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On 3/6/2025 12:05 PM, WM wrote:No. The first element is the empty set. It does not contain Bob. All following elements are the empty set equipped with further curlybrackets.On 06.03.2025 12:08, Jim Burns wrote:On 3/6/2025 3:54 AM, WM wrote:Am 06.03.2025 um 02:52 schrieb Jim Burns:It does not exclude Bob.<<JB<WM>>>><</JB<WM>>>>>
You need not the intersection however because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
>
No BECAUSE
that doesn't DEFINE Z₀
It defines Z₀ precisely.
In order to exclude Bob,
some clause must state or imply Bob isn't in Z₀
Induction says what's IN Z₀Therefore there is no Bob.
More description is needed.
That doe not help. Induction does not talk about endsegments.I generalize 'end segment':What does your inductive description permit in Z₀>
which should not be permitted in Z₀ ?
Anything.
Your description permits Bob, as long as
{Bob}, {{Bob}}, {{{Bob}}}, ... are also in Z₀
Induction excludes Bob.
An unreliable argument is unacceptable.Therefore I deleted your waffle.
{ } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀. There is no slot for Bob!Induction excludes Bob.Induction alone does not exclude Bob.
Induction alone does not exclude Bob.Waffle!
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