I'm a bit new to lambda calculus and was wondering about the equivalence of two expressions

$$(\lambda x.\lambda y.xy)\lambda z.z\overset{?}=(\lambda x.\lambda y.xy)(\lambda z.z)$$

Can anyone help out?

  • $\begingroup$ Please make sure that I didn’t introduce any errors when I converted to $\LaTeX$. $\endgroup$ Mar 2, 2013 at 17:56
  • $\begingroup$ No, looks good. Thanks. $\endgroup$ Mar 2, 2013 at 17:57
  • 1
    $\begingroup$ In $\lambda$-calculus the $\lambda$ symbol behaves similarly to a quantifier and its scope spans until the enclosing parenthesis or end of term. The two expressions are equivalent syntactically. $\endgroup$
    – dtldarek
    Mar 2, 2013 at 18:04
  • $\begingroup$ @dtldarek I'm not sure what you mean by syntactically - if they're equivalent syntactically, aren't they equivalent generally as well? $\endgroup$ Mar 2, 2013 at 18:37
  • 2
    $\begingroup$ I believe that @dtldarek is saying that the difference is purely cosmetic, like that between $\exists x\varphi(x)$ and $\exists x\big(\varphi(x)\big)$, and hence that the two are trivially equivalent. $\endgroup$ Mar 2, 2013 at 18:44

1 Answer 1


By convention the outer most parenthesis are dropped for minimal clutter. $$\color{red}{(\lambda x.\lambda y.xy)}\color{blue}{\lambda z.z}\iff\color{red}{(\lambda x.\lambda y.xy)}\color{blue}{(\lambda z.z)}$$ The same thing is done in algebra: $$\color{red}{(z)}\color{blue}{(x+y)}\iff \color{red}z\color{blue}{(x+y)}$$ In lambda calculus there is a similar "order of operations" as in conventional mathematics. Things to note are parenthesis are evaluated first.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .