I agree it is not a statement, meaning it is not about anything. — Fire Ologist

Yes, it seems to me that this is just another case of "philosophers" confusing themselves.

However you slice it, the intent of the "liar" determines whether he is lying. He either is or he is not. There is no both-and. The same goes for someone who claims to be speaking falsely rather than lying. Either they intend to speak falsely or they do not. Many "philosophers" mistakenly hold that sentences have meaning apart from speakers, and when one reifies sentences in this way they have taken the first step towards this sort of self-confusion. They strangely believe that a sentence can self-negate itself because they have taken their eye off the ball: the speaker.

For example, I can't conceive of anything as being other than it is, because as soon as I conceive it, it is what it is, and not something else. I cannot imagine something as being otherwise. This reminds of the law of identity, and it just might be. — Lionino

This is very close to the way that Aristotle defends the PNC in Metaphysics IV. Much of this is just a question of what we mean by 'logic'.

in "This statement is false" we're never saying what we're referring to. — leo

Right. The so-called "Liar's paradox" seems quite silly, akin to something a third grader thought up at recess.

I agree it's not much use to spend much time pondering about them — leo

Me too. :up:

I am thinking of what SEP calls, "Aristotle’s Challenge to the Opponent to Signify Some One Thing."

More:

The Aristotelian can counter that without those qualifications the dialetheist has not said anything meaningful at all. — SEP | 11. Dialetheism, Paraconsistency, and Aristotle

To which the dialetheist may simply say "so much for Aristotle". — Banno

I would suggest actually reading Metaphysics IV.

But what is second-order rules of discourse? — Lionino

The examples I gave were:

Even in English when we say, "If you make that claim you will be contradicting yourself," we are shifting between two different registers: first-order claims and second-order rules of discourse (i.e. Thou shalt not contradict thyself). — Leontiskos

Note, though, that, "You are contradicting yourself," or, "This is a contradiction," is a different genus, and deviates from first-order discourse, moving into the meta-language. — Leontiskos

So an example of a second-order rule of discourse is, "Thou shalt not contradict thyself."

It's a mistake to think that there are laws of logic that have complete generality - and must be obeyed in all circumstances.

...

Logic sets up systems in which some things can be said and others are ruled out, but natural language is far broader than that, allowing for the breach of any such rule. — Banno

Yet if what Aristotle does in Metaphysics IV is correct, then there is a logical law that cannot be breached, namely the law of non-contradiction. Or in other words, "logic" is not a purely formal exercise. It was created for a reason and that reason has implications for reality/metaphysics.

Elaborate. — Lionino

For example:

The English has to do with a relation between P and Q that transcends their discrete truth values. One way to see this is to note that an English speaker will be chastised if they use the phrase to represent a correlation that is neither causative nor indicative, but in the logic of material implication there is nothing at all wrong with this. — Leontiskos

"If the Baltic sea is salty, then the Eiffel Tower stands." According to material implication this is a perfectly good statement, but according to English it is foolish. There is nothing which surpasses this sort of statement according to material implication: the antecedent is true, the consequent is true, and therefore the implication is true. What more could we ask? But for the natural speaker what is lacking is a relation between the two things. What is lacking is a relation between the saltiness of the Baltic Sea and the standing-ness of the Eiffel Tower.

Passing to another kind? What kind? — Lionino

Further, I am of the opinion that speech about contradictions is always a form of metabasis eis allo genos. Even in English when we say, "If you make that claim you will be contradicting yourself," we are shifting between two different registers: first-order claims and second-order rules of discourse (i.e. Thou shalt not contradict thyself). — Leontiskos

In the example I gave, "First-order claims and second-order rules of discourse."

A first order claim in propositional logic is something like, "P is true," or, "Q is false." Sentences consist of propositional affirmation, negation, and logical operators. Note, though, that, "You are contradicting yourself," or, "This is a contradiction," is a different genus, and deviates from first-order discourse, moving into the meta-language.

The key is that in English we prescind from many things that material implication does not prescind from — Leontiskos

For example, one can assert the material implication (P→Q) for three reasons:

1. P is true and Q is true
2. P is false (and Q is true)
3. P is false (and Q is false)

In English, on the other hand, we only say, "If P then Q," when we believe that the presence of P indicates the presence of Q. The English has to do with a relation between P and Q that transcends their discrete truth values. One way to see this is to note that an English speaker will be chastised if they use the phrase to represent a correlation that is neither causative nor indicative, but in the logic of material implication there is nothing at all wrong with this.

So, what could one say about ¬(A→B) in English? — Lionino

As I alluded to in the other thread, material implication captures English usage only insofar as it guarantees that if the antecedent is true then the consequent will also be true. Similarly, the negation of a material implication says that if the antecedent is true then the consequent will be false, and this is vaguely similar to the denial of an implication in English except for the fact that the falsity of the consequent is not guaranteed in English.

