An analysis of truth and metaphysics

Michael

q ≔ the proposition that p
T(q) ≔ q is true

1. ∀p: T(q) ↔ p
2. ∀p: T(q) → ∃x(x=q)
3. ∀p: p → ∃x(x=q)
*4. ∀p: ¬∃x(x=q) → ¬p

4 appears to be consistent with anti-realist metaphysics; if the proposition "it is raining" does not exist then it is not raining.

To avoid this situation a further argument can be made:

5. ∀p: ¬T(q) ↔ ¬p
6. ∀p: ¬T(q) → ∃x(x=q)
7. ∀p: ¬p → ∃x(x=q)
8. ∀p: ∃x(x=q)

8 appears to be consistent with both realist and anti-realist metaphysics; for every "way the world is" there exists an associated truth-bearer. If truth-bearers are dependent on thought or speech then this would be anti-realist metaphysics and if truth-bearers are independent of thought and speech then this would be realist metaphysics.

However, it may be that one wishes to avoid anti-realist metaphysics without requiring that truth-bearers be independent of thought and speech, in which case we can reformulate 1:

9. ∀q: T(q) ↔ p

From this we derive only the truism that for every truth-bearer there exists that truth-bearer:

10: ∀q: ∃x(x=q)

In asserting 9 and rejecting 1 it then follows that for at least one "way the world is" there does not exist an associated truth-bearer (either because that "way the world is" is ineffable or because it just isn't spoken about):

*11: ∃p: p → ¬∃x(x=q)

This then allows for both realist metaphysics and anti-realist (or should we say, semi-realist) truth, i.e. that truth depends on thought or speech (and the "way the world is") but that "the way the world is" does not.

* 4 and 11 require free logic.

Banno

I'm tired, so I might not be seeing it right, but (2) looks troubling. Don't you need T("p") independent of the conditional in order to introduce the existential quantifier?

So
T("p")
_______
∃"p"

might be valid.

∀p: T("p") → ∃"p" (from 1, by existential introduction) — Michael

But in (1) you have ∀p: T("p") ↔ p, not T("p").

Also, ∃"p" is ill-formed, unless you are using Free logic.

jgill

T(x) ≔ x is true (definition) — Michael

Looks like you've defined a fixed point of some function. But I doubt that is what you mean?

Michael

↪Banno

According to existential introduction:

Q(a) → ∃xQ(x) (if John is bald then there exists at least one thing which is bald)

And surely:

∃xQ(x) → ∃x (if there exists at least one thing which is bald then there exists at least one thing)

And so:

Q(a) → ∃a (if John is bald then John exists)

Maybe my particular symbols aren't being used quite right, but surely the logic works? In ordinary language (and providing the complete account of existential introduction) it would be:

1. proposition "p" is true if and only if p
2. if proposition "p" is true then there exists some proposition which is true (existential introduction)
3. if proposition "p" is true then there exists some proposition
4. if proposition "p" is true then proposition "p" exists

4 must follow otherwise we would have the situation where, from 2, the truth of proposition "p" would entail that there exists some other (true) proposition. Or in the case of John being bald, that John being bald entails that there exists some other (bald) thing. Which seems absurd.

Michael

Looks like you've defined a fixed point of some function. But I doubt that is what you mean? — jgill

I'm just explaining what is meant by the predicate T. I could have written the argument as:

1. ∀p: "p" is true ↔ p

But that would require more typing.

Banno

↪Michael

From the other thread:

2. ∀p: T("p") → ∃"p" (from 1, by existential introduction)
— Michael

Again, this looks problematic. "p" is an individual variable, so ∃"p" is like ∃(a). You would have to move to a free logic and use ∃!"p"; but that something exists cannot be the conclusion of an argument in free logic. You have to assume that p is true and that there is a sentence "p".

This is now spread over two threads. https://thephilosophyforum.com/discussion/13170/an-analysis-of-truth-and-metaphysics — Banno

Michael

↪Banno

I'm pretty sure my comment above addresses that.

If John is bald then John exists
If the proposition "it is raining" is written in English then the proposition "it is raining" exists
If the proposition "it is raining" is true then the proposition "it is raining" exists

Banno

↪Michael

But. ∃"p" is ill-formed. "p" is an individual. So it's lie writing ∃(a).

If John is bald then John exists — Michael

It's "If John is bald then there is something that is bald".

Michael

It's "If John is bald then there is something that is bald". — Banno

Is there a difference between "if John is bald then there is something that is bald" and "if John is bald then there exists something that is bald"?

