Can you explain how this expression follows the grammar of the lambda calculus?

$$\lambda x.x((\lambda y.yy)x)x = λx.x(xx)x$$

I am not sure why we have the parentheses following the $.x$ (on both sides of the equation).


1 Answer 1


When in doubt, write down a lot of parentheses: $$\lambda x . ((x ((\lambda y . ((y y) x)) x)) x)$$ and $$\lambda x . ((x (x x)) x)$$ In $\lambda$-calculus there is the convention that $A B C$ is understood to be $(A B) C$, that is, application associates to the left. So, if somebody wants to $A (B C)$ instead, they have to put in some parentheses.

  • $\begingroup$ Thanks, after some back and forth with the lambda grammar definition, this makes perfect sense. $\endgroup$
    – matanster
    Dec 29, 2016 at 17:05

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