That would be my leaning too.

Yes, we say 'necessary' and 'sufficient' conditions. But that is not "necessarily implies' or 'necessarily leads to'.

If P -> Q, then P is a sufficient condition for Q and Q is a necessary condition for P, and that is not to suggest any modal operator. But we don't say that P necessarily implies Q.

curt — Leontiskos

Actually, your imperative "Then give your proof" was curt, especially as spoken to someone giving you correct information.

The proof is so simple that one would think that someone posting claims about the matter would know the proof or would just look it up, or even run it in their mind for five seconds, rather than a challenging "Then give your proof". People don't need to give such utterly basic proofs. It's like saying, "Then give your proof that 1+1=2". Moreover, as good as proofs had already been given in this thread.

Run it in your mind for five seconds: If A implies both B and ~B, then A implies a contradiction, so we infer that A is not the case. You really need to challenge people to prove that for you?

opposite assertions cannot be true at the same time — Aristotle on Non-contradiction | SEP

That is the law of non-contradiction. What I said is a more formal way of saying the same.

It is not trolling to point out an incorrect statement, and it not trolling nor handwaving to suggest that one can look in textbooks to see that the statement is incorrect.

Or look in any resource you like to see that '->' in logic is ordinarily understood as the material conditional and any textbook in symbolic logic will explicitly state that '->' is the sentential connective such that 'P -> Q' is interpreted as true if and only 'P' is interpreted as false or 'Q' is interpreted as true. And that is material implication.

Look in any textbook on symbolic logic.

So what?

'->' is ordinarily regarded as standing for material implication that is not "P necessarily implies Q" nor "P necessarily leads to Q".

Tones says that both propositions imply ¬A. — Lionino

Both propositions together imply ~A. The conjunction of the propositions implies ~A. The premise set {A->B, A->~B} implies ~A.

That is true if "both props" is understood as (A → B) ^ (A → ¬B — Lionino

Right.

You misunderstand. — Philosophim

I replied exactly to what you wrote. What you wrote is wrong.

A -> B

or

material implication

is not "A necessarily implies B".

Then give your proof. — Leontiskos

Are you serious? You don't know how to prove it yourself?

Proof:

(1) (A -> B) ... premise

(2) (A -> ~B) ... premise

(3) A ... toward a contradiction

(4) B ... from (1), (3) modus ponies

(5) ~B ... from (2), (3) modus ponens

(6) ~A ... from (4), (5) and discharge (3) by contradiction

Truth table:

A B......A->B......A->~B......~A
T T..........T............ F..............F
T F..........F.............T..............F
F T..........T.............T.............T
F F..........T.............T.............T

All rows in which both A->B and A->~B are true are rows in which ~A is true.

/

Do I get anything for doing your homework for you? Cash? Philosophy Forums poker chips? Internet Brownie points?

A -> B. But that's not imply. that's "Necessarily leads to." — Philosophim

Wrong. Material implication does not require necessity.

You think the two propositions logically imply ~A? — Leontiskos

They imply ~A.

When we talk about contradiction there is a cleavage, insofar as it cannot strictly speaking be captured by logic. It is a violation of logic. — Leontiskos

I don't know what you mean by 'cleavage' and 'captured' in this context. But in logic systems we can write contradictions. Indeed, we often intentionally prove contradictions from premises in order to refute the premises.

A classical definition says that two propositions are contradictory if the denial of either entails the affirmation of the other, and vice versa. So there are materially four different relations, given that each of the two propositions can be denied or affirmed. — Leontiskos

A contradiction is a formula of the form P & ~P, or in other contexts the pair {P ~P}.

We don't have to check four different things to see that formula, or in other contexts a pair of formulas, is a contradiction.

A set of formulas is inconsistent if and only if it proves a contradiction.

it is not possible to contradict a material implication — Leontiskos

Wrong. P -> Q is contradicted by ~(P -> Q).

Assume P and suppose Q. If an absurdity results on Q, then P and Q are contradictory. — Leontiskos

Yes, people say "contradictory", but as a terminological preference, I would say they are together inconsistent. A pair of formulas is a contradiction iff they are of the form P and ~P. A set of formulas is inconsistent if and only if it implies a contradiction.

