问题
I have the following lambda calculus:
1) λx . katze(x)(Garfield)
2) λP . λx . P(x)(tea)
3) λy . λx . likes(x, y)(Mia)
How do I reduce them with the Beta Reduction?
My solutions:
1) katze (Garfield)
2) tea
3) likes(Mia)
回答1:
When performing beta reduction, you substitute the bound variable to the lambda function with the value supplied. The notation for that is [param := value] and you pick up the first variable that is given.
In the case
λx . katze(x)(Garfield) -> katze (Garfield) the reduction is correct. We've substituted the x variable for Garfield and removed λx in the process leaving just the expression inside. Here are the steps that would be taken:
λx . katze(x)(Garfield)
= katze(x)[x := Garfield]
= katze(Garfield)
However, the other two are not correct. You are forgetting that you have a lambda function where the expression inside is another lambda function. Since you have a single input, you only have to reduce one function - the first one, leaving the other. You can think of it of peeling off the outer one and exposing the inner.
In the case of λP . λx . P(x)(tea) this can be better represented as (λP . (λx . P(x)))(tea) where now each lambda function is surrounded by brackets. Since we supply a single input tea, we only resolve the outer function with parameter P (leaving the brackets for some clarity):
(λP . (λx . P(x)))(tea)
= (λx . P(x))[P := tea]
= (λx . P(x))
= λx . tea(x)
Or without the brackets:
λP . λx . P(x)(tea)
= λx . P(x)[P := tea]
= λx . tea(x)
As for the final function, it still has the same problem that you are removing both functions, when only one input is given. The correct reduction steps are:
λy . λx . likes(x, y)(Mia)
= λx . likes(x, y)[y := Mia]
= λx . likes(x, Mia)
来源：https://stackoverflow.com/questions/59840554/beta-reduction-of-lambda-calculus