Does the proposition – (Not (Not A)) → A imply (A ʌ -A)?

[Ed3]

The Excluded Middle is: (Not (Not A)) → A. If this is not the case then is it true that the following proposition is also true: (A ʌ -A)?

I am working on a post which has Brouwer /Constructivism as major subjects.

Thanks Ed

[Carleas]

As I understand your question, it's \( \neg ( \, \neg (\neg A) \rightarrow A) \overset{?}{\rightarrow} (A \land \neg A) \)

i.e., if it's not the case that [not not A implies A], then [A and not A]. Is that right? I will need to think more about the answer to the question, I just wanted to make sure I have it right before I get going.

[James S Saint]

The Excluded Middle is: (Not (Not A)) → A. If this is not the case then is it true that the following proposition is also true: (A ʌ -A)?

I am working on a post which has Brouwer /Constructivism as major subjects.

Thanks Ed

The symbol "ʌ" means "and also".

So the question is: If a middle is allowed between A and not-A, then can one have both A and also not-A at the same time?

And the answer is that it is anyone's guess because if logic doesn't apply, then there are no consistent rules (because consistency in language is what logic is).

The new possible cases would be; that both A and also not-A exited, that neither A nor not-A existed, or that neither A nor not-A applied to the situation at all (such as the presence of an nonsensical statement; e.g. "this statement is false"). Take your pick

[Ed3]

Hi Carleas,

Yes. You understand the question properly, and I am eager to hear your response.

Thanks Ed

[Ed3]

Hi James,

I understand the symbol ˄ to mean the word "and" however your interpretation is perfectly suitable in my opinion. Additionally I have some of the same intuitive feelings that you have.

However, A ˄ -A can be used to prove anything, whereas simply renouncing the excluded middle, just makes mathematics more difficult rather than pointless.

In any case I would like to see a rigorous evaluation.

Thanks Ed

[Ed3]

Hi Carleas,

I'm sorry!!! I did screw up the representation of my question.

I should have written – (NOT (NOT A) → A) imply (A Ʌ -A)

It appears that you have written the correct form.

Apologies - Ed

[James S Saint]

Hi Carleas,

I'm sorry!!! I did screw up the representation of my question.

I should have written – (NOT (NOT A) → A) imply (A Ʌ -A)

It appears that you have written the correct form.

Apologies - Ed

Are you sure about that?

Your OP said what Carleas said, but different than the Title. The title needs another "Not" in front.

Your new statement doesn't seem to say what you intended.

If ~~A → A (which is does) then A and also ~A ???

You need "If ~~~A → A ..."

[James S Saint]

I'm sure that you need something different because this seems too simple. The constructionist requires that the conclusion object be constructed. My question is how much of the rest are you allowed to assume, because it takes almost nothing to construct the (A ^ ~A) from the "~~~A → A".

~~A → A, LEM

| (~(~(~A)) → A) → (~A ^ A), hypothesis
|| (~~~A → A) → (~A ^ A)
| (~A → A) → (~A ^ A)
|
|| ~A, given
|| A ^ A, given
|| ~A → A, given hypothesis condition
| ~A ^ A, concluded hypothesis equivalent condition
∴
(~(~(~A)) → A) → (~A ^ A)

[Ed3]

Hi James,

I can not keep up with the pace at which you post. This is a response to your second to last post.

I use the “–” sign and “NOT” interchangeably. (I wrote it the way that I did because I wanted to emphasize that I was talking about the Law of the Excluded Middle).

What I really screwed up, in the title, was the use of the parenthesis. The final “)” belongs after the arrow and not after the first “)”.

What is amazing is that Carleas caught my mistake from the content of the article. Additionally he could have read the title and simply thought “the old man is nuts” and never read the article.

Thanks Ed

P.S. I hope the proposition is false (and I think that it is) because it would ruin not only my upcoming post but almost everything else.

[James S Saint]

Well sorry Ed, now you have lost me.

Your OP said this:

As I understand your question, it's \( \neg ( \, \neg (\neg A)) \rightarrow A) \overset{?}{\rightarrow} (A \land \neg A) \)

i.e., if it's not the case that [not not A implies A], then [A and not A].

But now you seem to be saying something else.

The final “)” belongs after the arrow and not after the first “)”.

