Re: How many flips of DIAG are on the infintie list of infinite con flippers ?

From: |-|erc (h_at_r.c)
Date: 01/22/05 

Date: Sat, 22 Jan 2005 11:47:25 +1000

"The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote in > >>

> >> >> AntiDiag(F,j) = !(F(j,j)) [1]
> >> >>
> >> >> Of course this is not to be confused with the
> >> >> proposition
> >> >>
> >> >> (Aj)(Ei)(F(i,j) = AntiDiag(F,j)) [2]
> >> >>
> >> >> which for most F is true.

> >> >
> >> > More importantly
> >> > (Aj)(Ei)(F(i,k) = AntiDiag(F,k) 0<=k<=j [3]
> >>
> >> I'd phrase that as
> >>
> >> (Aj)(Ei)(Ek)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) [4]
> >
> > no. that's identical to your near meaningless 1st attempt.

> >> Or did you mean
> >>
> >> (Aj)(Ei)(Ak)(F(i,k) = AntiDiag(F,k), 1 <= k <= j) ? [5]
> >>
> >> Not that it matters.
> >
> > There you go. Completely different assertions actually.

> Do you know the difference between
>
> (Ai)(Ej)(F(i,j) = AntiDiag(j)) [6]
>
> and
>
> (Ei)(Aj)(F(i,j) = AntiDiag(j)) [7]
>
> ?
>
> If so, explain it to me. I'm curious.

6 - all reals in F have an identical digit to antidiag (same position, same value)

7 - antidiag is a member of F

That was a courtesy, remember its YOU who are making useless formula in this thread!
what is the difference between 2 and 4 ?

Herc

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