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Lambda calculus normal form

  1. May 3, 2004 #1
    Hello, I have a simple question.
    When reducing a lambda expression what happens in this case:

    (L x.y) (L z.z) (L z.z) -> y (L z.z) right?

    How to interpret the variable y in front? Is that normal form, or can
    I reduce it to just: y
    Or is the expression ilegal?
  2. jcsd
  3. May 3, 2004 #2

    matt grime

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    At the risk of being flamed, I'm a pretty experienced mathematician, and I' have no idea what you mean by any of those expressions. I'm not going to say they are non-standard, but I think you ought to at least indicate what they mean if you want an answer.
  4. May 3, 2004 #3
    It's Lambda Calculus. It's used in computer science as an alternative to Turing machines; instead of representing computation as the operation of a machine, we represent it as the evaluation of functions. Unless you've specifically studied computer science it's unlikely you would have ever even heard of it.

    I would try and help, but I'm only an undergraduate; one who hasn't had any formal lectures involving lambda calculus. If you don't mind waiting a year I can get back to you... :biggrin:
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