Re: because g⤨(g⁻¹(x)) = g(y) [1/2] Re: how

Am 30.04.2024 um 11:56 schrieb Moebius:

Am 28.04.2024 um 19:29 schrieb Jim Burns:
 

4.
If n can be counted to from 0
then n⋅2 can be counted to _from n_

 Since n⋅2 is just short for n+n (in my book).
 n+n can be counted to from 0 the following way: 0 -> 1 -> 2 -> .. -> n and further n -> n+1 -> n+2 -> .. -> n+n (just like we already did when counting from 0 to n).*)
 Hence
 

n⋅2 can be counted to from 0 _through n_

 Indeed!
 In other words,
 "If n is a flying rainbow sparkle pony
then n⋅2 is a flying rainbow sparkle pony."

*) Due to the Peano axioms with the usual (recursive) definition for addition (n+0 = n, n+s(k) = s(n+k)) we have: n+n = S(...(S(n)...) where n = S(...(S(0)...). For example: If n = 2, then (with 2 := S(1) and 1 := S(0)) n = S(S(0)) and hence n+n = S(S(0))+S(S(0)) = S(S(S(0)))+S(0) = S(S(S(S(0)))+0 = S(S(S(S(0))). If we consider S a "counting step" we may depict that state of affairs the following way: 0 -> 1 -> 2 -> 3 -> 4 (with 3 := S(2) and 4 := S(3)).