
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 2:43 PM


"William Hale" wrote in message news:billh04154C6A.13334504022013@news.eternalseptember.org...
In article <c4a53c56e96f4524b9ddb7ae8c75160a@r14g2000yqe.googlegroups.com>, CharlieBoo <shymathguy@gmail.com> wrote:
> On Feb 4, 9:32 am, billh04 <bill...@gmail.com> wrote: > > On Feb 4, 6:26 am, CharlieBoo <shymath...@gmail.com> wrote: > > > > > > > > > > > > > > > > > > > > > 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. > > > > But, in ZFC, the statement "Ax.not x e x" is true and the statement > > "Ex. x e x" is false, among many other such statement. Certainly, e(x, > > y) and e(x, x) must be a predicate in ZFC. How can it not be? > > In this case, because primitives of logical expressions must be > relations and ~e is not a relation.
I (1) don't make the assumption that primitives of logical expressions must be relations. I (2) assume you mean the relation "~e" to be the set of ordered pairs (x, y) such that x ~e y.
Since I (3) don't take logical expressions to be sets, I (4) certainly don't take logical expressions to be relations. I (5) would prefer to say that a logical expression may sometimes determine a set. But sometimes a logical expression won't determine a set (e.g., the logical expression "x ~e x" wont' determine a set.)
Thus, I (6) say that "x ~e x" is a wff, but "x ~e x" cannot be used to define a relation that corresponds to it.
> It depends on how you define wff, > including substitution for (aka interpreting) symbols for these > primitives.
I (7) don't include the substitution for symbols in my consideration of whether a statement or expression is a wff. For me (8), whether something is a wff is a syntactical question.
> How do you define it?
Do you want to know or do you want to see if I (9) know? If you want to see if I (10) know, I (11) will concede that I (12) am not an expert on wffs. I (13) suspect that you want to show that it cannot be explained even though people claim that a wff can be defined. ============================================================== Thirteen references to yourself. You really are an opinionated fuckwit, aren't you? *plonk*
 This message is brought to you from the keyboard of Lord Androcles, Zeroth Earl of Medway. When the fools chicken farmer Wilson and Van de faggot present an argument I cannot laugh at I'll retire from usenet.

