Re: The set of necessary FISONs

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 is true.without.exception.
⎜ ∀ᵖʳᵉᵈA: A(0) ∧ ∀k:A(k)⇒A(k+1) ⇒
⎝ ∀n:A(n)
  (1)
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_
((1) places no condition on the domain.)
You mention some of the exceptions,
so you (WM) are aware of this.
A more explicit version of induction
places such conditions on the domain that
(1) is true.without.exception.
Let ℕ′ refer to the domain
in which (1) is intended to be used,
and Aᴺᣟ refers to a predicate on ℕ′
That more explicit version of induction is
⎛ In a domain ℕ′ which is its.own.only.inductive.subset,
⎜ each inductive predicate Aᴺᣟ is true.without.exception.
⎜ ∀ℕ′: {S⊆ℕ′:0∈S∧∀k:k∈S⇒k+1∈S} = {ℕ′}  ⇒
⎜⎛ ∀ᵖʳᵉᵈAᴺᣟ: Aᴺᣟ(0) ∧ ∀k:Aᴺᣟ(k)⇒Aᴺᣟ(k+1) ⇒
⎝⎝ ∀n:Aᴺᣟ(n)
  (2)

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.

What you (WM) have been trying to do  is
prove that a predicate is inductive  and
conclude that
each inductive predicate is true.without.exceptions ==
the domain is its.own.only.inductive.subset.
That's not induction.
The arrows point in the other direction.

Z₀ being the unique intersection of inductive subsets
makes
Z₀ its.own.only.inductive.subset,
which makes
Z₀ the only correct set to use in a proof by induction.

>
It is proved to exist by induction, i.e.,
by the rule that 0 exists and with a also a'.

Z₀ is a subset of a set Z holding 0 and all the {a}
The existence 0 and all the {a} is not sufficient
to prove the existence of Z and Z₀
(Counter.example: in ST+F, 0 and {a} exist but not Z)
That's why we have the Axiom of Infinity.

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

What you use to prove that is
∀n:Aᴺ(n) ⇒ A(ℕ)
That's not induction.
It seems to follow from confusion over
the difference between a set and its elements.