The key is that in English we prescind from many things that material implication does not prescind from, such as the value of the consequent in that denial case. As another example, if an antecedent is false then the material implication is true, whereas this does not hold in English. At the end of the day the English sense of implication simply isn't truth functional. It is counterfactual in a way that material implication is not.

And what about the following formulas:

A→(B∧¬B);
A→¬(B∧¬B);
¬(A→(B∧¬B))? — Lionino

I think in examining these we are combining two confusing and non-translatable logical concepts: material implication and contradiction. Neither one translates well into English, and their combination translates especially badly.

Further, I am of the opinion that speech about contradictions is always a form of metabasis eis allo genos. Even in English when we say, "If you make that claim you will be contradicting yourself," we are shifting between two different registers: first-order claims and second-order rules of discourse (i.e. Thou shalt not contradict thyself).

...it just takes forever if the input is large. — Count Timothy von Icarus

However, if I includes enough nodes then all of the world's super computers running P(I) until the heat death of the universe still won't have been able to actually compute O yet. — Count Timothy von Icarus

If we have a large number of nodes with infinite time then P(I)=O will produce the ideal solution, it will be unique, and in that case what you conclude no longer holds:

So then, in a very important functional sense P(I) is not "the same thing as O." — Count Timothy von Icarus

If we don't have infinite time then there is no sense in which P(I) is the same thing as O, but if we have infinite time then there is an important sense in which P(I) is the same thing as O.

But my point is that it isn't plausible to compare 2+2=4 to an NP-Complete problem and then claim that 2=2=4 suffers from the same limitations as the NP-Complete problem. 2+2 isn't NP-Complete. The basic objection to your argument would be, "Well I agree that P(I) is not the same thing as O, but it doesn't follow that 2+2 is not the same thing as 4."

The phrase «A does not imply a contradiction» really means specifically «A being true, it does not imply a contradiction». I think this meaning is indeed encapsulated in A→¬(B∧¬B), especially when it can be translated as «A implies True». — Lionino

I don't see it this way. I think the phrase, "A does not imply a contradiction," either means, "A implies something and that something is not a contradiction," or else, "Whatever is implied by A, it is not a contradiction." Both of these are examples of meta-language, and neither is represented by A→¬(B∧¬B). The first means, "A implies B and B is not ¬A." The second means that whether or not A implies anything, it does not imply ¬A.

In any case we have to distinguish these two propositions:

• A→(B∧¬B)
• A→¬A

They are found in "S". Or you can just replace "S" with axioms of the theory. Axioms are naturally assumed.

...

I don't. I know that S and ¬P can't coexist. I know that S, so ¬P can't be the case. ¬¬P is P. — Lionino

See:

<Lionino's "reductio" seems to be ambiguous between senses (2) and (3)> — Leontiskos

So:

(S∧¬P)→(B∧¬B)
S
∴ P — Lionino

What is happening is apparently:

1. (P∧Q)→R
2. ¬R
3. ∴ ¬(P∧Q)

As noted earlier in the thread, a reductio is not representable in the object language, and therefore what you present is not a reductio in any formal sense.

The first thing to note is that the conclusion is ¬(P∧Q). In the first place P and Q can both be false. But if we add an additional condition that they cannot both be false, ¬(¬P∧¬Q), then to stipulate that one is true or false automatically determines the other value, and yet there is no principled reason to stipulate such a thing apart from mere stipulation. Or, if we stipulate that one is true, as you did, then we must accept that the other is false, even without the additional condition. Still, we have no reason to stipulate P instead of Q. There is no theory here or set of axioms, except in a purely stipulative and imaginary way. P and Q are exactly on a par as far as the formalization goes.

1. (P∧Q)→R
2. ¬R
3. ∴ ¬(P∧Q)
4. __Suppose P
5. __∴ ¬Q

From the supposition we learn (P→¬Q), at which point P must be further asserted beyond supposition if we are to actually arrive at ¬Q:

• (P→¬Q)
• P
• ∴ ¬Q

(To suppose P and to assert P are here two different things)

..But it always goes back to the question of why we preferred P in (4) rather than Q.

Note too how this is different from a reductio:

(S∧¬P)→(B∧¬B)
S
∴ P — Lionino

...insofar as your supposition produces no contradiction at all, and the thing you suppose is never rejected. This sort of underlines that you are merely supposing that something is true without any reason. This is what confused me in my initial reply. When you called your proof a reductio I assumed your supposition was being rejected.

