If (a→b)∧(a→¬b) is False, ¬A is False, so A can be True or not¹.
— Lionino
But (a→b)∧(a→¬b) being False simply means that A does not imply a contradiction, it should not mean A is True automatically.
— Lionino

Isn't this a fairly big problem given that (¬¬A↔A)? — Leontiskos

A formal problem, philosophical problem, or problem in not adhering to everyday discourse.

It's not a formal problem. I don't know what philosophical problem you suggest. And I've discussed symbolic logic and everyday discourse.

Isn't this a fairly big problem given that (¬¬A↔A)? I take it that this is the same thing I have pointed out coming out in a different way? Namely the quasi-equivocation on falsity? — Leontiskos

Compare:

• A→(B∧¬B)
• ∴ ¬A

With:

• A→(B∧¬B)
• ¬(B∧¬B)
• ∴ ¬A

Whether or not we affirm the negation of the consequent, the antecedent still ends up being false. In the first case the consequent is pre-false, in the second case our explicit negation makes it false. In the first case it is treated as "falsity incarnate," whereas in the second it is treated as a proposition. This demonstrates the analogical equivocity with respect to falsity that I pointed out earlier, with help from . I think Lionino was somehow seeing this through the overdetermination of the biconditional with respect to ¬¬A, but it's slippery to grab hold of.

Note that we could also do other things, such as treat the second premise as truth incarnate, but this is harder to see:

• A→(B∧¬B)
• ¬(B∧¬B), but now conceived as "true"
• ∴ ¬A does not follow

...that is, if we conceive of the consequent as a proposition and the second premise as truth incarnate, then ¬A does not follow from the second premise (or from the consequent, absent a premise that negates the consequent qua proposition).

-

Around a third of folk hereabouts who have an interest in logical issues cannot do basic logic. — Banno

And about two thirds of folk hereabouts who are good at manipulating symbols "have no way of knowing when their logic machine is working and when it is not" ().

Aristotle always wins in the long run, as he could both use the machine and understand when it was misbehaving. :wink:

As in the concept/meaning of self as "that which is purple and square" vs. "that which is orange and circular" or any some such? And this in relation to "there both is and is not a self"? — javra

This makes no sense to me.

Again, one perspective being the mundane physical world of maya/illusion/magic-trick and the other being that of the ultimate, or else the only genuine, reality to be had: that of literal nondualistic being. — javra

The notion of a self from the perspective of "the mundane physical world of maya/illusion/magic-trick" is not the same as the notion of a self from the perspective of "literal non-dualistic being", so you are not talking about one thing.

That said, the self has no definitive definition, so introducing such a thing in the context of discussing whether anything could be the same in different contexts or thought under different perspectives seems incoherent from the get-go.

Consider the following substitutions which do not suffer from such ambiguities: Render (A implies B) as "the presence of water implies the presences of oxygen" and (A implies notB) as " the presence of water implies the absence of oxygen": do the two statements not contradict one another?

Compare:

((A→(B∧¬B))
∴ ¬A

With:

((A→(B∧¬B))
¬(B∧¬B)
∴ ¬A

With:

((A→(B∧¬B))
¬(B∧¬B)
¬(B∧¬B) = "True"
∴ A does not follow

This demonstrates the analogical equivocity — Leontiskos

What is the definition 'analogical equivocity'? Most helpfully it would be something like:

An thing has analogical equivoicity if and only if [fill in definiens here, and such that the definiens doesn't use terminology that hasn't itself been defined or common anyway]

What is the definition 'analogical equivocity'? — TonesInDeepFreeze

It is the kind of equivocity present in analogical predication, where a middle term is not univocal (i.e. it is strictly speaking equivocal) but there is an analogical relation between the different senses. This is the basis for the most straightforward kind of metabasis eis allo genos. The two different senses of falsity alluded to above are an example of two senses with an analogical relation. I was using the term earlier because I believe @Count Timothy von Icarus has an understanding of it.

I know you want a strict definition, but the wonderful irony is that someone like yourself who requires the sort of precision reminiscent of truth-functional logic can't understand analogical equivocity or the subtle problems that attend your argument for ¬A. As we have seen in the thread, those who require such "precision" tend to have a distaste for natural language itself.

I apparently misinterpreted your post.

Anyway, a implies B and (not a or b) are synonyms in classic symbolic logic. They have the same truth table.

By saying "'Lassie is a dog; is true", they are adopting Dx as a premise. So, of course...