It shows that they cannot both be true, but it does not show that they cannot both be false — Leontiskos

Wrong. If one is true then the other is false. If one is false then the other is true. [EDIT:] Correction: Obviously, it is not correct that it is always the case that there is at lease one true statement in an inconsistent set of statements.

it does not show that they cannot both be false, and it does not show that the trueness or falseness of one results from the falseness or trueness of the other — Leontiskos

Wrong. If the members of a set of sentences that includes P are all true, and that set of sentences along with Q (that is not in the set) proves a contradiction, then Q is false. Put another way, we cannot derive a contradiction from a set of all true sentences. [EDIT:] Correction: Obviously, it is not correct that it is always the case that there is at lease one true statement in an inconsistent set of statements.

"The car is wholly green." "No, the car is wholly red."

This is a contradiction classically but not according to symbolic logic — Leontiskos

Classical logic includes propositional logic such as treated in symbolic logic.

"The car is green" and "The car is red" is not a contradiction. But if we add the premise: "If the car is red then the car is not green," then the three statements together are inconsistent. That's for classical logic and for symbolic rendering for classical logic too.

If this is about material implication then the answer is utterly simple:

A -> B
A -> ~B

are not together inconsistent, since they are both true when A is false, and the propositional calculus doesn't permit an inference of a contradiction from two formulas such that there is a model in which they are both true.

The digressions in this thread don't affect that very simple fact.

Are you referring to 'is' in terms of identity or value? — javi2541997

Identity. I don't know how to make it more clear than I already have.

x is x.

With models:

'=' is interpreted as {<x x> | x e U} where U is the universe for the model

Although every bill or note is represented by the payment of x5, it will depend on the value. So, x = x, doesn't equal to "is." — javi2541997

5 = 5

5 dollars not= 5 pounds

If you have 5 dollars and I have 5 pounds, then the number of your dollars is the number of my pounds. 5 is 5. But that does not entail that 5 pounds is 5 dollars.

The number of things is not the things.

'x = x' doesn't equal 'is'. Rather, '=' stands for 'is', so 'x = x' stands for 'x is x'.

None of that is a contradiction or problem.

Maybe it's the symbols. Maybe it's the words. — fishfry

The symbols are standard. The words are ordinary for logic and mathematics, or if personal, they're defined.

So maybe it's something else.

Most glaringly of all, what accounts for you recently claiming that I hadn't specified 'identity theory' when I had specified it multiple times in this thread, including multiple times addressed to you, and even twice quoted by you? Your claim is bizarre.

Explicity stated in any textbook in mathematical logic.

'=' is interpreted:

For any terms 'T' and 'S'

T = S

is true

if and only if

the denotation of 'T' is the denotation of 'S'.

Consult any textbook in mathematical logic. The ignorant, confused, arbitrary personal dictates of the crank don't count.

It is crystal clear that '=' is interpreted as 'is' in mathematics, since it is explicitly stated that '=' is interpreted as 'is' in mathematics. That the crank doesn't know anything about '=' in mathematics, or anything else about mathematics or logic doesn't entail that it is not the case that '=' is interpreted as 'is' in mathematics. Fortunately, what is the case about mathematics or logic is not affected by the crank's ignorance.

If A is false, then (A->B) & (A->~B) is true. So (A->B) & (A->~B) does not entail a contradiction. It's that simple.

Further Adventures Of The Crank:

Sales Clerk: That will be five dollars.

/ The crank puts four one dollar bills on the counter /

Crank: There you go, five dollars.

Sales Clerk: That's only four dollars.

Crank: No, it's five dollars.

/ The sales clerk counts the bills by hand /

Sales Clerk: You see, only four dollars.

Crank: Five in your mind is different from five in my mind. I pay based only on what is in my mind. I gave you five dollars, now would you please put that copy of 'Hegel For Dummies' in a bag for me, as I just paid you five dollars for it?

Sales Clerk [on microphone]: Security at register seven. Security at register seven please.

Mathematics adheres to the law of identity, since in mathematics, for any x, x=x, which is to say, for any x, x is x.

I have two different dollar bills in my pocket. They are not the same dollar bill. But they are equal in value.

(1)

bill 1 is not the same as bill 2
bill 1 does not equal bill 2
bill 1 is not identical with bill 2

Those are true and say the same thing as one another.

(2)

the value of bill 1 is the same as the value of bill 2
the value of bill 1 equals the value of bill 2
the value of bill 1 is identical with the value of bill 2

Those are true and say the same thing as one another.

