On Dec 31, 8:31 am, George Greene <gree...@email.unc.edu> wrote: > On Dec 29, 10:29 pm, Graham Cooper <grahamcoop...@gmail.com> wrote: > > > If you disagree with a numbered point then which one? > > Everybody disagrees WITH POINT 4, DUMBASS. IT DOESN'T FOLLOW > from anything. YOU CAN'T PROVE IT. We by contrast HAVE EASILY > proved that since for EVERY n, the nth position of the diagonal > DIFFERS from the nth R on the list AT Rn's nth position, > THE ANTI-DIAGONAL *IS*NOT*ON* the list. If it WERE on, it would > have to be on it *AT* some row n. But the anti-diagonal IS NOT > on the list at row n because Rn DIFFERS from the anti-diagonal IN > POSITION n. > > > > > > > > > > > AD METHOD (binary version) > > Choose the number 0.a_1a_2a_3...., where a_i = 1 if the i-th > > number in your list had zero in its i-position, a_i = 0 otherwise. > > > LIST > > R1= < <314><15><926><535><8979><323> ... > > > R2= < <27><18281828><459045><235360> ... > > > R3= < <333><333><333><333><333><333> ... > > > R4= < <888888888888888888888><8><88> ... > > > R5= < <0123456789><0123456789><01234 ... > > > R6= < <1><414><21356><2373095><0488> ... > > > .... > > > By breaking each infinite expansion into arbitrary finite length > > segments > > >  The anti-Diagonal never produces a unique segment > > (all finite segments are computable) > > >  The anti-Diagonal never produces a unique sequence > > of segments (all segment sequences are computable) > > MOST of the *infinitely*-long segment sequences ARE NOT computable. > The fact that all the FINITE ones are DOES NOT IMPLY that the infinite > ones are also. > The mystery is why you would think it would. >
Point 4 inductively follows from point 3.
< <sub-segment> <sub-segment> <sub-segment> >
this sequence of segments.
You opposing argument is that some sequence of sentences cannot be written down and placed on a stack.
INFINITE STACK OF INFINITE ESSAYS OF FINITE VOCABULARY in order to be equivalent to a BASE-|VOCAB| List of infinite strings.