Re: The set of necessary FISONs

On 27.02.2025 21:41, Jim Burns wrote:

On 2/27/2025 2:50 PM, WM wrote:

On 27.02.2025 19:19, Jim Burns wrote:

On 2/27/2025 5:45 AM, WM wrote:

On 26.02.2025 23:17, Jim Burns wrote:

 

This next bit you (WM) might like, for a change.
It looks like the pseudo.induction.rule which
you have been trying to use.

>
It is induction.

>
This is what you (WM) have called induction:
⎛ Each inductive predicate A

>
No, I call induction
a very restricted number of predicates.

I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).

Correct. But not necessary in its generality for my purpose.

>

</WM>
Date: Thu, 6 Feb 2025 17:55:57 +0100
Message-ID: <vo2pit$31hlr$1@dont-email.me>
 

If A(n) is useless for UA = ℕ,
then A(n+1) us useless too.
No reason to extend this simple concept.

 You extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
but you only claim it, you don't justify it.

I use Zermelo's approach without wich there is no set theory.

 

I do it order to avoid the following waffle:

 How very Orwellian of you.
I justify my claims. Doing so is 'waffle'.
You don't. Abstaining from doing so is
'mathematics' and 'logic' and 'geometry'.
 

What that version of 'induction' seems to say
is false if it's read literally.
It's false that
each inductive predicate is true.without.exception
_in each domain without exception_
Z₀ is a subset of a set Z holding 0 and all the {a}

>

By the same induction
I prove UF = ℕ  ==> Ø = ℕ.

>
What you use to prove that is
∀n:Aᴺ(n) ⇒ A(ℕ)

>
That is how Zermelo guarantees Z₀.

 Zermello's Infinity guarantees a superset Z of Z₀

How is that accomplished?

 From Z, it follows,
by Powerset and by Separation,
that Z₀ exists.
 ∀n:Aᴺ(n) ⇒ Aᴺ(ℕ)
is your fantasy.

It is Zermelo's approach.

 You would find your posts greatly improved
by criticizing (if you can) _our_ reasoning,

You deny Zermelo's approach. His Z is ensured by induction.

There is no difference in some cases like these:
When all n are added by induction to the empty set,
then we have constructed ℕ.

 When we have shown that there is

Gibberish! Simply agree that Z is ensured by induction.

the intersection of all inductive subsets of
an inductive set,
then we have constructed ℕ.

We don't even need the intersection if we reduce Zermelo's approach to Lorenzen's approach: I is a natural number, and if x is a natural numbers then x+1 is a natural number.

 In this context,
a 'construction' is a proof of existence.

Induction is a proof of Z.

 

When all n are subtracted by induction from ℕ
then we have created the empty set.
Do you agree?

 I am trying to reach some expressions
with which I can answer you and be understood.
I'm not there yet.

Simply say yes.
Regards, §WM