*Post by peteolcott**Post by peteolcott*This universal Truth predicate rejects as malformed any expressions of language that would otherwise show Incompleteness or Undefinability.

∀x ∈ L True(L, x) ↔ Theorem(L, x).

For all expressions x in language L, x is True in L if and only if x is a Theorem of L

Oh sweetie. Bless your heart.

*Post by peteolcott*True(F, G) ↔ Theorem(F, G)

False(F, G) ↔ Theorem(F, ~G)

Presumably, you intend "True()" and "False()" to be predicates in F, rather than from a higher metalanguage M, where truth is actually evaluated.

Simply labelling the provability predicate and its negation as "True()" and "False()" in F is not a proof that they correspond to actual truth and falsity as evaluated in M.

Labels are not proofs. Without a proof of equivalence, it would be just as meaningful to label your predicates "Fuzzy()" and "Smooth()", or "Predicate1()" and "Predicate2()", or "Intelligent()" and "Olcott()".

When we hypothesize that True(L, x) ↔ Theorem(L, x)

and thus False(L, x) ↔ Theorem(L, ~x) we find that we now

have a corresponding pair of predicates in L

No, there is only one predicate in L: Provable(). You just assigned it a second label in L "True()". You literally defined them to be equivalent in L.. They have exactly the same extensional meaning.

As I pointed out, this relabelling tells us nothing at all about actual truth, since that is evaluated in the metalanguage M.

EFQ

I keep seeing the validity of where you are coming from.

Theorem(L, x) ⊆ Provable(L, x)

All that I have done is taken the subset of Provable expressions x of language

L such that the Mendelson members of Γ are called the hypotheses or premisses

of the proof are empty and labeled this subset of expressions True(L, x).

This by itself really tells us nothing much at all.

This by itself is nothing more than a dictatorial decree.

The interesting part comes in when we test the hypothesis of:

∀x ∈ L True(L, x) ↔ Theorem(L, x)

Does the corresponding pair of predicates:

True(L, x) ↔ Theorem(L, x)

False(L, x) ↔ Theorem(L, ~x)

always reject expressions of language that have previously been shown

to "prove" Incompleteness and Undefinability ?

LP ↔ ~Theorem(L, LP)

Here is the simplest possible example:

LP ↔ ~Theorem(LP) // L is implied rather than specified to avoid clutter

If LP was a Theorem then since LP is ~Theorem(LP) then

Theorem(LP) would be contradicted therefore LP is not a Theorem.

If ~LP was a Theorem then since ~LP is Theorem(LP) then

Theorem(~LP) would be contradicted therefore ~LP is not a Theorem.

Since it is neither True nor False that LP is a Theorem,

then LP is decided to be a malformed logical proposition.

At least many of the set of "undecidable" decision problems

can be decided this same way.

Copyright 2018 Pete Olcott