Re: Axiom of Pairing, Scheme of Replacement from others

On Jan 24, 6:07 pm, "Stephen J. Herschkorn" <sjhersc...@xxxxxxxxxxxx>
wrote:

There was no need to go to the level of detail presented above (except
perhaps for the "[ | ]" notation). In fact, while I appreciate the
effort MB went to, presenting a proof in such low-level detail obscures
the issue at hand.

I felt the details were needed to preclude any objection as to my use
of variables. The proofs are famous and in standard textbooks, yet I
was being called upon to produce such proofs, so I didn't know that
some technical objection would not be raised, so I crossed t's and
dotted i's.

As to obscuring, my point was not to make the proof as easy as
possible, but rather to give it with all details that would preclude my
having to defend against technical objections or an objection that I
was handwaving through fine but crucial distinctions regarding
variables and things like that.

For my own purposes, I'd just say:

To prove separation from the formulation of replacement I gave, let the
phi in replacement be the formula: chi & u=y. The rest is an exercise
in obvious manipulation of formulas.

To prove the existence of an empty set from the formulation of
separation I gave, let the chi in separation be the formula: yey &
~yey. The rest is an exercise in obvious manipulation of formulas.

Axiom schema of replacement
For all P,
if P is a formula and
v does not occur free in P and
w does not occur free in P and
v is free for y in P and
w is free for y in P and
b does not occur free in P, the all closures of

AuezAvAw(P[v|y] & P[w|y]) -> v=w) -> EbAy(yeb <-> EuezP)

are axioms.

That is not how I understand repleacement.

My formulation is really just the one you find in Enderton, though mine
is more pedantic and finicky with details.

Your hypothesis has
uniqueness but not existence. Should it not be

Auez (Ey P & (AvAw(P[v|y] & P[w|y]) -> v=w)) -> EbAy(yeb <-> EuezP)
?

Maybe there is such a formulation in the literature; I don't know. But,
in the meantime, my formulation boils down to just what you find in
Enderton and many other textbooks.

Or, in notation I find a little bit clearer, let f be a formula
wherein w is not free. We have the universal closure of

[Ay (y in x -> E!z f)] -> Ew Az [z in w <-> Ey (y in x & f)]

That looks like it might work (I just glanced at it quickly). There are
many ways of formulating replacement, but the one I gave is correct
also and it is basically the one given in many textbooks.

Axiom schema of separation:

For all Q,
if Q is a formula and
b does not occur free in Q, then all closures of

EbAy(yeb <-> (yez & P))

are axioms.

/

Proof of separation from replacement (in tedious detail):

Let Q be a formula such that b does not occur free in Q.

Let P be the formula Q & u=y.

An instance of the axiom schema of replacement:

AuezAvAw((Q & u=y)[v|y] & (Q & u=y)[w|y]) -> v=w) -> EbAy(yeb <->
Euez(Q & u=y)), which is:

AuezAvAw(Q[v|y] & u=v & Q[w|y] & u=w) -> v=w) -> EbAy(yeb <-> Euez(Q &
u=y)).

The antecedent is a theorem of identity theory, since u=v & u=w implies
v=w.

So, by sentential logic, EbAy(yeb <-> Euez(Q & u=y)).

Here is where the omission in your statement of the replacement scheme
wreaks havoc.

There is nothing left out of my formulation of replacement that must be
there for a correct formulation. My formulation is correct. It's just a
bit more pedantic than such formulations as Endertons' but it is
basically the same formulation, just as his formulation (and dozens and
dozens of other formulations you can find in textbooks) do not have
this "existence clause" you seem to think is required.

And there is no "havoc". I'm just giving tedious details for a proof
that is mentioned, without tedious details, in such texts as Suppes's
'Axiomatic Set Theory'. Hey, there might be more elegant ways of moving
through the mundane details, but replacement does entail separation, by
the overall strategy of using the formula Q & u=y, as you can find in
textbooks, and (barring typos), my details are correct.

Proof of ExAy ~yex from the axiom schema of separation:

This part was unncessary.

No, it shows that replacement alone entails the existence of an empty
set, as follows: Replacement alone entails separation and separation
alone entails the existence of an empty set, so replacement alone
entails the existence of an empty set. Yes, we could go directly from
replacement to the existence of an empty set, but since I proved
replacement entails separation, I might as well kill two birds with one
stone: show separation entails empty set (thus that replacement entails
empty set) and also showing that not just in ZF, but in Z, there is no
need of an empty set axiom, as the author of the Wikipedia article is
incorrect that we can't prove the existence of an empty set from
replacement or separation (or whatever the exact statement the author
of the Wikipedia article made on that subject).

Among the relevent pages (if I recall correctly, one you linked to),
there is an auxilliary discussion (found under the tab 'discussion')
about the page itself.

I have looked at that tab, and I find no such discussion. Do you have a
more precise URL?

I'm sorry that I can't manage to stumble across that page as I did the
first time. It was under the discussion tab for one of the related
articles. It might not be a discussion with the same author as the one
for the page on replacement, so I might be mistaken as to the author.
Though, the discussion was quite parallel to this issue. I'll try to
come across it again (though, really, I very much do not relish the
thought of visiting Wikipedia pages on the subject of mathematics), and
I'll link you to it if I do.

Without even appealing to the semantic
principle of non-empty domain, the SYNTACTICAL rules of first order
logic allow the derivation I gave. Yes, I recognize that the syntactic
rules are meant to work with a semantics that presuposes non-empty
domains, but the question here was the PROVABILITY of the replacement
from separation and the provability of the empty set from separation.
And the proofs I gave are correct (adjusting for any possible typos).

No. See above.

You're just incorrect. You are demanding a clause that is not part of
the axiom schema. Then you are just saying my proof wreaks some
unspecified "havoc".