Banno

↪Michael

But you have "if John is bald then John exists". That's invalid ill-formed. ∃(a) is not a formula in first order logic, unless you move to free logic. Then it would be ∃!(a)

∃!(a) = ∃(x)(a=x) (dfn)

Michael

But you have "if John is bald then John exists". That's invalid ill-formed. ∃(a) is not a formula in first order logic, unless you move to free logic. — Banno

How is it ill-formed? It makes perfect sense to me:

If John is bald then something exists which is bald
If John is bald then something exists

Are you saying that the second sentence doesn't make sense? Or is false?

Banno

↪Michael

I'm saying it's not allowed in the rules of classical logic.

Banno

God exists, Whatever thinks exists, Fiction: Free Logic might help.

Agent Smith

Do metaphysical claims entail observables? The problem of evil, god - metaphysics - denied because his omnibenvevolence, omniscience, and omnipotence is contradicted by observation (rampant evil).

Are metaphysical claims amenable only to a priori proofs?

What theory of truth are we using for metaphysical claims? If the correspondence theory of truth then, they can be and should be verified/falsified.

Michael

I'm saying it's not allowed in the rules of classical logic. — Banno

Then what logic am I using when I say that if John is bald then John exists? Or that if the cat is on the mat then the cat exists? Because they seem like logical inferences to me. It would be strange to say that the cat is on the mat but there isn't a cat.

If your problem is with my (mis-)use of formal symbols then you can consider the argument in natural language as I started with here.

But then let's look at the argument using the complete form of existential introduction:

1. ∀p: T("p") ↔ p (premise)
2. ∀p: T("p") → ∃xT(x)
3. ∀p: p → ∃xT(x)
4. ∀p: ¬∃xT(x) → ¬p
5. ∀p: T("¬p") ↔ ¬p (premise)
6. ∀p: T("¬p") → ∃xT(x)
7. ∀p: ¬p → ∃xT(x)
8. ∀p: ∃xT(x)

The main conclusions being 4 (if there are no true truth-bearers then nothing is the case) and 8 (there is at least one true truth-bearer).

And so there is still the issue that either a) truth-bearers are dependent on thought and speech and so if something is the case then something true is thought or spoken or b) truth-bearers are independent of thought and speech.

If needed we can reformulate our initial premise and derive the slightly different conclusion that for all truth-bearers there is at least one true truth-bearer:

9. ∀"p": T("p") ↔ p
...
10: ∀"p": ∃xT(x)

Still working on how to properly formulate 11. Perhaps something like:

11. ∃p: p → ¬∃"p"T("p") (there is at least one case where if that thing is the case then there is no true proposition that it is the case).

Agent Smith

I'm saying it's not allowed in the rules of classical logic. — Banno

:up: Some of us are running an old logic module; others seem to have been habituated to double-think, and still others are at ease contradicting themselves, not because they're inured to it, but for the simple reason that it makes complete sense to them.

Classical logic is 2.5k years old - time for an upgrade, oui? Paraconsistent logic comes to mind, but that's just tinkering around with the rules of logic and it, for some reason, hasn't caught on among philosophers. I wonder why?

Michael

↪Banno

This (and the comments by @Snakes Alive in the discussion you linked to) would suggest that ∃x(x = q) is valid in first-order logic, and doesn't require free logic? So I can do T("p") → ∃x(x = "p").

I've updated my original post accordingly.

Actually, looking at this, it does appear that steps 4 and 11 (x does not exist) of my argument depend on free logic:

Free logic allows such statements to be true despite the non-referring singular term. Indeed, it allows even statements of the form ∼∃x x=t (e.g., “the ether does not exist”) to be true, though in classical logic, which presumes that t refers to an object in the quantificational domain, they are self-contradictory.

I'm happy with this, as I would say that "the ether does not exist" is in fact true.

My one initial concern is with whether or not modus tollens applies to ∀p: p → ∃x(x=q). I'll do some digging.

Michael

OK, I've done some further research and in classical logic ¬∃x(x=q) isn't allowed and in free logic T(q) ⊢ ∃x(x=q) is false.

I'll strike out the free logic stuff for now as the rest is still interesting and will come back to them when I've learnt more.

On that note though, what logic would you say ordinary language uses? Because in classical logic you can't say "if the cat does not exist then the cat is not on the mat" and in free logic you can't say "if the cat is on the mat then the cat exists."

It seems to me that in ordinary language we can say both, and so an ordinary language interpretation of my argument still holds.

TonesInDeepFreeze

what logic am I using when I say that if John is bald then John exists? — Michael

"John exists" is not expressed in mere predicate logic. You need modal logic for it.

