Re: New equation

On 2/28/2025 6:08 PM, Ross Finlayson wrote:

On 02/28/2025 10:46 AM, Jim Burns wrote:

Reliable inductive inferences
do not fail at being correct.
>
A reliable inductive inference
concludes, from a subset being inductive,
that that inductive subset is the whole superset.
It is a _reliable_ inductive inference
only where there is only one inductive subset:
the whole superset.
>
A failure of inductive inference would be
x and c such that  x ∈ {c} ∧ x ≠ c
>
Are you claiming you have such x and c?

There are plenty of Zeno's arguments
using induction that never get anywhere.

Let's play spot.the.supertask.
I claim that
I justify _something_ induction.like
without calling for a supertask.
You claim that
I don't deal with issues around supertasks.
As I see it,
I have dealt with those issues,
all of those issues, yours and any TBA,
by erasing supertasks from the discussion,
not even in it implicitly.
Have I failed at erasing supertasks?
Show me where.
----
I start with a broader class of
pre.inductive properties.
Some pre.inductive properties clearly
don't involve supertasks.
This should allow you to focus
your search for a supertask in a tighter area.
I define that
a property P is pre.inductive  iff,
if each set in set 𝒞 of sets has P,
then its intersection ⋂𝒞 has P
∀S∈𝒞:P(𝒞) ⇒ P(⋂𝒞)
For example,
'holding 7,11,13' is pre.inductive.
If each set of a set 𝒞 of sets holds 7,11,13
then ⋂𝒞 holds 7,11,13.
Another example is 'being inductive'.
No supertasks so far. Right?
Consider a set Z which has P
and P is pre.inductive.
The intersection ⋂𝒫ᴾ(Z) of subsets having P
has P.
The only subset of ⋂𝒫ᴾ(Z) which has P
is ⋂𝒫ᴾ(Z)
Consider (what seems to be) another property A
such that we prove the subset {i:A(i)} ⊆ ⋂𝒫ᴾ(Z) has P
There is only one subset of ⋂𝒫ᴾ(Z) which has P
{i:A(i)} = ⋂𝒫ᴾ(Z)
By definition of {i:A(i)}
∀n ∈ {i:A(i)}: A(n)
{i:A(i)} = ⋂𝒫ᴾ(Z)
∀n ∈ ⋂𝒫ᴾ(Z): A(n)
Therefore,
if property A has P, and Z has P,
then ∀n ∈ ⋂𝒫ᴾ(Z): A(n)
I don't see any supertasks.
It's your turn.
Spot the supertask.
If property P is being inductive,
and ℕ = ⋂𝒫ᴾ(Z)
then we have the familiar 'law' of induction
⎛ If A(0) ∧ ∀k∈ℕ: A(k)⇒A(k+1)
⎝ then ∀n∈ℕ: A(n)