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  1. 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.
  2. 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.

    View Postmojo666, on 08 January 2013 - 08:04 PM, said:


    (^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.

My Information

Member Title:
D.I.C Head
23 years old
March 25, 1991
Full Name:
Mihai Boisteanu
Programming Languages:
Java, C++

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