I like the idea of structuralism, and my own personal understanding of set theory is that we may think of the axioms as specifying structural relations rather than worrying whether there is some abstract world of which the axioms are true. In other words, from the axioms, we may consider those abstract relations that the axioms define, rather than worry about whether there is a particular abstract world in which the "objects" in those relations exist.

The question I have though is, even if we don't have to worry about what it means to say the objects exist, haven't we just kicked the can down the road to the question "In what sense do the relations, the structures exist?"?

If you replace numbers with sets, do you keep the same ontology? — Gregory

"Replace" might not be a good way of putting. A better way of putting it might be that sets "play the role" of numbers, or something like that.

And what ontology? Different philosophers and/or mathematicians have different conceptualizations of ontology for set theory.

What of structuralism? — Gregory

I addressed that:

Set theory does provide the structural relations we expect. Even though the objects have "extra-structural" properties (e.g. that 0 is the empty set and 1 is the set {0}), the structural relations are captured (e.g. that 0 < 1). — TonesInDeepFreeze

a set is something that contains and not something in its own right — Gregory

That's not the case in set theory.

Or maybe we can think of vacuity and unity as the bases. Then we have 0 and 1, the binary. But in set theory, with the pairing operation, we can define '1' from '0'..

df: {x y} = z <-> Ak(k e z <-> (k = x v k= y))

dr: {x} = {x x}

df: x = 0 <-> Ay ~xey

df: 1 = {0}

df: n and m are natural numbers -> (n < m <-> n e m)

/

Set theory does provide the structural relations we expect. Even though the objects have "extra-structural" properties (e.g. that 0 is the empty set and 1 is the set {0}), the structural relations are captured (e.g. that 0 < 1).

https://thephilosophyforum.com/discussion/comment/568807

With a Hilbert style system, the axiom we use to derive modus tollens is given in the intuitionistically invalid form:

(~P -> ~Q) -> (Q -.> P)

and then we derive the intuitionistically valid form:

(P -> Q) -> (~Q -> ~P)

The reason is that we can't derive the intuitionistically invalid form from the intuitionistically valid form.

inductive set well-ordered by ∈ — fishfry

I think you meant 'transitive set well ordered by ∈'.

My mistake.

My mistake. My proof does not use RAA.

I see that what I left tacit is not RAA, but just modus tollens:

Let f:X -> PX
Let S = {y e X | ~y e f(y)}
S e PX
S e ran(f) -> EyeX f(y) = S
Suppose EyeX f(y) = S [for -> introduction]
Let f(y) = S
y e f(y) <-> ~ y e f(y)
EyeX f(y) = S -> (y e f(y) <-> ~ y e f(y)) [-> introduction]
~EyeX f(y) = S [intuitionistically valid modus tollens]
So f is not onto PS

Another instance in which modus tollens does the same job as RAA.

I don't know whether this bears on anything here, but just in case, there is huge difference between:

(1) ~~P RAA premise ... contradiction ... infer ~P

and

(2) P RAA premise ... contradiction ... infer ~P.

(1) is not intuitionistically valid. (2) is intuitionistically valid.

I did bother. I read your post three times but couldn't figure it out.

RAA premise would not need to deny ~P. Rather, in this case, the premise is P.
— TonesInDeepFreeze

Exactly, if for some reason you want to label the RAA line "P" rather "~P". In a line properly signposted as RAA, and in a discussion in which someone had bothered to say

We don't need to suppose toward contradiction that there is a surjection.
— TonesInDeepFreeze — bongo fury

I don't know what you're saying.

it could make sense to display under that signpost (P or ~P depending on signposting preferences, or a form of words such as I chose so that the question didn't arise) the denial of what is to be shown. This denial will be the supposition toward a contradiction. What is to be shown is that S can't, without contradiction, be in the range of f. So the denial, the suitable RAA line, the supposition toward contradiction, if you or anyone did want to belabor the point, or understand the point about "not needing to suppose toward contradiction..." is indeed "S is in the range of f", and it might be interesting that this is taking the place of "f is surjective", in a proof by contradiction. — bongo fury

You lost me.

I can see it both ways.

Starting with an RAA premise provides a clear structure.