That edit would merely be;
\((\neg (\neg A) \rightarrow A) \overset{?}{\rightarrow} (A \land \neg A) \)
\((A \rightarrow A) \overset{?}{\rightarrow} (A \land \neg A) \)

Obviously that isn't true nor relevant. And I don't think that is what you wanted either. So n/m.

[Carleas]

I actually got the parentheses wrong too, though my verbal expression seems to have gotten it right. I've fixed it above, and I think we could take out some more parentheses to make it clearer (assuming negation precedes implication in the order of operations for logical operators).
\( \neg ( \, \neg \neg A \rightarrow A) \overset{?}{\rightarrow} (A \land \neg A) \)

I think James is right that we don't know what follows from this, although I don't think it's in principle undecidable. Rather, the statement is ambiguous: the \(\neg\) operator is a binary-logic operator. If \( \neg ( \, \neg \neg A \rightarrow A) \), it seems there must be a third truth-value besides true or false, and it isn't clear what the truth table for \( \neg \) looks like with respect to those three values.

On the other hand, if \( \neg ( \, \neg \neg A \rightarrow A) \), it doesn't seem that \(A \land \neg A \) is necessarily problematic. As I learned it, the proof that any statement follows from a contradiction relies on the law of the excluded middle. If that law doesn't apply, and \(\neg\) doesn't apply the way we think it does, we might be saved.

For example, if we have a three valued logic with values True, False, and Unknown, it might be that while A can't be both True and False, it can be True and Unknown. Then if we define \(\neg\) such that \(\neg True \rightarrow (Unknown \lor False)\), there is no contradiction for \(A \land \neg A \): \(A\) can be both True and Unknown.

So, I think the original question is just ambiguous. If the truth table for \(\neg\) is violated, we need a new truth table, but it's possible to make one that makes your statement consistent.

[Ed3]

Hi James,

I meant to ask the following:

Does the negation of the Law of the Excluded imply (A and Not A)?

If it does then it would be a giant disaster for Constructivism. (You can go to Wiki and search for the Principle of Explosion to see a proof that for all propositions, Q, (A and not A) imply Q). But I think you already knew that.

Thanks Ed

[Ed3]

Hi Carleas,

I think that you are right to claim that we would be required to use a multi valued logic if we deny the Law of the Excluded Middle. In practice I, personally, think in terms of the three values that you laid out. I.e. True, False and Unknown.

However Gödel had a proof that in fact the logic has to be infinitely valued.

I am not sure how to proceed here. I know that in classical math and binary logic we need to have the Law of the Excluded Middle. However in constructive math (which has a number of practical uses) and multi valued logic we can deny the Law of the Excluded Middle.

I just don’t want the denial of the Excluded Middle to imply (A and Not A).

If a proof can be constructed, preferably not using the Law of the Excluded Middle (basically I don’t want a proof by contradiction) I am still interested.

A simple counter example might be OK. I know that some Constructivists will still allow Not (A and Not A)

Thanks Ed

[James S Saint]

Hi James,

I meant to ask the following:

Does the negation of the Law of the Excluded imply (A and Not A)?

If it does then it would be a giant disaster for Constructivism. (You can go to Wiki and search for the Principle of Explosion to see a proof that for all propositions, Q, (A and not A) imply Q). But I think you already knew that.

Thanks Ed

Frankly Ed, I think it is a nonsense question. It is like asking;
If the hypotenuse of a right triangle is not equal to the square root of the sum of the squares, does 5 - 3 = 4?

I don't see how the lack of a law (or an existence) can imply the existence of anything .. other than irrationality. So the "rational" answer to the question would be "No. The lack of the LEM does not imply .. anything".

But the way to find out in constructive logic/math is to spell out the construction of A → ~(~A) and see if it necessarily requires the LEM without which (A ^ ~A) is necessary. But what is the acceptable constructivist proof that A → ~(~A)?

IF (A → ~(~A)) → ~(A ^ ~A) then

By contradiction:

IF ~(A ^ ~A) then (A ^ ~A) is invalid. -- the LEM
But
1.IF (A ^ ~A) then ((A ^ ~A) ^ ~(A ^ ~A)), subst of "A" with "(A ^ ~A)"
2..Assuming (A ^ ~A),
3..Then ~(A ^ ~A), via 1 & 2, the construction of the LEM.
4..Thus (A ^ ~A) is not true, via 3, a direct self-contradiction of the assumption.