(1) and (2) together is not a contradiction.

The valuation of a bill is a function, call it 'v'.

bill 1 does not equal bill 2
but
v(bill 1) equals v(bill 2)

Example from math:

Let v be the squaring function. So v(x) = x^2.

1 does not equal -1

but

v(1) = v(-1)

as

1^2 = -1^2

I was rushing. My mistake.

it makes no difference if it's an antecedent in a conclusion or as a premise. — Michael

No, it doesn't. It's called 'the deduction theorem'. For example:

P, Q, R |- S

is equivalent with

P, Q |- R -> S

P5 is an inherent contradiction — Michael

You haven't paid attention to my answer to that. Now, you're arguing by mere assertion and repeated mere assertion.

Moreover, if P5 is deemed in and of itself contradictory, then we don't need an argument with other premises to derive a contradiction. If P5 is, from the outset declared a contradiction, then we don't have to bother with the bloody lamp at all.

You're not thinking about my points as it seems you just lump them in with other people's arguments. You are skipping key points. You argue by question begging and mere assertion. And now you link me to a post from eleven days ago that I had already addressed in detail.

As to C6, you put the halving ad infinitum as an antecedent in a conclusion, which is fine. But it's equivalent to just making it a premise. And I mentioned that a while ago, and mentioned why it is more stark to make it a premise, but you ignored that too. Your terse argument that I've recently mentioned and your argument eleven days ago are essentially the same: It doesn't matter whether you put the halving ad infinitum as an antecedent in a conclusion or as a premise - it's logically the same.

Anyway, we're going in circles as you skip my arguments while reasserting your own. When I saw that initially you were taking care to make numbered arguments, I got interested and thought you might be open to more scrutiny. But I can see you're not, as you only keep repeating assertions and not actually thinking about the replies to them. It seems your primary interest is to persist that you are rigtht and not to truly think through objections. So, bye for now.

You're stuck thinking I'm making a certain kind of argument, but I am not. You're not thinking about what I've specifically written, as probably you take me to be making a version of other arguments around. If you're not going to take my arguments as given, then there's no rational inquiry to be had.

But I'll address these two:

"after I have completed the whole infinite sequence of jabs, i.e. at the end of the two minutes, is the lamp on or off?". — Michael

But completion is not in your premises.

I am no longer pushing the button at any time after 12:00. My infinite button pushes has allegedly ended. — Michael

That is begging the question. It is begging the question to rule that there can't be denumerably many tasks executed in finite time. You haven't proved: If there are denumerably many tasks executed in finite time then there is an end to their executions. So you have to either prove it or add it as a premise.

I use the phrase "metaphysical impossibility" rather than "logical impossibility" simply because it's the weaker claim. — Michael

It's not much of a claim if it is not defined.

P1. The lamp is turned on and off only by pushing the button
P2. If the lamp is off and the button is pushed then the lamp is turned on
P3. If the lamp is on and the button is pushed then the lamp is turned off
P4. The lamp is off at t0
P5. The button is pushed at successively halved intervals of time between t0 and t1
P6. The lamp is either on or off at t1

But actually, if we look at the mechanics of the inferences toward the contradiction, we see that we need that the lamp is never both on and off. So I would write.

P1. The lamp is turned on and off only by pushing the button
P2. If the lamp is off and the button is pushed then the lamp is turned on
P3. If the lamp is on and the button is pushed then the lamp is turned off
P4. The lamp is off at t0
P5. The button is pushed at successively halved intervals of time between t0 and t1
P6. At all times the lamp is either on or off and not both.

Then we derive a contradiction:

C1. The lamp is either on or off, and the lamp is neither on nor off.

Now, what premise do we delete to avoid the contradiction? Since none of them are logically true, and each is needed in the derivation of the contradiction, we may delete any one of them.

You answered pretty fast. That's your prerogative. But it make me wonder whether you're giving much thought to my remarks, as still it would be your prerogative not to. So I'll take the same prerogative.

The definition of a super-task is as you say. But your listed premises don't say anything about completion or ending.

One of the contradictions does; the state of the lamp at 12:00. — Michael

The contradiction is: The lamp is either On or Off T 12:00 and the lamp is neither On nor Off at 12:00.

But that contradiction comes from a set of premises, each of which is not logically true, and dropping any one of the premises blocks deriving the contradiction. It would help if you would at least tell me that you understand that.