Yes, that's right in there. I'm not sure how they are "slipping" it in. That's exactly the point they are making and trying to highlight. You seem to be missing the point of the example, yet you've hit on it right here.

I think you and may be supposing that the example is being called in to show something it is not being called in to show. The point being made might be more obvious in context, IIRC there is a section on how the old logic presupposes a sort of epistemic realism right before it, and to "common sense" form without realism often seems silly.

That's just a matter of defining the words. If 'dead' and 'living' are defined so that they are not mutually exclusive, then of course we don't make the inference. It's silly to claim that sentential logic is impugned with the example.

...when it comes to vamps it's deadly serious. :death: :death: :death:

Isn't this a fairly big problem given that (¬¬A↔A)? — Leontiskos

Yes, I am waiting for someone to clarify what is up with that.

I had written more under my own post but got myself confused also on whether the letter here is a statement or a variable.

¬(a→(b∧¬b)) → a — Leontiskos

Yes. It is what I say here.

But let's say (a→b)∧(a→¬b) is False, does that mean A is true? That is what the logical tables would say: — Lionino

But, as my second phrase in this post, I got confused as to whether ¬(a→(b∧¬b)) means a variable that is in contradiction with (a→(b∧¬b)) or simply that (a→(b∧¬b)) is False.

¬(a→(b∧¬b)) is only ever True (meaning A does not imply a contradiction) when A is True. But I think it might be we are putting the horse before the cart. It is not that ¬(a→(b∧¬b)) being True makes A True, but that, due to the definition of material implication, ¬(a→(b∧¬b)) can only be True if A is true. Likewise, what I say here in the first post:

and the way material implication works in classical logic is that, if the antecedent is false, the implication is always true — Lionino

The denial of the implication can only be True if the antecedent is True. If the antecedent is False, the denial of the implication is False.

So let me ask directly.

@TonesInDeepFreeze

Let a proposition P be A→(B∧¬B)
Whenever P is 1, A is 0.
In natural language, we might say: when it is true that A implies a contradiction, we know A is false.

Now a proposition Q: ¬(A→(B∧¬B))
Whenever Q is 1, A is 1.
Do you think it is correct to translate this as: when it is not true that A implies a contradiction, we know A is true?

Here is something I wrote when my internet was out, and before I was able to read this post of yours. I will read what you have written after posting this (and therefore there may be overlaps or redundancies):

———

If (a→b)∧(a→¬b) is False, ¬A is False, so A can be True or not¹. — Lionino

Note, too, how the LEM operates here.

On the LEM, (a→b)∧(a→¬b) must be either true or false, and therefore a→(b∧¬b) must be either true or false. If a→(b∧¬b) is true then ¬a must be true and (b∧¬b) must be indeterminate. It must be this way in order to avoid affirming the contradiction (and never mind the odd question of whether, in this case, saying “¬a must be true” is the same as saying “a must be false”).

But on the other side of the LEM, if a→(b∧¬b) is false then by material implication (a) must be true and (b∧¬b) must be false. This is different from (b∧¬b) being indeterminate. It is to affirm ¬(b∧¬b), which is the same as (¬b∨b) with an inclusive-or. The inclusive-or provides the second-order possibility of a true contradiction.

Another way this comes out is seen by comparing the truth table of (p→q) with the truth table of a→(b∧¬b). If you run the table “forward,” by using the value of (b) to compute the value of (b∧¬b), then (b∧¬b) is always necessarily false. But if you run the table “backwards,” by using the value of (p→q) to compute the values of (p) and (q), then in that singular case where the entirety of the conditional is false we get (a ^ ¬(b∧¬b)), which again results in the inclusive-or (¬b∨b).

In one way, in saying that the inclusive-or is true we are admitting the possibility that both disjuncts could be true, which would be impossible. Put differently, negating this conjunction results in an exclusive-or rather than the usual inclusive-or, just as affirming the special conditional of (a→(b∧¬b)) results in (a) automatically being false, even though there is obviously no rule that ((p→q)→¬p). Special cases and special exceptions are popping up.

