My FINAL Cantor thread!
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From: |-|ercules on 30 Jun 2010 18:16 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 30, 12:44 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> "George Greene" <gree...(a)email.unc.edu> wrote >> >> > On Jun 29, 10:49 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> Herc >> >> --> There IS NOT a computer program that lists the outputs of all computer programs! >> >> >> WRONG! > > You claim to think this is wrong, yet later, you yourself say, > >> Considering you would require a halt function to generate such a list, You CANNOT generate a full list of computable REALS. You CAN generate a full list of computable OUTPUTS. > > EXACTLY. YOU WOULD NEED a "halt function" -- or, equivalently, a LOOP > function -- > that is, you would need to be able to confirm that a TM was looping -- > AND WAS THEREFORE > NOT (any longer) in the process of computing a computable real -- in > order to generate this list. > >> it's as real as halt-omega, FOOL! > > "Halt-omega" IS NOT real, FOOL! > > If it halts then it halts AFTER A FINITE number of steps, NOT after > OMEGA steps! > And there IS NO TM that tells you whether other TMs halt or loop! > IF THERE WERE, THEN THERE WOULD ALSO be a TM that says, > "If my input TM loops, then halt, > but if it halts, then loop". And what would this TM do given ITSELF > as input?? > > >> Where are you going to get this list? > > You ARE NOT going to get it -- that's the whole point! Is that why you snipped what I was refuting? That the list is REAL, you can GET this list, but not from a computer. Your words dumbass. Herc
From: |-|ercules on 30 Jun 2010 18:25 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 30, 1:54 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> Which of these is a logical formula? >> >> 1/ ((a -> b) & (b -> c) -> (a -> c) >> 2/ (a = 2) -> (a > 1) > > Neither. > 1 is not well-formed because it has 4 left parentheses and only 3 > right ones. > 2 contains "=" which is NOT a logical symbol. You just reached troll status on Transfer Principle's scale, not accepting Peano Arithmetic as a sound theory. Herc
From: George Greene on 1 Jul 2010 01:22 On Jun 30, 6:11 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > HAHAHA. > Are you really that stupid to assume induction on both forms is the same too? I don't NEED to ASSUME! I KNOW what induction is! YOU DON'T! > > Does the list > 0.0 > 0.1 > 0.2 > ... > 0.9 > 0.10 > 0.11 > ... > 0.99 > 0.101 > ... > > use this induction schema too? A LIST *CANNOT*USE* "an induction schema" !! WHY DON'T YOU SIMPLY STATE which induction schema YOU are using??? THE STANDARD induction schema requires you to START with a ONE-place predicate that takes A NUMBER as a parameter. You can change that from a number to something else as LONG as you have a clear notion of what THE SUCCESSOR, THE NEXT element, of something, would be. And in any case, you have THE WRONG list. The ACTUAL list is ..0 ..1 ..2 ..3 ..4 ..5 ..6 ..7 ..8 ..9 ..01 ..11 ..21 etc. It does NOT have .10 after .9 ! The reason for this is that you need LOTS OF LEADING ZEROS as prefixes! You have to spell 10, 100, 1000 etc. FROM RIGHT TO LEFT. > > phi( <[1] 2 3 4...> ) & An ((phi ( <[1 2 ... n] n+1 n+2 ...>) -> phi( <[1 2 ... n n+1] n+2 n+3 ...> )) > -> > phi( <[1 2 3 4...]> ) THIS IS NOT an induction schema, DUMBASS. The induction schema JUST has phi(0), NOT phi (<[1] 2 3 4>). Moreover, the conclusion of an induction schema IS NOT phi(<[1 2 3 4...]>). IT IS, rather and INSTEAD, An[phi(n)]. THIS MATTERS because there MIGHT be MORE things in the domain THAN JUST natural numbers. The point is, you CANNOT conclude -- not by induction, anyway -- that something that is true for [1] AND [1 2] AND [1 2 3 ] AND [1 2 3 ] AND [1 2 3 4 ], .etc. IS ALSO true for [1 2 3 4 5 ...] TO INFINITY. THAT IS NOT WHAT INDUCTION *SAYS*, DUMBASS. What it DOES say is that something is true for 1 AND for 2 AND for 3 AND for 4 AND for 5 AND for 6 AND for 7 AND for 8 ... etc., BUT NOT *TO*INFINITY* -- RATHER, IT ONLY says this for infinitely MANY DIFFERENT *FINITE* things, DUMBASS! It does NOT say it for EVEN ONE INFINITE thing!
