# Lambda Calculus Equivalence

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?

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Please make sure that I didn’t introduce any errors when I converted to $\LaTeX$. –  Brian M. Scott Mar 2 '13 at 17:56
No, looks good. Thanks. –  John Roberts Mar 2 '13 at 17:57
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. –  dtldarek Mar 2 '13 at 18:04
@dtldarek I'm not sure what you mean by syntactically - if they're equivalent syntactically, aren't they equivalent generally as well? –  John Roberts Mar 2 '13 at 18:37
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. –  Brian M. Scott Mar 2 '13 at 18:44

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 "operation of orders" as in conventional mathematics. Things to note are parenthesis are evaluated first.

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