What is happening, I think, is that the two different kinds of falsity are coming into play, for there are two different ways of calling (b∧¬b) false. We can mean that it is false inherently, even without a negation (and this is reflective of the first argument I gave <here>).* Or else by saying it is false we can mean that we logically negate it, viz. ¬(b∧¬b) (and this is reflective of the second argument I gave <here>). When are we supposed to reduce a contradiction to its functional truth value, and when are we supposed to let it remain in its proposition form? The mind can conceive of propositions symbol-strings like (b∧¬b) and ¬(b∧¬b) in these two different ways, the manner in which we conceive of them results in different logical outcomes, and symbolic logic provides no way of adjudicating how the propositions symbol-strings are to be conceived in any one situation. When we introduce contradictions into the logic—even in minor ways such as in the consequents of conditionals—the logic becomes underdetermined. Different people are legitimately interpreting the propositions symbol-strings in different ways.

* This is reflective of what I have been calling “falsity incarnate”

You seem to be missing the point of the example — Count Timothy von Icarus

I am exactly on point. The paper says that symbolic logic permits a certain inference. But symbolic logic does not permit that inference. The paper's argument is specious exactly as I showed.

That's just a matter of defining the words. If 'dead' and 'living' are defined so that they are not mutually exclusive, then of course we don't make the inference. It's silly to claim that sentential logic is impugned with the example.

...when it comes to vamps it's deadly serious. :death: :death: :death: — Count Timothy von Icarus

Death is serious. The example is silly, for the reason you quoted.

Hopefully I'll have time at some point to address the logic question. I want to look at some materials and think through my reply. But first I need to clean up more of your misrepresentations and mischaracterizations:

I know you want a strict definition, but the wonderful irony is that someone like yourself who requires the sort of precision reminiscent of truth-functional logic can't understand analogical equivocity or the subtle problems that attend your argument for ¬A. As we have seen in the thread, those who require such "precision" tend to have a distaste for natural language itself. — Leontiskos

A rigorous definition is best, for obvious reasons. And giving one is intellectual considerateness. But, again you put words in my mouth, thus a strawman. I said a certain form would be helpful. I did not say that I "require" it or any specific degree of rigor let alone "precision".

And I don't have "distaste" for language. You're just making that up. I love language and revel in its expressiveness, freedom, complexities, nuances and even its vagaries. You don't know jack when you presume to mischaracterize my sensibilities.

Your method is to characterize me in a certain way, to wedge in a prejudice that my arguments are tainted by a point of view, even though I don't have that point of view. You are arguing by characterization of an interlocutor, actually characterization.

You lied that I don't distinguish between formal language and natural language. I had said at least a few times that they don't agree in certain respects. Depending on the conversation, they don't agree on "if then".

You would do well to knock it off with trying to impugn me with lies about what I've said a sloppy mischaracterizations of my point of view and sensibilities.

No, you aren't, your post clearly demonstrated the point went right over your head. I even quoted you on the exact point where you say the same thing they are pointing out is counterintuitive. The fact that you think this is "slipping" something in or a "gotcha," demonstrates that you have missed the intent.

By saying "'Lassie is a dog; is true", they are adopting Dx as a premise. So, of course,

Ax(Dx -> Fx)
Fs
Dx
Therefore Dx

is valid.

This is counterintuitive. That's what the example is there to illustrate; it's right their in the text. Then you go off "challenging" them to show something they've never asserted.

The intent in that part of the paper was to make symbolic logic look like it says that this argument is valid:

Dogs have four legs, and Lassie has four legs, therefore Lassie is a dog.

But symbolic logic does not say that argument is valid.

I made a typo. It should be this:

Ax(Dx -> Fx)
Fs
Therefore, Ds

is not valid. And the paper is correct in saying it is not valid. But the paper is incorrect in saying that symbolic logic says it is valid.

In other words:

All dogs have four legs.
Lassie has four legs.
Therefore, Lassie is a dog.

is not valid. And the paper is correct is saying it is not valid. But the paper is incorrect in saying that symbolic logic says it is valid.

The paper says that symbolic logic says it is valid on the basis that "Lassie is a dog" is true.

That is not correct, since if we are using "Lassie is a dog", then it has to be a premise in the argument, so the argument that the paper puts on symbolic logic would actually be:

Ax(Dx -> Fx)
Fs
Ds
Therefore Ds

which is valid.

/

If you think the paper is correct in saying that symbolic logic regards this as valid

Ax(Dx -> Fx)
Fs
Therefore, Ds

In other words:

All dogs have four legs.
Lassie has four legs
Therefore, Lassie is a dog.

then you may prove that symbolic logic says it's valid. That is, show a derivation in symbolic logic that starts with the premises

Ax(Dx -> Fx)
Fs

and ends with the conclusion

Ds

Hint: You won't be able to do it. Morever, using the method of counter-example, symbolic logic proves it is not valid.