Or Perhaps easier to read:
IF the lack of the LEM is the case then the LEM exists, in which case the lack of the LEM is not the case.

That should be a constructivism proof of the LEM. That proof did not require the LEM as a principle in use. But the lack of the LEM did not demand (A ^ ~A), quite the opposite. The lack of the LEM demands the LEM -- ~(A ^ ~A).

(A ^ ~A) → ~(A ^ ~A) → LEM

So again the answer is "No."

[Carleas]

Ed, I take it you have a specific project on which you're working that requires denying the LEM, but where \(A \land \neg A \) would break something. But I wonder if it would really be so bad if \(A \land \neg A \). Without the LEM, I don't think the Principle of Explosion applies.

The proof for the PoE as I learned it goes like this:
1. \(A \land \neg A\)
2. \(A \)
3. \(A \lor B \)
4. \( \neg A\)
5. \(\therefore B\)

But that relies on the LEM: the 'or' in line 3 is eliminated by contradiction to prove B. If \( A \land \neg A \), we can't use \( \neg A\) that way. So even if \(\neg ( \neg \neg A \rightarrow A) \rightarrow (A \land \neg A) \), none of the usually nasty consequences seem to follow. Especially in this case, where we've explicitly denied that the negation of ~A implies A.

So, I'm not sure what follows from \(\neg ( \neg \neg A \rightarrow A) \) (because it seems to break the very logic required to answer that question), but at the same time if something seemingly nasty does follow, it probably isn't as nasty as it seems (because the logic it needs to have any consequence is broken).

Or is there a way to prove the PoE without using contradiction?

[Ed3]

Hi James,

I will respond after I complete my upcoming post, because some of what you commented on is best dealt with in that post.

Ed

[Ed3]

Hi Carleas,

That was freaking brilliant.

Thanks Ed

[Ecmandu]

This is really a matter of "what is beyond?"

Then getting into the additional muck of what is beyond beyond?

So let's say there's a tree.

What is beyond that tree? If nothing, then nothing exists besides the tree (lack of tree), and there is no beyond, there is no possible negation, from which to discern tree from not tree, such as a sidewalk, which is not tree.

So there is both a tree and not a tree, not tree, is "beyond"

You always need a beyond to see something, and a beyond is always -something

So what is beyond beyond?

How can we see beyond, which we know is there, without something beyond that?

You have a couple options.

You see the beyond by virtue of the non beyond: I.e the tree, or, and this is correct, you see the beyond by virtue of the tree, the non tree beyond, and a third variable which is beyond both of them...

Think of it this way...

We see the tree, we see the stars, and beyond those stars, there is another beyond that allows us to see the first beyond.

Not only does this imply a and not a, it implies another not a that forces the first not a to be a middle.

[Ecmandu]

Another way to look at this is the river situation.

You can't stand in the same river twice.
Then how can we ever get to a river? How do we name them?

It is both a river and not a river.

The same is true of trees etc...

It is both a tree and not a tree.

You can't even look at the same "a" twice, it is both, it's identity and not its identity.

The middle of the a and not a is a balance of acuity...

Not too far or too close..

It is this middle, which allows us to observe identity...

Far from there being a law of no middle, the middle is required for identity.

[Jakob]

Since "A" >< "A", ""A"="A"" always begs the question, and dissecting its implications will result in ambiguity of terms.
This is how such flat logic works - actual statements are discovered through inconsistencies in the abstract.
The abstraction that is the law of identity doesn't apply to the real world, hence the open ended nature of formal logic.

[Meno_]

But the law of the excluded middle has very important philosophical implication, therefore, so does the law of identity.

Existentialism has direct connections to the law of the excluded middle. The philosophy in that sense can derive the logic which underlies the existential argument. Historical inevitability reduces to that logic.

[Jakob]

But the law of the excluded middle has very important philosophical implication, therefore, so does the law of identity.

I opposed both, precisely because they have been the ruin of thought.
Neither applies to reality. Both are means to make something different than reality.

Existentialism has direct connections to the law of the excluded middle. The philosophy in that sense can derive the logic which underlies the existential argument. Historical inevitability reduces to that logic.

In real existence, all statements of true fact are subjective statements of states. And such statements will always contradict other occasions where they have been made.
Phenomena have not yet been allowed into logic. Aristotle was really, a really dry sheet of paper.