Re: 1-1/2+1/3-1/4+1/5-1/6+1/7

From: Han.deBruijn@xxxxxxxxxxxxxx
Date: Fri, 8 Feb 2008 13:15:46 -0800 (PST)

On 8 feb, 00:57, G. Frege <nomail@invalid> wrote:

On Thu, 07 Feb 2008 22:49:56 +0100, G. Frege <nomail@invalid> wrote:

Ooops... Of course

s[A] := {s(x) : x e A} (A c N)

"the image of A under s"
~~~ ~~~~

Am I allowed to hook in, with another of my small but certain steps?

Let's repeat the definition of S(). I hope this is the latest version:

y is in S(x) iff there is some n e N such that for all m e N ,
with m > n , y is in S_m(x) .
Here, S_m(x) = s o s o ... o s(x), where there are exactly
m compositions. (m e N)

Lemma ( with () -> [] ) :

y is in S[x] iff there is some n e N such that for all m e N ,
with m > n , y is in S_m[x] .
Here, S_m[x] = s o s o ... o s[x], where there are exactly
m compositions. (m e N)

Proof:

Because S[x] = { S(y) , y e x } and S(y) according to the above.

Is this correct?

Han de Bruijn
.

Follow-Ups:
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: G . Frege
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: Virgil
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: Jesse F. Hughes

References:
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: Jesse F. Hughes
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: Han de Bruijn
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: G . Frege
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: Han de Bruijn
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: Jesse F. Hughes
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: Han de Bruijn
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: G . Frege
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: Han de Bruijn
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: Virgil
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: G . Frege
- Re: 1-1/2+1/3-1/4+1/5-1/6+1/7
  - From: G . Frege

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