Not starting with RAA, but instead talking about an arbitrary function from S to PS, is a little more arch, and reveals the hammer blow more by surprise.

I wonder why Cantor didn't start with RAA.

In any case, the choice is not to be confused with eschewing intutionistically invalid RAA, since it is intuitionistically valid.

And, of course, I wouldn't even think of denying the claim that S is not in the range of f.
— TonesInDeepFreeze

Except in a line properly signposted as RAA. — bongo fury

There's a double negative in what you're saying. RAA premise would not need to deny ~P. Rather, in this case, the premise is P.

Anyway, that's not the beginning of my proof.

yours begins (read as a proof by contradiction) by denying a more specific claim of failure of surjectivity: the claim that such sets as, in particular, S will fail to be in the range of f. — bongo fury

In the beginning, I didn't deny any claim claim whatsoever.

And, of course, I wouldn't even think of denying the claim that S is not in the range of f.

In that way, fishfry's RAA is deferred in my proof to later.
— TonesInDeepFreeze

Yes. It's still a proof by contradiction, just not so upfront. — bongo fury

Yes, that is fair to say. Notice, I didn't say that RAA would be avoided in a (natural deduction style) proof. I only mentioned a particular opening RAA premise, viz. "There is a surjection" and that we don't need to adopt it as an RAA premise.

And it's still a proof by contradiction. — bongo fury

It has an RAA nested within. Of course.

RAA and modus tollens are basically the same.
— TonesInDeepFreeze

Another reason not to expect an important contrast in your reworking. — bongo fury

As far as I can tell, the difference is merely stylistic. I said that in my previous post.

/

For reference:

Let f:X -> PX
Let S = {y e X | ~y e f(y)}
S e PX
S e ran(f) -> EyeX f(y) = S
Let f(y) = S
y e f(y) <-> ~ y e f(y)
So f is not a surjection

Of course the premise "A Republican wins" restricts. The impression that there is not good reason to believe "If Reagan doesn't win then Anderson wins" comes from (1) Overlooking that it "If Reagan doesn't win then Anderson wins" is merely a conditional and not a statement about Anderson winning, and (2) overlooking that "If Reagan doesn't win then Anderson wins" is equivalent with "A Republican wins", so whatever the bases are for believing "A Republican wins" are the same bases for believing "If Reagan doesn't win then Anderson wins".

McGee's error is the claim that the conclusion doesn't have as good a reason to believe as the premises. That is the explicit error in his argument about the MP example.

Indeed Carter is left out with the premise "A Republican wins", but stating that Carter is left out and that therefore the conclusion is clouded is not in and of itself a refutation. The follow-through is that Carter being left out is just "A Republican wins" which is equivalent with the conclusion, so whatever basis there is for "A Republican wins" is bases for the conclusion.

The fact that there is more reason to believe "If Reagan does not win then Carter wins" than there is reason to believe "If Reagan does not win then Carter wins" is McGee's red herring.

There is a difference between (1) What is the best way to set up inferences about the election? and (2) Does the conclusion of the particular MP mentioned have as great a reason for believing as the premises have?

The key to refuting McGee is not (1). It's (2).

mental state — Assumptions and the Supposed Counterexamples to Modus Ponens, D. E. Over, Analysis, 1987

McGee also erred by dragging in what people believed. (Probably, nudging in "people believed" is part of the sleight of hand.) What people believed is irrelevant to the example. But we can still couch his argument without concern for what people believed and stick with "reason to believe" only.

/

Another article about the puzzle went into a bunch of stuff about subjunctive mood. For me, that 's a wrong tack: (1) We can rephrase the MP without subjunctive and (2) The puzzle is dissolved much more easily, trivially, anyway.

Incorrect: We should not use 'least' if we don't mean quantity.

It is typical of cranks unfamiliar with mathematical practice to think that the special mathematical senses of words most conform to their own sense of the words or even to everyday non-mathematical senses. The formal theories don't even have natural language words in them. Rather, they are purely symbolic. Natural language words are used conversationally and in writing so that we can more easily communicate and see concepts in our mind's eye. The words themselves are often suggestive of our intuitions and our conceptual motivations, but proofs in the formal theory cannot appeal to what the words suggest or connote. And for any word such as 'least' if a crank simply could not stomach using that word in the mathematical sense, then, if we were fabulously indulgent of the crank, we could say, "Fine, we'll say 'schmleast' instead. 'schmardinality' instead'. 'ploompty ket' instead of 'empty set' ... It would not affect the mathematics, as the structural relations among the words would remain, and the formal symbolism too.
— TonesInDeepFreeze — Metaphysician Undercover

That incorrectly makes it appear that I said, "Incorrect: We should not use 'least' if we don't mean quantity."