The other contradiction is the inherent contradiction of an endless sequence of operations coming to an end. — Michael

(1) The premises don't say it comes to an end. It would help if you would at least tell me that you understand that.

(2) It is begging the question merely to declare it is a contradiction that denumerably many tasks can be executed in finite time. Indeed, the argument itself doesn't declare that it is a contradiction. Rather, the argument derives a contradiction from that premise along with other premises.

Again, the example I gave:

AxEy yex

ExAy(yex <-> ~yey)

entails a contradiction, but it doesn't entail that either of the above is itself a contradiction.

Even most minimally:

P
~P

entails a contradiction, but that doesn't entail that either P or ~P is a contradiction.

It would help if you would at least tell me that you understand this.

Just incase you missed my edit to my previous post:

Let's even assume for the sake of argument that this wizard will only appear with a probability of 0.5, and that this is determined only at exactly 12:00, i.e after the performance of the supertask. It must already be possible for the supertask to be performed for him to even appear, and so his appearance cannot retroactively make the supertask possible, even if half the time it resolves the secondary contradiction regarding the state of the lamp at 12:00. — Michael

There's no just in case that you missed my own post. You missed that I said I don't argue in any such way that is knocked down as a straw man as you have.

/

The link doesn't go to a defininition. It merely says that metaphysical possibility may be logically possibility and that there's another notion that the article describes ostensively. So is it just the same as logical possibility, and if not what is a proper definition that is not merely ostensive?

Finding some way to resolve the former does not retroactively resolve the latter.

Finding some way to resolve the former does not retroactively resolve the latter. — Michael

So, yes, clearly taking your prerogative to answer so quickly did result in your not even taking note of what I said, let alone taking a moment to understand it.

I am not "finding some way to resolve the former [to] retroactively resolve the latter." You can read my post again to see my explanation.

I didn't say they end.

Having the operation be to push a button, and having this button turn a lamp on and off, is simply a way to make this inherent contradiction even clearer. — Michael

Again, the contradiction comes from the conjunction of the premises. It is not a given that it is a contradiction in and of itself that infinitely tasks are executed in finite time. It's quite unintuitive that infinitely many tasks an be executed in finite time, but to show that doesn't entail that it is a contradiction in and of itself.

what is the reasoning behind the claim that if some wizard steps in at 12:00 to magically turn the lamp on — Michael

I never claimed any such thing. It's a straw man, even if unintentional.

What I said is that if we drop the premise that the lamp is only turned on by the button, then we don't get the contradiction. The point of that is that we are not logically obliged to reject only one certain premise. Note that I am not committing the fallacy of not addressing Thomson's premises. Rather, I am pointing out that rejecting one of the premises to avoid contradiction must then allow rejecting any other premise to avoid contradiction.

When I saw that you had put care into your numbered arguments, I surmised that you were interested in pursuing rigor. So in that regard, I'm examining all your reasoning.

Again (new numbering):

(1) Since none of the premises are logically true, but they yield a contradiction, we may reject any one of them to avoid contradiction. We are not logically bound to reject the one that happens to be least intuitive.

(2) Infinitely divisibility of time does not entail executability of denumerably many tasks in finite time, even though, executability of denumerably many tasks in finite time entails infinite divisibility of time.

(3) We don't have a satisfactory definition of 'metaphysical possibility' here.

(4) The argument is more complicated than appears with only a cursory look, since it involves the modality of 'possible'.

(5) If I'm not mistaken, Thomson does not conclude that time is not infinitely divisible. So, heuristically, we may wonder why that is if your conclusion actually follows as ineluctably as you claim.

(6) I wonder why you don't note my point about continuousness and density, which I mentioned to help sharpen your argument.

Symbols need explanatory words to go with them. — fishfry

(1) The symbols I used are common. The formulas I gave are not complicated. If one knows merely basic symbolic logical notation, then one can read right from my formulas into English. For example:

AxEy yex

reads as

For all x, there exists a y such that y is a member of x.

For example:

(P(x) & x=y) -> P(y)

reads as

If P holds for x, and x equals y, then P holds for y.

And now that I've provided a one-stop list of the most common symbols, it's even easier.

(2) I did give lots of explanations in certain contexts.

(3) You complain about the length of my posts, but also say I should give more explanation. You can't have it both ways. And you're hypocritical since your own posts are often long, and often enough have not merely a few symbols.

