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Re: This is False. 0/0 {x | x ~e x} e {x | x ~e x} A single Principle
to Resolve Several Paradoxes
Posted:
Feb 4, 2013 7:26 AM
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On Feb 3, 11:53 pm, camgi...@hush.com wrote:
> On Feb 4, 2:19 pm, Charlie-Boo <shymath...@gmail.com> wrote:
>
>
>
> > > RELATION
> > > p(a, b, e)
>
> > If wffs are built on relations then { x | x ~e x } is not a wff
> > because ~e is not a relation.
>
> if e(x,y) is a predicate
> then not(e(x,y)) is a predicate
And more importantly not(e(x,x)) is a predicate (diagonalization.)
Yes, that is Naïve Set Theory, which is correct. But the IF fails.
"e(x,y) is a predicate" is not correct due to diagonalization. There
is no Russell Paradox, only Russell's Diagonalization.
If e(x,y) were a predicate then not(e(x,x)) would be a predicate but
because of diagonalization it is not.
What I believe about Prolog consists only of its being a database and
query language similar to predicate calculus (aka FOL.)
Then I take it that:
if [ not [ e rs rs ]] [ e rs rs ] Line 1 X => rs
if [ e rs rs ] [ not [ e rs rs ]] Line 2 X => rs
[ not [ e rs rs ]] = = [ e rs rs ] Def = = , if
~P = = P [ e rs rs ]
=> P
Then some axiom about ~P = = P.
In Combinatory Logic:
M x => N x x Def M
M M => N M M x => M
There is no M M = = N M M because it is executed (no Turing Machine
can halt both yes and no because it stops as soon as it halts either
cf Rosser 1936.) M M and N M M occur at different points in time!
The system simply changes its mind as it goes along learning using AI
techniques. The circle is amicable.
In CBL:
M # P(x) / Q(a,b) means program/wff M enumerates/represents set P and
is written in programming language/logic Q meaning Q(x,y) iff program/
wff M outputs at some point/is provable on input y. Then Q(M,x) = =
P(x) (all x). Abbreviated M#P/Q means P=Q(M).
P(x) / Q(a,b) means (exists M) M # P(x) / Q(a,b) There is a program/
wff that enumerates/represents set P.
Let SE(x,y) iff set x contains element y.
Then the Russell Paradox is:
~SE(x,x) / SE(a,b) There is a set of all sets that do not contain
themselves.
M # ~SE(x,x) / SE(a,b) Let M be the set of all sets that do not
contain themselves.
SE(M,x) = = ~SE(x,x) Def # /
SE(M,M) = = ~SE(M,M) x => M
P = = ~P SE(M,M) => P
This is encapsulated in axiom (lower level theorem) - ~P/P where - E
means expression E is not true and P/Q abbreviates P(x)/Q(a,b) or
P(x,x)/Q(a,b) if P is 2-place. Then if P/Q then P differs from Q and
we prove truth, provability and unrefutability distinct (Godel,
Rosser, Smullyan) - see 18 Word Proof in FOM July 2010.
C-B
> http://en.wikipedia.org/wiki/First_order_logic#Formation_rules
> Negation. If f is a formula, then f' is a formula.
>
> ---------------------------------------------
>
> In BLOCK PROLOG Russell's Set would be:
>
> if [ not [ e X X ]] [ e X rs ]
> if [ e X rs ] [ not [ e X X ]]
>
> I'm trying to forward MODUS PONENS to a contradiction!
>
> Herc
> --www.BLoCKPROLOG.com
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