Is that a thing? — bongo fury

For some mathematicians its a stylistic preference.(I'm not sure, but I think maybe my version is Cantor's version.)*

proving (still by contradiction) a stronger denial of surjectivity than mere failure of surjectivity. — bongo fury

The proofs prove the exact same result - nothing more nothing less, except for a stylistic choice. Both proofs are correct and intutionistically valid.

EyeX f(y) = S
— TonesInDeepFreeze

which (I guess?) follows from

f(y) = S
— TonesInDeepFreeze — bongo fury

That's backwards.

EyeX f(y) = S
Let f(y) = S

That's existential instantiation.

EyeX f(y) = S is itself an RAA premise within the proof. In that way, firshfry's RAA is deferred in my proof to later.

I didn't write EyeX f(y) = S as a separate line, since I didn't belabor certain obvious steps; it's not a fully formal proof.

/

RAA and modus tollens are basically the same. RAA as a rule in natural deduction(or a derived rule from a Hilbert style system) while modus tollens does the job as an axiom in a Hilbert style system.

/

* Re Canor's diagonal proof, if I recall correctly, he doesn't start with a premise that there is a surjection and then derive a contradiction to infer there is not a surjection. Rather, he reasons about any arbitrary enumeration and shows that it is not a surjection. And I've read certain mathematicians say that they prefer not to set it up as an RAA. Same with the infinitude of the primes and other results. I don't know any philosophical reason for that; I take it as a stylistic choice.

The definition of 'precedes' ('less than') had been given by fishfry many posts ago. To have missed it is to have not paid attention to the posts.

Df. If x and y are ordinals, then x precedes y (x is less than y) iff x is an element of y.

Th. If y is an ordinal, then for all x, if x is a member of y, then x is a subset of y.

Th. If y is an ordinal, then y = {x | x is an ordinal & xey}. I.e., every ordinal is the set of its preceding ordinals.

/

Ordinals are called 'ordinal numbers'. But that is not needed for the actual formal results in set theory. We could just as easily always say 'ordinal' instead of 'ordinal number'. However, 'ordinal number' does reflect that there are operations on ordinals - ordinal arithmetic, ordinal multiplication, and ordinal exponentiation.

'is a number' is not a predicate of set theory. Saying 'ordinal number' rather than 'ordinal' does not confer any special property of being a "number"; it's just suggestive phrasing and does not add any inferential force.

/

There is no set of all the ordinals.

/

'is an ordinal' was defined, at least a few times, already in this thread.

/

fishfry is using 'cardinally equivalent' to mean 'equinumerous' (also 'equipollent'). the term 'cardinally equivalent' does not depend on having previously defined 'is a cardinal' nor 'the cardinality of'.

Again:

df. x and y are cardinally equivalent iff there is a bijection between x and y
[neither 'ordinal' nor 'cardinal' are mentioned in that definition]

Probably a more common terminology is:
df. x and y are equinumerous iff there is a bijection between x and y

df. x is a cardinal iff (x is an ordinal & ~Ey(yex & x and y are equinumerous))

df. card(y) = the cardinal x such that x and y are equinumerous

th. Ax(x is a natural number -> x is an ordinal)

th. {x | x is a natural number} = {x | x is an ordinal & x is finite

/

IMPORTANT SUGGESTION:

One should always understand that English terminologies such as 'ordinal number', 'equinumerous', etc. are only nicknames for actual formal symbols in set theory. The nicknames are suggestive of certain intuitions and motivations, but the nicknames do not confer any deductive effect. The only things that definitions that provide only abbreviation of actual formulas written solely in the primitive language.

It is a common error for people who are not familiar with formal theories to think that we can make deductions from the themes suggested in the nicknames; for example, saying "Well, if they're ordinal numbers then they must have all the properties of other numbers such as natural numbers, rational numbers, et. al}. That's wrong.