So I don't understand why fishfry would deny the plain record of the postings here.

Not only did I indeed state the axioms of identity theory several times, I stated them directly to fishfry, and he even quoted my statement of them.

Later he denied that I stated them, even though I had stated them not very many posts prior.

What explains fishfry's bizarreness?

From a couple of years ago, not involving fishfry. Probably there are others over the last couple of years. For simplicity, I didn't mention the quantification (in the meta-theory) over formulas:

x=x ... reflexivity
with
(x=y & Px) -> Py ... indiscernibility of identicals (aka substitutivity)

is a complete axiomatization of identity theory — TonesInDeepFreeze

From this thread, not addressed specifically to fishfry:

Identity theory is first order logic plus:

Axiom: Ax x=x

Axiom schema:
For all formulas P,
Axy((x=y & P(x)) -> P(y))

Semantics:

For every model M, for all terms T and S,
T = S
is true if and only if M assigns T and S to the same member of the universe. — TonesInDeepFreeze

The first mention that extensionality provides only a sufficient condition (which fishfry now regards as an epiphany for him), though not addressed specifically to fishfry, but there were at least a few others addressed specifically to him:

With identity theory, '=' is primitive and not defined, and the axiom of extensionality merely provides a sufficient basis for equality that is not in identity theory. Without identity theory, for a definition of '=' we need not just the axiom of extensionality but also the 'xez <-> yez' clause. — TonesInDeepFreeze

In this thread, not addressed specifically to fishfry.

Eventually, mathematical logic provided a formal first order identity theory:

Axiom. The law of identity.

Axiom schema. The indiscernibility of identicals. — TonesInDeepFreeze

In this thread. Comments on the subject, not addressed specifically to fishfry:

https://thephilosophyforum.com/discussion/comment/911857

In this thread, addressed specifically to fishfry:

identity theory (first order) is axiomatized:

Axiom:

Ax x = x (law of identity)

Axiom schema (I'm leaving out some technical details):

For any formula P(x):

Axy((P(x) & x = y) -> P(y)) (Leibniz's indiscernibility of identicals) — TonesInDeepFreeze

In this thread, addressed specifically to fishfry.

As I said much earlier in this thread, it is the first order theory axiomatized by:

Axiom:

Ax x = x (law of identity)

Axiom schema (I'm leaving out some technical details):

For all formulas P(x):

Axy((P(x) & x = y) -> P(y)) (indiscernibility of identicals) — TonesInDeepFreeze

In this thread, fishfry himself quoting me:

The identity relation on a universe U is {<x x> | x e U}. Put informally, it's {<x y> | x is y}, which is {<x y> | x is identical with y}.

Identity theory (first order) is axiomatized:

Axiom:

Ax x = x (law of identity)

Axiom schema (I'm leaving out some technical details):

For any formula P(x):

Axy((P(x) & x = y) -> P(y)) (Leibniz's indiscernibility of identicals)
— TonesInDeepFreeze — fishfry

In this thread, addressed specifically to fishfry:

Start with these identity axioms:

Ax x=x (a thing is identical with itself)

and (roughly stated) for all formulas P(x):

Axy((P(x) & x=y) -> P(y) (if x is y, then whatever holds of x then holds of y, i.e. "the indiscernibility of identicals") — TonesInDeepFreeze

In this thread, addressed specifically to fishfry:

So identity theory has axioms so that we can make inferences with '='.

The axioms are:

Ax x=x ... the law of identity

And the axiom schema (I'm leaving out technical details):

For all formulas P:

Axy((P(x) & x= y) -> P(y)) ... the indiscernibly of identicals — TonesInDeepFreeze

In this thread, fishfry again quotes me stating the axioms:

And '=' has a fixed interpretation (which is semantical, not part of the axioms) that '=' stands for identity.

So identity theory has axioms so that we can make inferences with '='.

The axioms are:

Ax x=x ... the law of identity

And the axiom schema (I'm leaving out technical details):

For all formulas P:

Axy((P(x) & x= y) -> P(y)) ... the indiscernibly of identicals

— TonesInDeepFreeze — fishfry

Then, in this thread, fishfry denies the plain record of the posting, denying that I had never neve said what I mean by identity theory, even though he had not very long before quoted me stating the axioms:

I stated explicitly several times that that is what I mean by 'identity theory'.
— TonesInDeepFreeze

You never said that LOL! — fishfry

Not quite my point, but thanks. — fishfry

At least, if you are ever interested in a formula, but you don't know the use of the symbol, there you have it. Of course, if you're not interested in formulas, though they are the most exact and often the most concise communication, then I can't help that.

