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new unknown: diagonalization-lemma looks like obvious-formalization,
but understanding it further helps with issue with its use

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great we can maybe study-repeat it

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The diagonal lemma shows that in theories that can represent
computability, all formulas have a fixed point.

i don't think i agree wih that, maybe relates to formal definition of theory

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what would it mean to not have a fixed point?

it w

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so

selecting not working, but new unknown diag lemma: not obvious
formality. states every formula has fixed point.

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wasn't using reference equality

diag lemma says g b g same g both sides
it asserts not that b and g can exist but that a g exists for every b.
strong claim

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we expect diag lemma to cross logics

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ybe mistake how absorb mistake

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new unknown: "formula" in diag-def

1: The diagonalization of X is the formula (∃x)(x=⌈X ⌉ ∧ X )

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the proof is broken into two lemmas
the first defines "diag(n)" (diag-def)
the second asserts formulas have fixed points

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(it might be helpful to learn the formula syntax that uses an and?
symbol and looks a little like substitution

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i figured it but itKs not stabilized in thinking

Add star  mailbombbin<[email protected]>    Sat, Sep 16, 2023 at 4:34 PM
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X is a formula containing x
diag(n) = ⌈(∃x)(x=⌈X⌉∧X)⌉
the expression means that -- waaaait !
I think it is saying that ... ohhhh

it's stating that x is X, and that the contents of X hold true, both
not sure why it says ∃x instead of something like ∃X but by the left
side they're equivalent

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(so lemma 1 is talking about fixed points)

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lemma 1 says that if you can serialize unary formulas you can
serialize the expression of them having a fixed point in this way,
(and that they are "computable" new-unknown ctxual might just mean you
can solve it)

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so: lemma 1 says fixed point expression (oops)

...the proof of lemma 2 parts looks obvious but don't see how
conclusion arises, have not connected all parts. some not understood,
looks relates "theorem copy-paste not work with whole at once

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Assume that D represents diag in T and let F be the formula (∃y)(D(x,y) ∧ B(y)).

i think this says that f says there exists a fixed point y in B

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thought: i'm guessing godel issolving his own expression here
general proof of fixed-pointness, if youhave a metamath youcan solve
the expression ofpassing theexpfession toitself, for the parameter.
requiresconcievingof where self-refis and whereparameter is andmoving
unknown to one side

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wait so um

x = f(

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(we woulduse diff terms for describing domain andrange naybe confused

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i don't believeitithink metasummarywrongor misinterpreted

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i'm havingtrouble understanding itbrelatedto retaning andcomparing the
summariesformed after considering theparts

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> Assume that D represents diag in T and let F be the formula (∃y)(D(x,y) ∧ 
> B(y)).

> Choose G ≡ (∃x)(x=⌈F⌉∧F) as the diagonalization of F and let n and g be the 
> Go ̈del

new unknown: guessing that triple bars are formal definition, 1 check

> numbers for F and G, respectively. By definition of diag we know diag(⌈F⌉) = 
> ⌈G⌉ and

> thus D(n_, g_ ) must be valid in T .

> Furthermore G ≡ (∃x)(x=n ∧ (∃y)(D(x,y) ∧ B(y)) is logically equivalent to the 
> formula

> (∃y)(D(n_ ,y) ∧ B(y)). Because of the functionality of D and the validity of 
> D(n_,g_) this

> formula is equivalent to D(n_, g_) ∧ B(g_), which in turn is equivalent to 
> B(g_).

substituted g for y which was F which expands to similarity containing B?

> Thus G is logically equivalent to B(⌈G_⌉) in T and hence |=T G_ ⇔ B(⌈G_⌉).

> Note that the diagonal lemma holds for Peano Arithmetic, as diag is 
> representable in any theory that can represent the computable functions.

it's notable godel and peers must have been holding famikiar intuition
to develop and argue this. used structures to describe/simplify
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let's rephrase for clarity

> diag(⌈X⌉) = ⌈(∃x)(x=⌈X⌉∧X)⌉

...
> Let f and g be the Go ̈del numbers for F and G, respectively.
> Let F be the formula (∃y)(y=diag(x) ∧ B(y)).

issue
- Show quoted text -

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more study8ng methinks unless typing easier and can pin to one part

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i don't know what D(x,y) means when diag has only one variable
it looks like it relates to the outside vs inside of theexprrssion
like one is n and the other diag(n)

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> diag(⌈X⌉) = ⌈(∃x)(x=⌈X⌉∧X)⌉
> Assume that D represents diag in T and let F be the formula (∃y)(D(x,y) ∧ 
> B(y)).
> Choose G ≡ (∃x)(x=⌈F⌉∧F) as the diagonalization of F and let f and g be the 
> Go ̈del
> numbers for F and G, respectively. By definition of diag we know diag(⌈F⌉) = 
> ⌈G⌉ and
> thus D(f_, g_ ) must be valid in T .
> Furthermore G ≡ (∃x)(x=f ∧ (∃y)(D(x,y) ∧ B(y)) is logically equivalent to the 
> formula
> (∃y)(D(f_, y) ∧ B(y)). Because of the functionality of D and the validity of 
> D(f_, g_) this
> formula is equivalent to D(f_, g_) ∧ B(g_), which in turn is equivalent to 
> B(g_).
- Show quoted text -
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> diag(⌈X⌉) = ⌈(∃x)(x=⌈X⌉∧X)⌉
> Assume that D represents diag in T and let F be the formula (∃y)(D(x,y) ∧ 
> B(y)).
new unknown: D(x,y) ≡? diag(x) = y
- Show quoted text -

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- how did y turn to g_
- how is B(g_) isolated out of ∧
- Hide quoted text -

thinking about diag(n)
this stating that when a formula is substituted into itself, it is true
it is itself a formula that converts this statement to a godel number

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