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Mike+9 So damn sexy...

Posts I've Made

In Topic: Lambda calculus reductions

Posted 10 Jan 2013

Got it. Thanks a lot for the help. I had the exam today and I did pretty good.
In Topic: Lambda calculus reductions

Posted 8 Jan 2013
Yeah, the last step is should practically be 2 steps, I got that now. What I tried to say is that I first try to replace y in the first (^y. x), and since there is no y there I should get:
```
(^x. x)
```
But your way with the alpha conversion first is way better and easier to understand.

Thanks a lot for the help! I've also read some tutorials and some courses online and got the hang of it.

mojo666, on 08 January 2013 - 08:04 PM, said:
Quote
(^y. x (^x. x))[(^y. y x)/x]

Here I am supposed to substitute x with (^y. y x). The book states that in cases like this I should rename some variables so they do not affect the expression. But I'm really confused and don't know which one. I would go for the y first because if I would substitute the first x with the expression after it that y would become bound.
In this example, renaming is not necessary. You only need to rename if the substituting expression contains a free variable that would become bound upon performing the substitution. The only free variable in (^y. y x) is the x and it will not be bound to anything when you perform the substitution.
So, when I want to replace x with (^y. y x) I would have to replace both x's in the main expression? This would be one of the things I still do not know for sure.
In Topic: Count to a million

Posted 20 Dec 2012

11330
No bots here!
In Topic: See any good movies recently?

Posted 20 Dec 2012

I've seen Intouchables recently and it was a good movie from my point of view.
Not really 2012 new but still a good movie.
In Topic: question !

Posted 20 Dec 2012

Ok, and what is the question?