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## Re: Become an Objectivist in Ten Easy Steps

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• ... What do these step mean??? Where is this value expression defined? What does (6) mean?? In particular x=x is always True (according to the same
Message 1 of 9 , Jul 16, 2002
On Tue, Jul 16, 2002, Ofir Carny wrote about "RE: Become an Objectivist in Ten Easy Steps":
> (6) |- rational(y) & value(y, v) ==> v [Rational Value Theorem]
> (7) |- egoist(x) <==> value(x, x = x) [definition of egoism]

What do these step mean??? Where is this "value" expression defined?
What does (6) mean??

In particular "x=x" is always True (according to the same reasoning in
statement 3), so in (7) we actually have that x is an egoist if and only
if "the value of x is True", or something like that. What the heck does
that mean? Why write "x=x", when "x=x" is simply a tautology, always True,
and as we see later we indeed use that fact?

Sorry, but this whole "proof" looks like hogwash to me...
(but maybe I'm missing something...)

> (8) |- not(egoist(y)) ==> not(value(y, y = y)) [from (7) instantiating x = y]
> (9) |- not(egoist(y)) ==> value(y, not(y = y)) [by not-propagation]

I'm still waiting to hear what this "value" expression does, and how come
you can propegate a NOT into its second argument...

> (10)|- not(egoist(y)) ==> not(y = y) [by (9) and (6)]
> (11)|- not(egoist(y)) ==> false [by (10) and (3)]

Ok, so *finally* y=y is written as True. why was it necessary to write it
as y=y in the first place?? just to make it look more complicated??

This proof appears to me as valid as the following proof, that dogs can
fly:

(1) dog(x) <=> bark(x) & has_a_tail(x) [definition]
(2) dog(y) [hypothesis]
(3) arctan(1)=PI/4 [from math class]
(4) has_a_tail(y) [by (1) and (2)]
(5) can_fly(x) <=> altitude(x, arctan(1)=PI/4) [definition of flying]
(6) has_a_tail(y) & altitude(y, v) ==> v [Having-a-Tail Theorem]
(7) not(can_fly(y)) => not(altitude(y, arctan(1)=PI/4) [instantiating x=y]
(8) not(can_fly(y)) => altitude(y, not arctan(1)=PI/4) [not propegation]
(9) not(can_fly(y)) => not (arctan(1)=PI/4) [by (8), (6) and (4)]
(10) not(can_fly(y)) => false [by (9) and (3)]
(11) can_fly(y) [by (10) ad absurdum]
(12) dog(y) => can_fly(y) [by (2) and (11) by ==> introduction]

For those wondering how come I "proved" dogs can fly, well, it all rests
on my silly made-up definitions and theorems, like "definition of flying",
"having-a-tail theorem", and "not propegation" :)

--
Nadav Har'El | Tuesday, Jul 16 2002, 7 Av 5762
nyh@... |-----------------------------------------
Phone: +972-53-245868, ICQ 13349191 |A man is incomplete until he is married.
http://nadav.harel.org.il |After that, he is finished.
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