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Math Help - Lambda Conversion (alpha and beta conversion)

  1. #1
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    Lambda Conversion (alpha and beta conversion)

    Can any one please help:


    Lambda Conversion (alpha and beta conversion)-alpha_beta-conversion.png
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  2. #2
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    Re: Lambda Conversion (alpha and beta conversion)

    Alpha-conversion needs to be performed to avoid variable capture. This happens when reducing (\lambda x.\,M)N and the term N has some free variable y that would become bound if N is substituted for x in M because there is a \lambda y somewhere inside M. To avoid capture of y, \lambda y in M has to be renamed into some \lambda z. For example, in problem (a), x would be captured if substituted for y in \lambda x.\,xy, so this bound x has to be renamed.

    Consider problem (b). I'll omit some parentheses and will write \lambda xy for \lambda x\lambda y.

    (\lambda xy.\,x(\lambda x.\,xy))(\lambda xy.\,yx) -> (substituting (\lambda xy.\,yx) for x)
    \lambda y.\,(\lambda xy.\,yx)(\lambda x.\,xy) -> (need to rename y because the argument has y free)
    \lambda y.\,(\lambda xz.\,zx)(\lambda x.\,xy) -> (substituting (\lambda x.\,xy) for x)
    \lambda yz.\,z(\lambda x.\,xy)
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