
Re: If ZFC is a FORMAL THEORY ... then what is THEOREM 1 ?
Posted:
Oct 11, 2012 4:03 PM


On Oct 12, 1:23 am, George Greene <gree...@email.unc.edu> wrote: > On Oct 9, 6:01 pm, Graham Cooper <grahamcoop...@gmail.com> wrote: > > > So the AXIOMS are a *CONSTRAINT* what formula are allowed. > > No. *NOTHING* is "allowed". "Allowed" is not a relevant status. > The formulas come from a firstorder LANGUAGE with a SIGNATURE. > The signature is a list of predicate and function NAMES, each of > which has an arity (a number of arguments). ALL of the wellformed > formulas in the relevant > firstorder LANGUAGE are built out of logical symbols predicates, > applied to arguments that are built out of these functions. > > > [2] > > > > (P>Q)>((P^R)>Q) IS A TAUTOLOGY  and there are MANY SIMPLE > > > "algorithms" that PROVE THAT THAT IS A TAUTOLOGY! > > > only in FOL. > > You are LYING. Tautologies are from *0*OL. > > > Axioms (R here) are 2OL or 3OL (Regularity) > > NO, you are lying AGAIN. The VERSION of set theory that we are > talking about > here IS *1ST*order ZFC! The LANGUAGE that all these formulas > (including > all the axioms) come from is a *1ST*order language! The LOGIC in > which > we are deriving these proofs IS *1ST*order logic! > > >You need ZFC to tell you what sets DO exist!  This validity ONLY > > >tells you that a certain kind of set canNOT exist! > > > ALL(set) mem e set IFF p(mem,set) > > AND EXIST(anotherset) mem e anotherset > > > This is what ZFC does, limits the relation 'e'. > > Exactly. It *defines* e. > With no axioms, e wouldn't mean anything at all. > With no axioms, you would HAVE NO WAY OF TELLING whether > the "e" in ~Er[ xer <> ~xex ] meant > "_ shaves _ " (in which case the sentence would mean "there is no > barber > who shaves all and only those barbers who don't shave themselves") or > "set _ is a member of set _ " (in which case the sentence would mean > there > is no set whose members are all&only those sets that are not members > of themselves). > > In all these theories, the axioms constrain the meaning or > interpretation > of the functions and predicates occuring in them. > > > [4] > > What's the difference between Predicate Calculus > > and Naive Set Theory. > > Predicate calculus is a calculus. It's a logic. It is defined by its > INFERENCE RULES. It's used to prove/derive/infer THEOREMS > *from*prior* assumptions. > SOME theorems (like P V~P) can be inferred or derived (USING THESE > RULES, AND USING THE ALGORITHMS DEFINED for applying these rules) FROM > NO prior assumptions. But the formulas so proved/derived/assumed need > to be in a first > order LANGUAGE, first. > > Naive set theory is a theory. WITH AXIOMS. > > > They are both ZEROAXIOM logic systems. > > NO, THEY ARE NOT. Naive Set Theory IS NOT a "logic system". > It's a A THEORY. WITH SOME AXIOMS. ONE OF ITS AXIOMS IS > the NAIVE setCOMPRENSION axiom, ExAy[ yex <> phi(y) ]. > This is an axiomschema saying that you get an axiomatic automatically > proventrue theorem confirming the existence of a set corresponding > to EVERY predicate phi(.) that is definable in the language. > In particular, it says you get a set when phi(y) means ~yey. > SO THIS THEORY IS INCONSISTENT. > This theory is *IN* predicate calculus. It's UNDER predicate > calculus. > It's worked with USING predicate calculus, IN THE FRAMEWORK of > predicate calculus. > > > If NST is inconsistent, what rules in Predicate Calculus > > are you using to ensure ZFC does not inherit NST inconsistency? > > We are using THE SAME rules of predicate calculus FOR BOTH > ZFC and NST. > There is NO inheritance because ZFC *replaces* NST's > NAIVE setCOMPREHENSION axiom with the axiom of > SEPARATION. > > It is the naive setcomprehension axiom that causes the problem. > That axiom is inconsistent/paradoxical in itself. > It simply CANnot be the case that EVERY predicate defines a set. > You need a more complicated, less naive set of axioms to define sets. > Zermelo gave a first one in 1903. > Frankel filled in an obvious hole later (That is how we got ZF). > The axiom of choice seemed like an obvious hole too, but that > got very complicated for about 30 years from 193666.
So this formula is barred!
ExAy[ yex <> phi(y) ].
Now you're saying Predicate Calculus (with only infererence rules over predicates which you wrongly call FOL)
P > P ~P v P (P ^ Q) > P .... and a few more like that..
et voila!
NOT( EXIST(INFERENCE) INFERENCXE= ExAy[ yex <> phi(y) ] )
******************
OK so you DEFINE THE MEANING OF 'e' purely for use at the AXIOM LEVEL.
and you STILL GET THE PARADOX:
E(rs) A(x) x ee rs <> !(x ee rs)
in Predicate Calculus.
Predicate Calculus WORKS ALL OF THIS OUT, it does resolution, consistency checks, et al (using FOL INFERENCE RULES (ranging over variables))
PC is consistent because it derives NOT[ EXIST(rs) A(x) x ee rs <> !(x ee rs) ]
and prevents E(rs) going on the LHS of any further consistent inferences.
BUT, JUST TO BE DOUBLE SURE... the SETS REALLY DO EXIST IN THE AXIOMS LEVEL by BARRING rs from instantiating with 'e' too.
since 'e' is now a reserved letter, (implicit by being a new symbol in the axioms)
***********
WRONG GEORGE!
You're flicking in and out of FORMULA SYNTAX CALCULUS 1OL, 2OL, 3OL INFERENCE vs MODUS PONENS NAIVE SET THEORY CONSISTENCY BY VERIFICATION THEORY CONSISTENCY BY STRATIFICATION THEORY CONSISTENCY BY FORMULA RESTRICTION (of the term 'e')
You have no LOGIC THEORY, you have an ORACLE, you have DERIVED SETS with no RUSSELL SET
for the 20TH TIME HERE:
E(X) NOT(X=X)
is in the language of predicate calculus.
YOU DON"T HAVE A TRANSITIVE CLOSURE OF LOGIC BASED ON INFERENCE RULES
and if you did THEY CERTAINLY WOULDN'T BE FOL THEY WOULD NECESSARILY RANGE OVER FORMULAS.
Herc

