> On Sep 6, 6:34 pm, galathaea gmail.com> wrote:

>> On Sep 6, 6:03 pm, Tegiri Nenashi gmail.com> wrote:

>>> On Sep 6, 3:31 pm, galathaea gmail.com> wrote:

>>>> if i understand what you mean by lambda quantifier
>>>> (as in the standard form on which eval's work through application
>>>> e.g. lambda calculus lambdas
>>>> with their well known universal properties)
>>>> then i would expect the exponential to be the corresponding operation
>>>> in other words
>>>> given a objects X and Y
>>>> a function f from X to Y is an element of the exponential object

>

>>>> X
>>>> f elementOf Y

>

>>>> and the evaluation at an element of X is given by

>

>>>> X
>>>> eval: Y x X --> Y

>

>>>> (such that a variety of commutative diagrams obtain)

>
> Could you please be more specific what algebraic structure you have in
> mind? Specifically, what are the elements ("element of" prompts that
> perhaps sets)? Given two sets X and Y, they define the set of
> functions X->Y; is that the operation you suggested?

> If my interpretation is wrong, may I ask for amn example? In the
> analogies that I brought in the root message I could, for example,
> explain how universal quantifier works on domain of three elements
> x={1,2,3}, that is how we do conjunction P(1) ^ P(2) then amend the
> expression to (P(1) ^ P(2)) ^ P(3) this is how Ax.P(x) becomes a
> proposition. Can you draw a similar venue for lambda abstraction?

>>> In binary form -- exp(x,y) --
>>> exponentiation is not associative, that would also disqualify it.

>

>> why do you think an operation must be associative?

>
> Because all the other analogies (sum, product, and predicate logic
> quantifiers) have this property?