Re: [External] : Re: How to make M-x TAB not work on (interactive) declaration?

[Jean Louis]

Thanks, I could follow the previous.

* Yuri Khan <yuri.v.khan@gmail.com> [2023-01-17 19:00]:
> School had also said multiplying by 1 has the same effect as not
> multiplying at all. 

Not in absence of factors!

> That is, a * 1 = a. Also, school had said that for every non-zero a,
> if a * b = a * c, then b = c, hadn’t it? Somewhere around the time
> you learned to solve equations. It was called canceling.

I can see how you try to make it logic, but I do not ask for
mathematics out of context of Lisp. I am asking why is in Lisp so?
Common Lisp, Emacs Lisp, Guile, Newlisp, elk, with difference that
Emacs Lisp (-) ➜ 0 but in others arguments missing, with Picolisp
where (-) ➜ nil, (*) ➜ nil, (+) ➜ nil, but `apply' works.

Why would Emacs Lisp have (-) ➜ 0 and other Lisps not?

If `apply' works in Picolisp without problem, is the usage of such
functions really the reason for yielding identity for (*), (+), (-)?

> Let’s look at that a * 1 = a. On the left, we have the product of two
> numbers. On the right, we have one number, and if we squint at it like
> this, we can say it’s a product of one number. So these are two
> products that have a common element. We can cancel it. (Assuming it’s
> not zero. But we know that a * 1 = a holds for any a, including
> non-zeros.)
> 
> Now, on the left, we have 1. On the right… we have a product of no
> numbers. And there is an equality sign in between.

I have followed your guidance, but your guidance speaks of existence
of something to cancel, and I speak of absence of arguments!

Not even identity theory does not speak of identity elements in
absence of everyting else!

-- 
Jean

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