So, again, it is incorrect to think that 'cardinally equivalent' has some kind of "ontological" connection with 'is a cardinal' or 'the cardinality of'.

Again, three separate things (and the suggestiveness of the English word 'cardinal' in each of them does not confer any deductive force):

x and y are equinumerous [fishfry is calling that 'x and y are cardinally equivalent']

x is a cardinal

x is the cardinality of y [or said by, 'x = card(y)']

and:

th. x is a cardinal iff Ey x = card(y) [and 'x is a cardinality' just means 'x is a cardinal']

/

Every set is an object. Every cardinal is a set. Every cardinal is an object.

But 'object' is not a term in the language of set theory. It's just a word we use to talk about the things (objects) that are ranged over by the quantifier in the primitive language.

When talking about Z set theories, one can use 'set' and 'object' interchangeably since Z set theories don't "talk about" anything other than sets.

df: S is equinumerous with T <-> there is a bijection between S and T

theorem: For every S, there is a unique T such that T is an ordinal & S and T are equinumerous & no member of T is equinumerous with S

df: card(S) = the unique T such that T is an ordinal & S and T are equinumerous & no member of T is equinumerous with S.

We say card(S) is the cardinality of S.

df: k is ord-less-than j <-> k e j

When context is clear, we just say "k is less than j" or "k < j" or "k e j".

df: k is ord-least in S <-> (k e S & ~Ej(j e S & j ek))

When context is clear, we just say "k is least in S".

df: k is the least ordinal such that P <-> (k is an ordinal & Pk & ~Ej(Pj & j ek))

df: S is a cardinal <-> (S is an ordinal & there is no ordinal T less than S such S and T are equinumerous)

theorem: S is a cardinal <-> Ex S = card(x)

/

There is no circularity there.

If one has a definition of 'inherent' then they can add it.

'inherent' has not been given a mathematical definition. Dispute about it can go on and on and on, and in circles, for as long as people have the opportunity and willingness to dispute about it.

But notions that have been given mathematical definition include:

ordinal
well ordering
ordinal less-than
least ordinal in a set
least ordinal with a formula defined property
bijection
equinumerous
isomorphism
order type
injection
cardinal
cardinality
cardinal less-than

Definitions adhere to forms that ensure ensure eliminability (formulas with defined terms can be set back to formulas without the defined terms) and non-creativity (formulas that weren't already provable aren't made provable with the introduction of defined terms). By adherence to the forms for definitions, the definitions are never circular.

And new definitions can be provided - anyone is free to introduce a new term and define it.

Set theory is not properly critiqued by acting as if some undefined terms controls results in set theory.

https://thephilosophyforum.com/discussion/comment/568451

We would continue to prove that the uncountability of Pw implies the uncountability of R:

It suffices to prove that the interval [0 1] is uncountable.

We have the theorem Ax Px 1-1 with 2^x.

So P^w 1-1 with 2^w.

And we prove 2^w injects in [0 1]

k = the-least-ordinal_x such that Px <-> (k is an ordinal & Pk & Ah(h e k -> ~Ph))

We don't need to suppose toward contradiction that there is a surjection.

Let f:X -> PX
Let S = {y e X | ~y e f(y)}
S e PX
S e ran(f) -> EyeX f(y) = S
Let f(y) = S
y e f(y) <-> ~ y e f(y)
So f is not a surjection

https://sites.duke.edu/wsa/papers/files/2011/05/wsa-defenseofmodusponens1986.pdf

I just now read that. My argument is basically the same as theirs.

From Donkeys and Elephants to Lungfish and Porpoises.

McGee has another supposed impeachment of MP.

https://sites.duke.edu/wsa/papers/files/2011/05/wsa-defenseofmodusponens1986.pdf

"I see what looks like a large fish writhing in a fisherman's net a ways off. I
believe
If that creature is a fish, then if it has lungs, it's a lungfish.
That, after all, is what one means by "lungfish." Yet, even though I
believe the antecedent of this conditional, I do not conclude
If that creature has lungs, it's a lungfish.
Lungfishes are rare, oddly shaped, and, to my knowledge, appear only in
fresh water. It is more likely that, even though it does not look like one,
the animal in the net is a porpoise"

c = that creature
Fx <-> x is a fish
Lx <-> x is a lungfish
Hx <-> x has lungs

(1) Fc -> (Hc -> Lc) [McGee believes]
(2) Fc [McGee believes]
(3) Hc -> Lc [McGee does not believe]

But, again, what people do or do not believe is not relevant. What is relevant is (1) for validity, whether truth is preserved, or I suppose (2) for "strict validity", whether "good reason to believe" is preserved.