TonesInDeepFreeze

Classical logic is 2.5k years old — Agent Smith

Classical logic is about 140 years old.

Yet again you shoot your mouth off not knowing what you're talking about.

bongo fury

what logic am I using when I say that if John is bald then John exists?
— Michael

"John exists" is not expressed in mere predicate logic. You need modal logic for it. — TonesInDeepFreeze

Why not (with "J" for "is John" and "B" for "is Bald"),

∃x (J(x) ∧ B(x)) => ∃x (J(x))

?

TonesInDeepFreeze

↪bongo fury

Yes, of course, we can do that.

[You know all this; I'm writing it for benefit of those who don't:]

You can have the predicate 'is John', which is something different from just the name 'John'.

Simplest example from mathematics:

'0' is an operation symbol, defined

0 = x <-> Ay ~yex.

But we don't write:

Ex 0

And we don't write:

Ex x

They are not well formed. A quantifier is concatenated with a formula, and a mere term is not a formula.

Meanwhile, for any term T whatsoever, it's a logical theorem:

Ex x = T

which includes:

Ex x = x

To put it another way, there is not "existence predicate" in predicate logic. We go to modal logic for that.

But we could define a predicate symbol:

Mx <-> x = 0

And we can say:

ExMx

Michael

↪TonesInDeepFreeze

Yes, I think I clarified my position here .

TonesInDeepFreeze

↪Michael

I added that modal logic is the main arena for this.

Michael

I think the below should avoid the need for that.

T(q) ≔ q is true
P(q) ≔ q is a proposition

1. T(q) ↔ p
2. T(q) ⊢ ∃xT(x)
3. ∃xT(x) ⊨ ∃xP(x)
4. p → ∃xP(x)
5. ¬∃xP(x) → ¬p

Does this logic work?

If it is raining then some x is a proposition. If no x is a proposition then it is not raining. These could be seen to be problematic conclusions, as it suggests either Platonism (of propositions) or some form of antirealism.

TonesInDeepFreeze

↪Michael

There is one premise there:

Tq <-> p

Following that, I don't see a problem with the logic. But you use vacuous quantification with

ExTq

and

ExPq

So, though there is no mistake in the logic, I don't see any point in it.

Michael

↪TonesInDeepFreeze

I have since edited it to ∃xT(x) and ∃xP(x) as I believe that's the correct application of existential introduction?

And the point is the conclusions on lines 4 and 5. It's easier to understand in ordinary language:

1. "it is raining" is true iff it is raining
2. "it is raining" being true entails that some x is true
3. Some x being true entails that some x is a proposition
4. If it is raining then some x is a proposition
5. If no x is a proposition then it is not raining

As I said above, it seems to suggest either Platonism (of propositions) or some form of antirealism. That's the point of the argument.

TonesInDeepFreeze

↪Michael

My mistake: The logic is not correct. Line 3 (whether original or reviesd) is a non sequitur.

1. Tq <-> p ... premise
2. Tq -> ExTx ... EG
3. ExTx -> ExPx ... non sequitur (prob you have an unstated premise in mind)
4. p -> ExPx ... sentential logic, but relies on non sequitur in step 3
5. ~ExPx -> ~p ... sentential logic, but relies on non sequitur in step 3

Michael

The logic is not correct. Line 3 (whether original or reviesd) is a non sequitur. — TonesInDeepFreeze

Line 3 is ∃xT(x) ⊨ ∃xP(x).

That some x is true semantically entails that some x is a proposition, given that truth is predicated of (and only of) propositions.

Maybe it's simpler to just understand T(q) as "q is a true proposition". If some x is a true proposition then some x is a proposition.

TonesInDeepFreeze

That some x is true semantically entails that some x is a proposition, given that truth is predicated of (and only of) propositions. — Michael

Truth is semantic. My point is that you are missing the premise:

Ax(Tx -> Px)

Michael

↪TonesInDeepFreeze

I did say that "maybe it's simpler to just understand T(q) as 'q is a true proposition'." So that gives us:

T(q) ≔ q is a true proposition
P(q) ≔ q is a proposition

1. T(q) ↔ p
2. T(q) ⊢ ∃xT(x)
3. ∃xT(x) ⊨ ∃xP(x)
4. p → ∃xP(x)
5. ¬∃xP(x) → ¬p

3 follows from the two definitions.

1. "it is raining" is a true proposition iff it is raining
2. If "it is raining" is a true proposition then some x is a true proposition
3. If some x is a true proposition then some x is a proposition
4. If it is raining then some x is a proposition
5. If no x is a proposition then it is not raining

An analysis of truth and metaphysics

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