Re: universal quantification, because g⤨(g⁻¹(x)) = g(y) [1/2] Re: how

On 5/14/2024 4:15 PM, Ross Finlayson wrote:

They're not all quite so strong,
the many, many examples
of the balking and clamming,
the actually quite a few very many,
though, these are pretty good.

You don't want to talk about
what I want to talk about.
And there's nothing wrong with that.
Really.
However, it's just as true
in the other direction.
Date: Sat, 11 May 2024 19:47:38 -0400
Message-ID: <a4700775-be6c-46db-ad41-361eee6a3b67@att.net>
<JB<RF>>

The case is that induction goes through,
an inviolable law you call it:
does it go all the way through?
Does it complete?

It is complete.
There is no completing.activity,
so I wouldn't say it completes.
Compare to right triangles:
Are all the squares of two shorter sides
summed to the square of the longest side?
That's a tricky question to answer because
there is no summing done.
That relationship between the sides
is simply something true about right triangles.
And it is complete == it is true for each.
We don't typically ask the tricky question
about right triangles.
We ask the tricky question about cisfinite induction
because we imagine it as a process,
which we don't for right triangles.
Cisfinite induction is NOT a process.
Cisfinite induction is an argument,
completely correct or completely incorrect.
</JB<RF>>