I stated the axioms of identity theory in multiple posts. Not funny, but true.
— TonesInDeepFreeze

Not true, but funny. — fishfry

Not not true, but not funny.

Say we accept that Thomson's lamp entails a contradiction; the lamp can neither be on nor off at 12:00.

I take this as proof that having pushed a button an infinite number of times is metaphysically impossible. — Michael

(1) If a set of premises G entails a contradiction, and for any member P of G we have that G\{P} does not entail a contradiction, then we are logically free to reject any member of G. None of the premises are logically true (except "Either On or Off and not both" as conclusion from a definition "'On' means 'not Off'"), so we can reject any of them. For example, we could reject "The lamp does not change from Off to On, or from On to Off, except by pushing the button."

(2) We still don't have a satisfactory definition here of 'metaphysically impossible'.

(3) For what it's worth, if I'm not mistaken, Thomson does not conclude the time is not infinitely divisible, but rather he weighs in against the notion of super-tasks.

As a coda to Thomson's argument, you rely on the premise "If time is infinitely divisible then the super-task is possible.' [not a quote of yours]

But we may reject that premise.

It seems to me that "the lamp super-task is executed" entails "time is infinitely divisible". But the converse - "time is infinitely divisible" entails "the lamp super-task is executed" - at least requires an argument.

And, it seems to me, that analysis is even more difficult because it involves modalities. First, we have to distinguish between "time is infinitely divisible" and "it is possible that time is infinitely divisible". Second, it's not really, "the lamp super-task is executed" but "it is possible that the lamp super-task is executed". The argument needs to checked whether the modal inferences are correct.

You seem to take this as proof that having pushed a button an infinite number of times is metaphysically impossible only if the premises are true. — Michael

No, I'm not claiming that.

let's say that our button is broken; pushing it never turns the lamp on. In such a scenario we can unproblematically say that the lamp is off at 12:00. But this does not then entail that it is possible to have pushed the button an infinite number of times. — Michael

Of course.

we can imagine a lamp with two buttons; one that turns it on and off and one that does nothing. Whenever it's possible to push one it's also possible to push the other, and so if it's possible to have pushed the broken button an infinite number of times then it's possible to have pushed the working button an infinite number of times. Given that the latter is false, the former is also false. — Michael

Whatever the validity of that, I don't see the point of it here.

Having pushed a button an infinite number of times is an inherent contradiction — Michael

It's not a contradiction in and of itself. Rather, it is inconsistent with the other premises (especially that the infinite number of executions occurs in finite time).

Having the button turn a lamp on and off, and the lamp therefore being neither on nor off at the end, is only a way to demonstrate the contradiction; it isn't the reason for the contradiction. — Michael

You would need to tell me the difference between a demonstration of a contradiction and the reason for a contradiction. For example, if someone asks "What is the reason that the unrestricted comprehension schema with the separation schema yields a contradiction?" then my best response would be to show a demonstration.

is also why Benacerraf's response to the problem misses the mark. — Michael

As far as I can tell, it's off-base because it doesn't address the premises of the lamp.

And related arguments against Thomson are that it is not problematic that the premises don't provide for concluding whether the lamp is Off or On at 12:00. But that misses the point that it is not that it is problematic that the lamp's state is undetermined, but rather that Thomson's argument shows that the lamp is neither Off nor On. Not that it is undetermined what the state is, but rather that is determined that the state is neither Off nor On.

/

I don't know enough about coding to have a comment on your pseudocode.

/

A while back you gave the argument in quite succinct form:

"P1. The lamp is turned on and off only by pushing the button
P2. If the lamp is off and the button is pushed then the lamp is turned on
P3. If the lamp is on and the button is pushed then the lamp is turned off
P4. The lamp is off at t0
P5. The button is pushed at successively halved intervals of time between t0 and t1
P6. The lamp is either on or off at t1"

But you add to that argument two things:

(a) We must reject P5.

But each of the premises is required for the contradiction, so we can reject any one of them rather than P5. Granted, P5 does stand out as the candidate we would intuitively reject, but it is not logically required that it is the one we reject. For example, famously:

(U) ExAy yex

(S) ExAy(yex <-> ~yey)

yield a contradiction.

