Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can someone explain this combinator? I understand $\lambda x. x$, but I don't understand $\lambda x. x x$ From what I've gathered, this means given x, return the application of x to x. I don't understand the application of x to itself part. For example, given $x = 2 + y$, would $\lambda x. x x$ result in $y^2 + 4y + 2$?

share|cite|improve this question
up vote 8 down vote accepted

2 + y is a number, not a lambda calculus term so you cannot use that here.

you can apply it to the identity and it will reduce like this:

$$(\lambda x. x x) (\lambda y. y) \to (\lambda y. y)(\lambda y. y) \to (\lambda y. y)$$

rewritten in terms of combinators this is: $\omega I \to II \to I$

notice what happens if you apply it to itself?

$$(\lambda x. x x) (\lambda x. x x) \to (\lambda x. x x)(\lambda x. x x) \to \ldots$$

rewritten in terms of combinators this is: $\omega \omega \to \omega \omega \to \ldots$

it reduces to itself in an infinite loop.

share|cite|improve this answer
Thanks! That was an great explanation of what I was misunderstanding. – J.R. Garcia Jan 3 '12 at 21:02
I think you mean "you can apply it to the identity"? – joriki Jan 3 '12 at 21:06
@joriki, corrected. Thanks. – user16697 Jan 4 '12 at 7:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.