# lambda calculus

Lambda calculus  (often referred to as $\lambda$-calculus) was invented in the 1930s by Alonzo Church, as a form of mathematical logic dealing primarly with functions and the application of functions to their arguments. In pure lambda calculus, there are no constants. Instead, there are only lambda abstractions (which are simply specifications of functions), variables, and applications of functions to functions. For instance, Church integers are used as a substitute for actual constants representing integers.

A lambda abstraction is typically specified using a lambda expression, which might look like the following.

 $\lambda\;x\;.\;f\;x$

The above specifies a function of one argument, that can be reduced by applying the function $f$ to its argument (function application is left-associative by default, and parentheses can be used to specify associativity).

## Examples

We can specify the Church integer $3$ in $\lambda$-calculus as

 $3=\lambda f\;x\;.\;f\;(f\;(f\;x))$

Suppose we have a function $\operatorname{inc}$, which when given a string representing an integer, returns a new string representing the number following that integer. Then

 $3\;\operatorname{inc}\;\texttt{"0"}=\texttt{"3"}$
 $\displaystyle\operatorname{add}$ $\displaystyle=$ $\displaystyle\lambda\;x\;y\;.\;(\lambda\;f\;z\;.\;x\;f\;(y\;f\;z))$ $\displaystyle\operatorname{add}\;2\;3$ $\displaystyle=$ $\displaystyle\lambda\;f\;z\;.\;2\;f\;(3\;f\;z)$ $\displaystyle=$ $\displaystyle\lambda\;f\;z\;.\;2\;f\;(f\;(f\;(f\;z)))$ $\displaystyle=$ $\displaystyle\lambda\;f\;z\;.\;f\;(f\;(f\;(f\;(f\;z))))$ $\displaystyle=$ $\displaystyle 5$
 $\displaystyle\operatorname{mul}$ $\displaystyle=$ $\displaystyle\lambda\;x\;y\;.\;(\lambda\;f\;z\;.\;x\;(\lambda\;w\;.\;y\;f\;w)% \;z)$ $\displaystyle\operatorname{mul}\;2\;3$ $\displaystyle=$ $\displaystyle\lambda\;f\;z\;.\;2\;(\lambda\;w\;.\;3\;f\;w)\;z$ $\displaystyle=$ $\displaystyle\lambda\;f\;z\;.\;2\;(\lambda\;w\;.\;f\;(f\;(f\;w)))\;z$ $\displaystyle=$ $\displaystyle\lambda\;f\;z\;.\;f\;(f\;(f\;(f\;(f\;(f\;z)))))$ $\displaystyle=$ $\displaystyle 6.$

## Russell’s Paradox in $\lambda$-calculus

The $\lambda$-calculus readily admits Russell’s Paradox  . Let us define a function $r$ that takes a function $x$ as an argument, and is reduced to the application of the logical function $\operatorname{not}$ to the application of $x$ to itself.

 $r\;=\;\lambda\;x\;.\;\operatorname{not}\;(x\;x)$

Now what happens when we apply $r$ to itself?

 $\displaystyle r\;r$ $\displaystyle=$ $\displaystyle\operatorname{not}\;(r\;r)$ $\displaystyle=$ $\displaystyle\operatorname{not}\;(\operatorname{not}\;(r\;r))$ $\displaystyle\vdots$

Since we have $\operatorname{not}\;(r\;r)=(r\;r)$, we have a paradox.

 Title lambda calculus Canonical name LambdaCalculus Date of creation 2013-03-22 12:32:39 Last modified on 2013-03-22 12:32:39 Owner ratboy (4018) Last modified by ratboy (4018) Numerical id 9 Author ratboy (4018) Entry type Definition Classification msc 03B40 Related topic CombinatoryLogic Related topic ChurchInteger Related topic RussellsParadox Defines pure lambda calculus Defines lambda abstraction Defines lambda expression