Re: Simple enough for every reader?

On 2025-05-27 15:24:50 +0000, WM said:

On 27.05.2025 14:27, Mikko wrote:

On 2025-05-26 13:44:06 +0000, WM said:
 

{1} has infinitely many (ℵo) successors.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors. For every n that can be defined.
 Do you accept this proof?

 No. A proof should start with a clear presentation of the premises.

 A proof as simple as this one should be understood as it stands. You only try to suppress it.
 

Then a sequence of sentences should follow, each with an indication
of how they follow from the previous sentence, and which earlier
sentences of the proof are also needed for the inference.

 {1} has infinitely many (ℵo) successors. If you don't understand, please ask.
 For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors. If you don't understand, please ask.

The indication that a sentence is a premise is still absent.
The indication that and how a sentence is a conseqence of earler sentences
is still absent.
The conclusion that

For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.

is not particularly interesting but it is an obvious consequence of the
assumption that

For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors. If you don't understand, please ask.

But an acceptable conclusion does not make the proof acceptable.
--
Mikko