I really don't know where you are coming from. You can just look in a
book like Suppes's 'Axiomatic Set Theory'. (Suppes has urelements, so
he has an empty set axiom; but his empty set axiom does not come into
play in the proof that replacement entails separation.)

Or, if you don't like my fussy formulations, just look at Enderton's
formulation of replacement. Then plug in the formula Q & u=y as I did
(adjusting for choice of variables), then do the exercise for yourself
to see that replacement entails separation.

Let us review here. I paraprhase.

SJH 1: Replacement + existence of empty set implies separation.
MB 1: You do not need the empty set for such a proof.
SJH 2: Well, here are reputable sites which explicitly show why you do.

Wikipedia a reputable source for mathematics?!!! Oh come on now. (And
no, I'm not going to debate the point. If you take Wikipedia as a good
place to research mathematics, then suit yourself.)

Moreover, it's not a question of showing how to use the existence of
the empty set in proving separation from replacement. One can
explicitly do that. But that doesn't entail that one can't ALSO show
separation from replacement withOUT first proving the existence of the
empty set.

The author of that Wikipedia article does NOT prove the indpedendence
of separation from replacement. And he can't, because separation is not
iindependent of replacement.

MB 2: It's not true. It's not. It's not. It's not. [See that post -
it really reads that way.]

No, it reads 'That is incorrect' to different assertions and some
assertions repeated.

SJH 3: Just saying that is not enough. Prove it.

No, you said more than that. You gave an admonition, with me held as an
example, about people making unsupported claims. My purpose was just to
put the brakes on the UNSUPPORTED and INCORRECT claim that you were
conveying from the Wikipedia article that replacement doesn't entail
separation. The Wikipedia article is incorrect that replacement does
not entail separation. And the Wikipedia article purports to explain,
but it does NOT PROVE that separation is independent of replacement.
And you're complaining about ME not substantiating?! Sheesh!

Instead of including your snide comment that made me out to be
intractable for not provding a proof, you could have just said, "Okay,
what's your proof?", and then I would not have occasion for expressing
my umbrage about your comment.

MB 3: Here's a [faulty] proof. You never said you wanted one before.
Are you calling me stupid?

"Are you calling me stupid?" Please do not put such supposed
paraphrases in my mouth. I never made such a comment. I said that you
had insinuated that I was intractable. I said nothing about stupidity.

In this forum, or in any rational discussion of mathematics (and other
subjects), if one is going to claim something discussed elsewhere is not
true, then one should give a justification of the claim.

If asked to do so, then one upholds a standard of intellectual
responsibility by providing a proof or mentioning where one can be
found. But at the point before one has been asked for a proof, one is
not required to provide a proof just to register a comment that
something stated is not correct.

I really figured that with you knowledge and expertise (I'm not being
sarcastic, you do know a lot of stuff, though you have lapsed and
gotten confused about this particular point), you'd take my 'That is
incorrect' comments as a spur just to look in some standard textbooks.
There you will find the trick for proving separation from replacement
and, if you do the exercise of the mundane predicate logic, then you
will see that it is not required to mention anything at all about the
existence of the empty set.

Such a
justification can be either an explicit proof (or outline of one) or a
citation. There is no need to wait for an invitation.

I don't have to drag a commonly known proof into my post just to put on
record that I am denying what a Wikipedia article is incorrectly
claiming. If you ever want a proof from me, all you have to do is ask.

And your
detection of insinuations are your own inference.

Oh come on, "MB's repetitions demonstrate how easy it is to claim
something is not true" is not dripping with insinuation that I am
intractable for not having given you a proof right from the start?

That much said, I repeat what I said at the beginning of this post.

I value MB's contributions to this discussion, and I hope he will
continue.Respectfully,

And I value your contributions. But for whatever reason, you've got a
blind spot on this particular issue:

1. The Wikipedia does not prove the independence of separation from
replacement. Just purporting to explain why the existence of the empty
set is needed for proving separation from replacement is not a proof
that replacement does not entail separation.

2. The Wikipedia article is incorrect in claiming that replacement does
not entail separation. I proved that replacement does entail
separation. And if you don't like my fussy formulas, you can look at
many a textbook, or you can do the details yourself from the main hint.

3. You are incorrect that my formulation is incorrect. My formulation
is basically that of Enderton and many other textbooks. I use the 'P[v
| y]' notation where other authors would use the less fusssy 'P(y)'
notation, but otherwise, my formulation amounts exactly to what
Enderton's formulation amounts to.

4. The axiom schema of replacement does not require such an existence
clause as you mention. Just look at Enderton's or Suppes's or many
other textbook formulations.

MoeBlee

.

Relevant Pages

  • Re: Axiom of Pairing, Scheme of Replacement from others
    ... separation follows from the axiom of replacement together with the axiom ... empty set is required. ...
    (sci.math)
  • Re: Axiom of Pairing, Scheme of Replacement from others
    ... separation follows from the axiom of replacement together with the axiom ... An empty set axiom is not needed for that. ...
    (sci.math)
  • Re: ZFC in another shape.
    ... Replacement -> Empty Set ... Replacement entails separation. ... this strong Replacement with power do entail finite union. ...
    (sci.math)
  • Re: Axiom of Pairing, Scheme of Replacement from others
    ... separation follows from the axiom of replacement together with the axiom ... An empty set axiom is not needed for that. ...
    (sci.math)
  • Re: Restricting Separation
    ... restriction on separation would eventually mutilate the mathematical ... It seems to me, your new axiom of separation, with side parameters x1, ... ... and replacement, and consider these axioms replacing usual separation and replacement to ... accessed the Xi dynamically inside the members where a copy of the ...
    (sci.logic)
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