From: |-|ercules on 1 Jul 2010 04:42 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 30, 6:11 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> HAHAHA. >> Are you really that stupid to assume ...? > > I don't NEED to ASSUME! You're right there! Stupid is as stupid does. > > I KNOW what induction is! > > YOU DON'T! > >> >> Does the list >> 0.0 >> 0.1 >> 0.2 >> ... >> 0.9 >> 0.10 >> 0.11 >> ... >> 0.99 >> 0.101 >> ... >> >> use this induction schema too? > > A LIST *CANNOT*USE* "an induction schema" !! > WHY DON'T YOU SIMPLY STATE which induction schema YOU are using??? > > THE STANDARD induction schema requires you to START with a ONE-place > predicate > that takes A NUMBER as a parameter. You can change that from a number > to something > else as LONG as you have a clear notion of what THE SUCCESSOR, THE > NEXT element, > of something, would be. > And in any case, you have THE WRONG list. > The ACTUAL list is > .0 > .1 > .2 > .3 > .4 > .5 > .6 > .7 > .8 > .9 > .01 > .11 > .21 etc. > > It does NOT have .10 after .9 ! > The reason for this is that you need LOTS OF LEADING ZEROS as > prefixes! > You have to spell 10, 100, 1000 etc. FROM RIGHT TO LEFT. >> >> phi( <[1] 2 3 4...> ) & An ((phi ( <[1 2 ... n] n+1 n+2 ...>) -> phi( <[1 2 ... n n+1] n+2 n+3 ...> )) >> -> >> phi( <[1 2 3 4...]> ) > > THIS IS NOT an induction schema, DUMBASS. > The induction schema JUST has phi(0), NOT phi (<[1] 2 3 4>). > Moreover, the conclusion of an induction schema IS NOT > phi(<[1 2 3 4...]>). IT IS, rather and INSTEAD, > An[phi(n)]. > THIS MATTERS because there MIGHT be MORE things in the domain THAN > JUST > natural numbers. > > The point is, you CANNOT conclude -- not by induction, anyway -- that > something that is true for > [1] AND > [1 2] AND > [1 2 3 ] AND > [1 2 3 ] AND > [1 2 3 4 ], .etc. IS ALSO true for > [1 2 3 4 5 ...] TO INFINITY. > THAT IS NOT WHAT INDUCTION *SAYS*, DUMBASS. > What it DOES say is that something is true for > 1 AND > for 2 AND > for 3 AND for > 4 AND for 5 AND for 6 AND for > 7 AND for > 8 ... etc., BUT NOT *TO*INFINITY* -- RATHER, IT ONLY says this for > infinitely MANY DIFFERENT *FINITE* things, DUMBASS! > It does NOT say it for EVEN ONE INFINITE thing! In the words of the commander in the starting scene of Gladiators, "You'd think the savage would know when he is beaten." phi(1) & An (phi(n) -> phi(n +1)) -> An phi(n) Let phi(n) = property-all-sequences holds for n. phi(1) & phi(2) & phi(3) = property-all-sequences holds for 1, 2 & 3. <=> < [1 2 3] 4 5 6...> An phi(n) = property-all-sequences holds for all digits <=> <[1 2 3 4...]> That is impossible to dispute, your argument could now become that property-all-sequences is merely equivalent to property-all-sequences-up-to In the words of Buzz Aldrin, UP TO INFINITY AND BEYOND! Herc
From: |-|ercules on 1 Jul 2010 04:47
"|-|ercules" <radgray123(a)yahoo.com> wrote > In the words of Buzz Aldrin, > > UP TO INFINITY AND BEYOND! > Buzz LIGHTYEAR! Buzz LIGHTYEAR! Herc |