# Re: [isabelle] domain of function

`There is of course a third (Z) option: modelling partial functions as
``deterministic relations,
``ie ('a * 'b) set. This is similar to the 'a => 'b option model in
``most respects, but is
``slightly more natural when defining finite functions and has the
``advantage of inheriting a lot of
`machinery from the relational super-type.

`You can reduce this model (and those mentioned by Clemens) to the
``arbitrary-off-domain model by defining an application operator, in
``this case
`
f.x = (THE y. (x, y) : f),
but you now have the advantage of an explicit record of the domain.
On 19/02/2007, at 9:29 PM, Clemens Ballarin wrote:

Dear Hidetsune,

`unfortunately, modelling partial functions with arbitrary is rather
``counter-intuitive and weak. Since arbitrary is an (unspecified)
``element from the underlying type, it is not possible to determine
``the domain of such a function. The function could be undefined at
``point x, or it could map x to a value that coincides with the
``(unspecified) value of arbitrary.
`

`If you want to deal with partial functions seriously, you will have
``to make the domain explicit. I can think of two ways of dealing
``with this. You can either make the function a pair (f, A) where A
``is the domain of f. Or you let f be of type 'a => 'b option, and x
``is in the domain of f iff f x = None. Both are precise but not
``convenient to deal with, I'm afraid.
`
Clemens

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