Obviously the strength of reason to believe the conclusion Hc -> Lc follows the strength of reason to believe the premise Fc. But we need to show that mathematically. I haven't finished it, but getting close:

prob(Fc -> (Hc -> Lc)) = 100%, since it's true by definition

Let prob(Fc) = x

Let prob(Hc) = y

Let prob(Lc) = z

We also have Lc -> Fc [implicit in the problem; I wish I could dispense with it though]

So prob(Lc) = prob(Fc & Lc) = x*prob(Hc | Fc)

~Hc and Lc are mutually exclusive so:

So prob(Hc -> Lc) = prob(~Hc v Lc) = prob(~Hc)+prob(Lc) = (100-y)+(x*prob(Hc | Fc))

So prob(Hc -> Lc) goes up and down as prob(Fc) goes up and down.

The possibility of the porpoise is not part of this. Just like the possibility of Carter was not part of the earlier problem. The porpoise is Carter! (And, as we know, Paul is the walrus.)

You are terribly confused. You asked me to prove there is not bijection between a countable set and an uncountable set. And I told you where to find the proof. But you say you already know about it. So there was no point in asking me.

But you say there are infinitely many more odd numbers than natural numbers. So YOU prove that. Not mind pictures, but mathematical proof.

If you really do want to understand, then get a book on set theory and start reading it from page 1. I could type proofs and explanations for you all day, but if you don't have the background for it, then it's a waste. Any proof I give you will depend on proofs and definitions previous to that proof, on and on backwards until we reach the axioms. So that is utterly impractical in a thread. The reasonable and enlightened way is to get a book and read it from page 1, from the axioms through the proofs.

By definition, there is no bijection between a countable set and an uncountable set.

By theorem, there is no bijection between N and R. The proof has been given thousands of times in textbooks, articles, and on the Internet (it's even outlined in the article linked to above). You really are not familiar with the proof?

I really want to know — Gregory

Then open an Internet search engine and type 'proof uncountability reals'.

Impressionistic descriptions are fine for stoking creativity in mathematics and sometimes for making certain mathematical concepts intuitive. But they are not mathematical demonstrations.

But there are infinite more natural numbers, just as with the reals. — Gregory

Please. If you have a rigorous definition of "infinite more" different from set theoretic "greater cardinality" then fine, state your definition, and your claim could be right relative to that definition. Meanwhile the card(N) = card(set of odds) is a theorem of set theory.

Is the point that there are far more infinities of reals than infinity of naturals vs the odd? — Gregory

I can't parse that.

I can imagine putting any two infinities one to one — Gregory

It is a theorem of set theory that no set is 1-1 with its power set. It's a theory of set theory that N is not 1-1 with R.

If you have an alternative theory, then state your axioms. What you merely imagine is not mathematics.

Beating a dead horse about donkeys, elephants, and red herrings.

Directly responding to the clip and its one sentence intro:

Vann McGee claims that modus ponens "is not strictly valid" in an article from 1985

MP is valid. McGee does not say MP is not valid. He says it is not "strictly valid". What does "strictly valid" mean? He mentions "good reason to believe". So the only way I can think of regarding "strictly valid" is "If there is good reason to believe the premises then there is good reason to believe the conclusion". He says, of the only non-logical premise, that we have good reason to believe "A Republican will win the election". But "A Republican will win the election" is equivalent to "If it's not Reagan who wins, it will be Anderson". So whatever the good reason we have to believe "A Republican will win the election" is the same good reason we have to believe "If it's not Reagan who wins, it will be Anderson". So this instance of MP that McGee claims shows that strict validity fails, is actually an example in which strict validity succeeds. It's that simple. McGee is refuted.

