Post by Eugene 2.0 on Oct 6, 2020 17:29:54 GMT
If there are two (or more) identical or indiscernible objects, then if we need one of the object of such a type, then all of identical objects must have been passed as relevant.
There's no ways to understand hiw much identicals are there. Total number aka ordinal number of things might help us in this, but it failed in calculating the structure: if two certain identicals are in some order (presumed that the identicals have transcendental abilities and as so they can be transcendentally ordered), no ordinals can say us about.
All the identicals of such type satisfy these conditions:
1) â={A, B, C, ..., Z} U ê=(μє{A, B, C, ..., Z}) → â
2) ă={A, A, ..., A} U ĕ=(πє{A, A, ..., A}) → ă
3) (μ → â) → (π → ă)
4) ~ă → ~[(μ → â) & π]
5) ~ă → μ v ~(â & π)
So, not ordered identicals imply the chance of it – of ordered identicals, that shows why ordering identicals is always possible to do by another type of calculation.
If so, then to satisfy a proposition like:
*) [р → (ф'' є {А, B, C,...,Z})]
There's no ways to understand hiw much identicals are there. Total number aka ordinal number of things might help us in this, but it failed in calculating the structure: if two certain identicals are in some order (presumed that the identicals have transcendental abilities and as so they can be transcendentally ordered), no ordinals can say us about.
All the identicals of such type satisfy these conditions:
1) â={A, B, C, ..., Z} U ê=(μє{A, B, C, ..., Z}) → â
2) ă={A, A, ..., A} U ĕ=(πє{A, A, ..., A}) → ă
3) (μ → â) → (π → ă)
4) ~ă → ~[(μ → â) & π]
5) ~ă → μ v ~(â & π)
So, not ordered identicals imply the chance of it – of ordered identicals, that shows why ordering identicals is always possible to do by another type of calculation.
If so, then to satisfy a proposition like:
*) [р → (ф'' є {А, B, C,...,Z})]
& [(ф'' є {А, B, C,...,Z} → p]
by putting not ordered identicals instead of ф' => ф'' and vice versa, are satisfied and not satisfied the formula at the same time.