters in body
Oct
26
answered How to prove $p \vee q \vdash \neg(\neg p \wedge \neg q)$?
Oct
23
comment Formal notion of computational content
Without starting to write an answer - because I don't know what you already know - I can point to this recent extensive book, which is exactly about working out the computational aspects of logical systems.
Oct
22
comment Can you cancel out a term if equal to zero?
No, in this case you cannot have $x=300$ and $y=500$, as you've already established $x=y$. For OPs equation $x(x−y)=(x+y)(x−y)$, the "canceled" version $x=x+y$ happens to be a perfectly valid statement (which however just isn't derivable via cancellation).
Oct
21
revised Demonstrating the equality of two power sets
added 5 characters in body; edited title
Oct
21
comment calculus first impressions
@Avitus: Disputable. Maybe apart from the function-concept.
Oct
21
comment calculus first impressions
In 27 days, 11 am.
Oct
19
comment $\sqrt{7}$ ,$\sqrt {7 - \sqrt {7}}$,$\sqrt {7 - \sqrt {7 + \sqrt {7}}}$,$\sqrt{7-\sqrt {7 + \sqrt {7 - \sqrt {7}}}}$, …
You know $2$ is a fixed point of $\sqrt{7-\sqrt{7+x}}$. You must show that (starting from $\sqrt{7}=2.65..$) we have $\left|\sqrt{7-\sqrt{7+x}}-2\right|<\left|x-2\right|$. Btw., once you got your answer, I'd say you should then take the magic "$7$" out of the game and find out for which numbers this holds. I'd guess for any number bigger than $4$.
Oct
19
revised $\sqrt{7}$ ,$\sqrt {7 - \sqrt {7}}$,$\sqrt {7 - \sqrt {7 + \sqrt {7}}}$,$\sqrt{7-\sqrt {7 + \sqrt {7 - \sqrt {7}}}}$, …
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Oct
18
comment Mathematics only with physics? What about biology and chemistry?
@Nico: I think the questioner asks for more than intuition. Otherwise, for example, if you identity and model causes for behavior of animals in the wild, the eventual real world distribution and fluctuations of population sure give "intuition" for how the solutions to the respective diff. equation you set up should look like. That's essentially the point I make above: If you got a definitive mathematical problem, and you want to solve it by arguing with physics/chemistry/biology, how do you want to obtain a quantitative solution without mathematically formulating the natural science you use.
Oct
18
comment The smallest girl in the world
In an international environment, I'd go for metric system. I, for one, would pay 5000 Vietnamese dong not having to google if your baby is actually small.