In math we have functions, as we do in programming. The functions of programming can be either lazy or eager, depending on the programming language. Are the functions of math lazy or eager?

In programming, the issue is quite important because it has a significant impact on one of the key equations of the lambda calculus, the so-called

*beta-law*, which says that we can substitute (i.e. inline) an argument for the parameter of the function.

`(fun x. M)N ≡ M[N/x]`

This law lies at the heart of all compiler optimisations.

The equation is always valid in call-by-name languages such as Algol. In lazy languages such as Haskell it still holds in the presence of some effects such as exceptions or non-termination, but it does not hold in the presence of stateful assignments. In an eager languages such as OCaml it is also invalidated in the presence of non-termination or exceptions. It is therefore common that in call-by-value languages the argument (

*N*above) must be

*a value*, i.e. a language constant, a variable, or a lambda expression, usually written as

*v*.

`(fun x.M)v ≡ M[v/x]`

Of course, substitution is also an important rule of equational reasoning, which is a big part of mathematical calculations. But which rule do we use in math, the by-name, lazy or the by-value (eager) substitution? And how can we tell?

Lets look closer at when the beta law fails in an eager functional program. If the argument is not a value (N≠v) but it is equivalent to a value (N≡v) then the rule holds. It only fails when the argument is not equivalent to a value, for example because it runs forever (it 'diverges'). An example of such a term is Ω =(fun x.x x)(fun x.x x). Or, more mundanely, Ω = while true do skip. An example of failure of the beta law is

`(fun x.k) Ω ≢ k[Ω/x]`

where k is a language constant. If we run the left-hand side as a program it will first run the argument Ω, which will result in a nonterminating computation. On the other hand, the right-hand side is just the term consisting of the constant k because substitution happens at a syntactic level. So its value is, trivially,

*k.*The two cannot be equivalent.

In maths we don't usually deal with functions such as Ω but we can still have expressions that don't denote a value, such as for example 1/

*m*when

*m*is 0. Both Ω and 1/

*m*are 'undefined' in this sense, that they don't denote a value.

The following example will test whether maths is lazy or eager. Consider the constant function

*c*(

*m*) = 1

The question is whether the equation

*m*⨉

*c*(1/

*m*) = 0

has a solution (

*m*=0) or not.

I have not run any referenda on the matter, but from my own understanding of mathematical practice and from pub chats with mathematician pals I would say that

*m=*0 is

*not*considered a valid solution for the above. A function such as the constant function

*c*above is defined over

*values*and not over

*terms.*

Perhaps it could be otherwise, but it ain't. Meanwhile, in Haskell

let c(x) = 1 in let m = 0 in m * (c (1/m))

So Haskell is impure. And not even mathematically so.