In other words:

All dogs have four legs.
Lassie is a dog.
Therefore Lassie has four legs.

is not valid.

You sure about that one?

No, that was also a typo. I corrected it all in the post as it stands now.

Deleted by me. Recomposing below.

Does it not?
¬(A → (B ∧ ¬B))→A is valid
¬(A → (B ∧ ¬B))↔A is valid

Let a proposition P be A→(B∧¬B)
Whenever P is 1, A is 0.
In natural language, we might say: when it is true that A implies a contradiction, we know A is false.

Now a proposition Q: ¬(A→(B∧¬B))
Whenever Q is 1, A is 1.
Do you think it is correct to translate this as: when it is not true that A implies a contradiction, we know A is true? — Lionino

I'm recomposing.

Sure, but that's not really what the example is there to assert, as is clear from the rest of the paragraph. They mentioned replacing the fact about dogs 2+2 = 4 in the next line. It's "if a statement is true, then that statement is implied by any statement whatever," which is straightforwardly counter intuitive.

(1) A -> (B & ~B) implies ~A

(2) ~(A→(B &~B)) implies A

"when it is not true that A implies a contradiction, we know A is true?"

No that is not a correct translation of (2). The translation universally quantifies over contradictions, which (2) does not.

So what is the proper translation of 2?

The example is incorrect, no matter what its purpose is.

"if a statement is true, then that statement is implied by any statement whatever." — Count Timothy von Icarus

I would not say that.

P may be true in some interpretations and not in others. If in a given interpretation, P is true, then per that interpretation, for any statement Q, Q->P is true.

~(A→(B &~B)) implies A

Translation:

It is not the case that if A then B & ~B
implies
A.

We can't say:

For all contradictions, if A does not imply the contradiction then A is true.

(1) True in what interpretations?

(2) "Winston Churchill was French" does not imply a contradiction. But that does not imply that "Winston Churchill was French" is true.

More precisely, "Winston Churchill was French" does not imply a contradiction. But that does not imply that "Winston Churchill was French" is true in all interpretations, but it does imply that "Winston Churchill was French" is true in at least one interpretation.

It is not the case that if A then B&~B implies A. — TonesInDeepFreeze

Is this supposed to be the translation of ~(A→(B &~B)) implies A?

I then edited it with better demarcations. Said another way:

"It is not the case that if A then both B and not-B" implies "A".

It is not the case that if A then both B and not-B — TonesInDeepFreeze

Isn't this the same as "if A then contradiction"?

I gave a direct translation, symbol by symbol to word by word.

The formula has a subformula that is a contradiction, but the formula doesn't itself say that it is a contradiction. The word 'contradiction' is not in the formula, and the formula doesn't quantify over contradictions. Again:

The formula does not say:

For all contradictions, if A does not imply the contradiction then A is true.

"Winston Churchill was French" does not imply a contradiction. But that does not imply that "Winston Churchill was French" is true.

More precisely, "Winston Churchill was French" does not imply a contradiction. But that does not imply that "Winston Churchill was French" is true in all interpretations, but it does imply that "Winston Churchill was French" is true in at least one interpretation.

With symbols:

If P does not imply a contradiction, then it is not implied that P is true in all interpretations, but it is implied that P is true in at least on interpretation.

No, your reading of it is incorrect because you seem to think it is saying:

All dogs have four legs
Lassie has four legs
Lassie is a dog

...is valid in symbolic logic. It doesn't say that. It says the exact opposite, that this is not valid.

What it turns to point out in the proceeding paragraph is different:

Since it is true that Lassie is a dog, "dogs have four legs" implies that Lassie is a dog. In fact, "dogs do not have four legs" also implies that Lassie is a dog! Even false statements, even statements that are self-contradictory, like "Grass is not grass," validly imply any true conclusion in symbolic logic. And a second strange principle is that "if a statement is false, then it implies any statement whatever." "Dogs do not have four legs" implies that Lassie is a dog, and also that Lassie is not a dog, and that 2 plus 2 are 4, and that 2 plus 2 are not 4.

i.e., as you put it: if "P is true, then per that interpretation, for any statement Q, Q->P is true."

Perhaps it could be written clearer, but the point is not what you seem to think it is, which would be a silly mistake to make, but it's simply an example to draw out one variety of paradox of material implication.