Re: Request for Review/Tutorage of Amateur Proofs

On Mar 13, 11:43 am, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
By induction, ~(KS=00). QED.

What?! WHAT "induction"? You didn't do any induction.

I will use s[x] for s_x (latex: s_{x}) for better clarity and n for
intersection between two sets.

For 0_r e N, we have KB[0_r]=s[0_r] ^ ~(s[0_r]=0).
For i>0_r ^ ieN, we have KB[i]=s[i] ^ ~(s[i]=0). Now KB[i] n B[i+1] =
s[i+1] = KB[i+1], and from L3 ~(s[i+1]=0), so KB[i+1]=s[i+1] ^ ~(s[i
+1]=0). By induction, ~(KS=0). QED
.