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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?

Brian M. Scott
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John Roberts
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  • Please make sure that I didn’t introduce any errors when I converted to $\LaTeX$. – Brian M. Scott Mar 02 '13 at 17:56
  • No, looks good. Thanks. – John Roberts Mar 02 '13 at 17:57
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    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 02 '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 02 '13 at 18:37
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    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 02 '13 at 18:44

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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.

Ali Caglayan
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