What makes the Hamiltonian Path problem intractable is precisely the extremely large number of operations and this can be true for any program provided it has enough steps. — Count Timothy von Icarus

An NP-Complete problem is, among other things, one that has no known polynomial time algorithm/solution. The point being that your P(I) = O is an approximate solution, not a deterministic solution. If O is only an approximation of a solution then of course it deviates from the ideal solution, and from an isomorphic relation.

Yup. We agree there, and that's basically what I mean with the story. It's just an introduction to a thought with a funny conclusion, not an argument or anything of that sort. — Moliere

Okay, fair enough. :up:

A more current but exactly the same example is Chidi from The Good Place :D — Moliere

I was told to watch it by all sorts of people but never did. :grimace:

Thanks that's very high praise :) -- It's all just me working out my own thoughts that I'm willing to share, though, and it's part of what I consider to be in fair trade: I like to read others' thoughts, and so share in kind. — Moliere

Makes sense.

Parables are hard anywhere I think. What makes them difficult is what also makes them attractive. I'm very much attracted to stories, though, because I think they set out nuances better thanwell even though the difficulty is that the nuances aren't specified and there's a certain amount of interpretation that has to go into them. — Moliere

Yep, and probably also because it is impossible to express all the nuance of certain things. In that case to even try is to show that you don't understand what you're dealing with.

Though maybe the distinction is between the sublime and the humorous? — Moliere

Yes, and that line can get fuzzy, too.

I never thought to interpret Balaam's Ass like that, though, which adds an interesting layer: "Get out of your head, dork!" is the kind of message I imagine which unites these. — Moliere

Haha - well the interesting thing about "the old book" is the presuppositions that are brought to it. I don't wish to reduce the value to those presuppositions, but when a text is approached as sacred or inspired it eo ipso comes to possess an unmatched power to express nuanced ideas, such as parables. This is something like Kierkegaard's idea that the believer measures himself against the infinite, and for that he stands taller.

The problem shows up because logicians, who tend to be the folks most interested in this problem, only look for formal solutions. But the issue is that "eternal implication," or "implication occuring outside time" is assumed. We can think of computation abstractly, but it remains defined by step-wise actions. Yet these abstractions are taken to be "real" as opposed to merely tools.

However, in the brain or in digital computers two things hold:

1. Computation always occurs over time.
2. Computation involves communication and can be thought of in terms of communication models (some very good work on this has been done and the two end up being almost the same thing, "information processing" indeed.) — Count Timothy von Icarus

Right, and this is related to my claim:

If this is right then (b∧¬b) introduces instances of formal equivalence that are not provable. — Leontiskos

I believe that given the way formalized logic works, there can be sentences which are formally equivalent and yet underivable from one another. According to Sime one implication of this can be seen in terms of Peano arithmetic (link).

Why? Because deduction/computation, be it in computers or humans, always involves communication and must occur over some region of space-time, not "all at once and all in one place." Aristotle gets at this in his essentially processual conception of demonstration in the Posterior Analytics. — Count Timothy von Icarus

Going back to Meno, if argument was not temporal then we could presumably never gain new knowledge. The other interesting question is how to account for forms of non-temporal knowledge.

So then, in a very important functional sense P(I) is not "the same thing as O." — Count Timothy von Icarus

But probably only because it is NP-complete. When P is not NP-complete it is a more difficult question whether P(I) is the same thing as O. P(I)=O and 2+2=4 are very different in that sense.

Something that I read recently, very interesting, and I can't remember where, on the topic of logic, is that syllogisms can be said to be question begging (this is a point that has been made by philosophers in the past).
"All men are mortal; Socrates is a man; therefore, Socrates is mortal" is of no value, since we could not know that the premise, "All men are mortal" is true unless we already knew that Socrates is mortal. So we learn nothing from the syllogism. — Lionino

This is really the problem of knowledge as expressed in places like the Meno:

I know what you want to say, Meno. Do you realize what a debater's argument you are bringing up, that a man cannot search either for what he knows or for what he does not know? He cannot search for what he knows—since he knows it, there is no need to search—nor for what he does not know, for he does not know what to look for. — Meno, 80e, (tr. Grube)

Aristotle applies his notions of act and potency to basically say that in knowing something partially we can come to know it more fully. When the mind engages in argument this is what it is doing, according to Aristotle. We are unfolding implications previously unseen.

Metabasis eis allo genos is a complicated topic. I expressed it this way originally:

Every time we make an inference on the basis of a contradiction a metabasis eis allo genos occurs (i.e. the sphere of discourse shifts in such a way that the demonstrative validity of the inference is precluded). — Leontiskos

Note that this is a sufficient condition and not a necessary condition. The same thing can be expressed in terms of the "meta-language":

One is a statement in the meta-language and the other in the object language. They are different levels of statement. — TonesInDeepFreeze

Whenever some logical move requires recourse to the meta-language, we are involved in metabasis. <The three senses> of interpreting a contradiction that I set out are all utilized in the service of a metabasis. This sort of ambiguity always attends metabasis. Sorting out the ambiguity requires us to go beyond the object language at hand.