Opinion polls taken just before the1980 election showed the Republican Ronald Reagan decisively ahead of the Democrat Jimmy Carter,

That's false. But nevermind, we can take it as a hypothetical given. And it doesn't even matter anyway. All we need is the background assumption that people had good reason to believe that a Republican would win. Carter is a red herring.

with the other Republican in the race, John Anderson, a distant third. Those apprised of the poll results believed,

What people believe in premise and conclusion is irrelevant. People can err by believing the premises of a valid argument but not the conclusion. The strict validity of MP couldn't depend on the empirical fact of what erring humans believe.

with good reason:

So it should be "Those apprised of the poll results had good reason to believe".

[1] If a Republican wins the election, then if it's not Reagan who wins it will be Anderson.
[2] A Republican will win the election.

[1] is a logical truth and [2] is a non-logical premise.

In the context, [2] is equivalent to "Either Reagan will win or Anderson will win".

Yet they did not have reason to believe
[3] If it's not Reagan who wins, it will be Anderson

If they had good reason to believe [2] then they had good reason to believe [3].

McGee did not show that strict validity of MP failed.

I just realized I made a really rookie mistake in some of my attempts several posts back.

prob(x) is not presumably the poll rating of x.

For example, if my memory is in the ballpark, the day before the election, Biden was given about an 85% chance of winning but he only had about 52% in the polls.

But my latest posts don't assign prob(x) but only rank the probabilities as given.

prob(R) > prob(C) > prob(A)

And we don't even need that!

Whatever the prob(R v A) is, it's no greater than prob(~R -> A), as indeed prob(R v A) = prob(~R -> A).

And we don't even need that!

All we have to do is see that whatever the strength of reason for believe for R v A, it's no greater than the strength of reason for belief of ~R -> A. How could that not be? R v A is equivalent with ~R -> A. It's that simple!

n the absence of the article — Banno

I said that I only have the clip from the article to reference plus the locution 'strictly valid'.

Anything I say is in that context alone. If there is more in his article that qualifies the context, then that would be another story.

if Reagan did not win, it would have been Carter — Banno

Not by the premises of the argument.

The point is not to challenge the premises, but rather to show that the conclusion has as great a reason for belief as the premises. That's all that's needed to refute McGee. And it's trivial. It only looks hard because he razzle dazzles us with a distraction.

Here's the new puzzle for me:

Based on the level of McGee's research in logic and his associations, he must be extremely intelligent and knowledgeable. Not a nano-mote of doubt that I could not possible fathom all the logic he knows. So how could he have made such a rank mistake?

I should have approached the problem more systematically from the start. I thought that the explanation would have to be at a high level involving intensionality and then probability. But then I saw that it is bare bones trivial. That's the magician's trick. He distracts you with a bunch of razzle dazzle hiding the explanation that is the one right in front of your nose.

Then I went on to win the Tour de France, the Indy 500, the Pulitzer Prize, and the Best Pecan Pie Award at the Polk County Fair, all in one week. But everybody already knows all about that, so enough about me.

My analysis doesn't need to say anything about Carter.

All I need to point out is that the conclusion of the example has strength of reason to belief not less than the strength of reason to believe the premises.

But I did mention that the strength of reason to believe ~R -> C is greater than the strength of reason to believe ~R -> A, though that is not needed to refute McGee.

'incomplete' is not part of my analysis.

MP is valid.

McGee claims MP is not "strictly valid" which I can only take to mean that MP does not preserve strength of reason to believe. But, contrary to his sleight of hand, his example shows MP preserving strength of reason to believe.

I agree.

"false -> P" is true.

And again, it's not even about MP:

S v H is equivalent to ~S -> H.

Everything is immaterial vs
Some things are material

Everything is material vs
Some things are immaterial

Everything is immaterial vs
Some things are material and some things are immaterial

Everything is material vs
Some things are material and some things are immaterial

Everything is immaterial vs
Everything is material

Which of those are being debated?

What is the definition of 'is material'?

Or if Idealism and Materialism can't be shoehorned into strict Pro and Con that way, then there could be just one particular representative proposition chosen as the Pro with Con as its negation.

Or there could be two propositions or sets of propositions, not necessarily one the actual negation of the other, and each party supports their own proposition and attacks the other party's proposition.