lambda calculus - if more parameters needed

Question

Lambda Calculus Question:

TRUE = lambda x y . x
FALSE = lambda x y . y
1 = lambda s z . s z
2 = lambda s z . s (s z) ...

BoolAnd = lambda x y . x y FALSE
BoolOr = lambda x y. x TRUE y
BoolNot = lambda x . x FALSE TRUE 

If I want to know the result of BoolNot 1:
BoolNot 1
(lambda x . x FALSE TRUE)(lambda s z . s (s z))
(lambda s z . s z) FALSE TRUE
(lambda x y . y) (lambda x y . x)

Here needs x and y 2 parameters, but only have 1 here, How can I continue this calculus?

score 1 · Accepted Answer · answered Sep 06 '16 at 19:22

1

λ x y. E is "shorthand" for λx. (λy. E).

Thus,

(λ x y. y) (λ x y. x)

==> (λx. (λy. y)) (λ x y. x)

==> λy. y

That is, the identity function.

answered Sep 06 '16 at 19:22

molbdnilo

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Thanks, but could you tell me what does λy. y mean? – Andy Sep 06 '16 at 22:36
@Andy I'm not sure what you're asking, but `λ` is the lowercase greek character "lambda", if that's what you're wondering. – molbdnilo Sep 07 '16 at 07:47

score 0 · Answer 2 · answered Sep 06 '16 at 19:03

Think that you apply one argument at the time and at each step you have a function taking one less. It doesn't make sense to do (not 1) but the result is the identity function since true becomes the variable not used and thus it will take another argument y and evaluate to y

lambda calculus - if more parameters needed

2 Answers2