But to dispel the contradiction it is not logically required that we reject (U) to keep (S) when we could reject (S) to keep (U). We may have reasons for preferring that we reject (U) rather than (S), but that is not a demonstration that (U) is logically impossible or even that it is false.

(b) If time is infinitely divisible, then the super-task may be executed.

But that is not logically true either. It may be the case that time is infinitely divisible but still the super-task cannot be executed. Moreover, the modality "may be" slips in there, so the argument requires that it is made clear that the modal inference is permitted.

The argument shows that the premises entail a contradiction, so at least one of the premises must be rejected.
— TonesInDeepFreeze
Which one do you think should be rejected? — Ludwig V

I don't proffer an opinion on that. But I can see that presumably the most likely candidate is "At 11:00 the button is pushed to turn the lamp On, at 11:30 Off, at 11:45 On, and alternating in that way ad infinitum." At least intuitively it is the ripest and lowest hanging fruit. Or put pejoratively, at least intuitively it is the sore thumb.

But logically we may reject any of them. None of them are logical truths (though, "At all times, the lamp is either Off or On and not both" would be logically true as a conclusion from defining 'On' as 'not Off'.

@fishfry

Nearly all of these text symbols are quite common:


~ ... it is not the case that

-> ... implies

<-> ... if and only if

& ... and

v ... or

A ... for all

E ... there exists a/an

E! ... there exists a unique

Axy ... for all x and for all y [for example]

if P(x) is a formula, then, in context, P(y) is the result of replacing all free occurrences of x with y [for example]

= ... equals

< ... is less than

<= ... is less than or equal to

> ... is greater than

>= ... is greater than or equal to

+ ... plus

- ... minus

* ... times

/ ... x divided by y

^ ... raised to the power of

! ... factorial

e ... is an element of

0 ... the empty set (also, zero)

w ... the set of natural numbers [read as 'omega']

{x | P} ... the set of x such that P [for example]

{x y z} ... the set whose members are x, y and z [for example]

<x y> ... the ordered pair such that x is the first coordinate and y is the second coordinate [for example]

(x y) ... the open interval between x and y [for example]

(x y] ... the interval between x and y, including y, not including x [for example]

[x y) ... the interval between x and y, not including x, not including y [for example]

[x y] ... the closed interval between x and y [for example]

U ... the union of

P ... the power set of

/\ ... the intersection of

x u y ... the union of x and y [for example]

x n y ... the intersection of x and y [for example]

x\y ... x without the members of y [for example]

|- ... proves

|/- ... does not prove

|= ... entails

|/= ... does not entail

PA ... first order Peano arithmetic

S ... the successor of

# ... the Godel number of

Z ... Zermelo set theory

ZC ... Zermelo set theory with the axiom of choice

ZF ... Zermelo-Fraenkel set theory

ZFC ... Zermelo Frankel set theory with the axiom choice

Z\I ... Zermelo set theory without the axiom of infinity

(Z\I)+~I ... Zermelo set theory with the axiom of infinity replaced by the negation of the axiom of infinity

Z\R ... Zermelo set theory without the axiom of regularity

ZF\R ... Zermelo-Fraenkel set theory without the axiom of regularity

ZFC\R ... Zermelo Frankel set theory with the axiom choice without the axiom of regularity

p ... possibly

n ... necessarily

when needed for clarity, ' ' indicates an expression not its referent ('Sue' is a name, Sue a person)

I stated explicitly several times that that is what I mean by 'identity theory'.
— TonesInDeepFreeze

You never said that LOL! — fishfry

I stated the axioms of identity theory in multiple posts. Not funny, but true.

I thought the = of set theory is the = from the underlying logic. But now you say it's not. — fishfry

I did not say that it's not.

I'll say again:

First order logic with identity provides:

(1) law of identity (axiom)

(2) indiscernibility of identicals (axiom schema)

(3) interpretation of '=' as standing for the identity relation (semantics)

Set theory takes (1) - (3) and adds:

(4) extensionality (axiom)

Got it.

I'm not rejecting anything.

I'm saying:

(1) What is the proof of C2 and C3 from the premises? (Though we don't need it, if we adopt my rP6.)

(2) Instead of rejecting infinite divisibility, we may reject other premises instead.