
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


On Feb 3, 11:53 pm, camgi...@hush.com wrote: > On Feb 4, 2:19 pm, CharlieBoo <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 2place. 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.
CB
> 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