This sounds like the "Scandal of Deduction," and it actually holds not just for syllogisms but for all deterministic computation and deduction. From an information theoretic perspective, because the results/outputs of computation and deduction always occur with a probability equal to 100% it follows that they are not informative. Everything contained in the conclusion must be contained in the premise; we learn nothing from deduction. — Count Timothy von Icarus

Yes, I was thinking about this as well.

1. (φ^~φ) means explosion
2. (φ^~φ) means reductio-rejecton
3. (φ^~φ) means false — Leontiskos

It seems plausible that:

1. (φ^~φ) takes on the meaning of <explosion> as the antecedent of a modus ponens
2. (φ^~φ) takes on the meaning of <reductio-rejecton> as the penultimate step of a reductio
3. (φ^~φ) takes on the meaning of <false> as the consequent of a modus tollens

It's as if (φ^~φ) can be whatever we need it to be for our current purposes, and this should not be surprising.

Note:

• <Sense (2) is quite different from sense (3)>
• <Lionino's "reductio" seems to be ambiguous between senses (2) and (3)>
• <My term FALSE is meant to capture sense (3)>

meta-language — Leontiskos

Another interesting point goes to natural language. "A→(B∧¬B) means ¬A."

Compare:

1. (φ^~φ) means explosion
2. (φ^~φ) means reductio-rejecton
3. (φ^~φ) means false

Without recourse to the meta-language, there is no way to adjudicate. I think this goes back to 's point.

Banno asked a good question:

So, what is a "direct proof"? I gather you think using MT is direct, but RAA isn't? WHat's the distinction here? — Banno

(i.e. What is the difference between a direct proof like modus tollens and an indirect proof like reductio ad absurdum?)

I said:

Modus tollens requires no "and-elimination" step. Is that a good way to put it in your language? — Leontiskos

Put differently:

One is a statement in the meta-language and the other in the object language. They are different levels of statement. — TonesInDeepFreeze

A direct proof requires no recourse to the meta-language. When the reductio identifies a contradiction it is dipping into the meta-language. That exchange earlier with Tones was about whether the reductio is truth-functional. It turns out that you cannot represent a reductio in the object language.

Another way to put it is that in modus tollens we have two premises whereas in reductio ad absurdum we have a premise and a supposition, and the difference between a premise and a supposition only exists at the level of the meta-language.

Edit: Indeed, this is instructive given that the unique <modus tollens> we are considering also <uniquely requires recourse to the meta-language>. No other modus tollens requires recourse to the meta-language. Nevertheless, the recourse that it requires is different from the recourse that a reductio requires. <If we avoid the meta-language we will only continue banging our heads against the wall>.

(@Lionino)

I didn't suppose ¬P. — Lionino

Sorry I misread a quote from above. You are right. You supposed S.

I know that S follows from the axioms of the theory. Not an assumption.
Conclusion: P. — Lionino

You know equally well that ¬P follows. Conclusion: ¬S.

You are importing "the axioms of the theory." They are nowhere to be found. They are background conditions, absent from your proof.

(S∧¬P), S does. — Lionino

And note that you supposed ¬P (which is the same as preferring S). Either way its a random pick for the second assumption. The point here is as I have said:

This is the formal conclusion, before the and-elimination step of the reductio takes hold:

A→(B∧¬B) {Assumption}
A {Assumption}
∴ (¬A ∨ ¬(A→(B∧¬B)))

...which is the same as, "∴ (A ∨ (A→(B∧¬B)))" We are merely picking an assumption to be true or false, for no reason.

Whether we call (1) a supposition or (2) a supposition is arbitrary, and purely stipulative. There is no formal reason to draw one of the disjuncts of (3) rather than another. — Leontiskos

We do because S fully follows from the axioms of a theory. — Lionino

Which premise do you think provides us with such information?

(S∧¬P) does not favor S over ¬P in the case of a contradiction. If a contradiction follows from (S∧¬P) it is no more rational to reject ¬P than to reject S.

I think you are referring to background conditions that are not formally present, hence my point. If we really had a set of axioms and a theory instead of a single assumption, then you could say that it follows from a theory. In this case we do not have that.

From within a theory T, the statement S is known to be true. We don't know whether P is true or not.
We verify that S∧P implies a contradiction.
Thus P cannot be the case within that theory T.
That is paramount for proofs from contradiction in mathematics. — Lionino

I edited that post a bit, perhaps after you read it. For example:

The issue is that we don't know S is true any more than we know that ¬P is true. All we really know is that S and ¬P are inconsistent. Given their inconsistency, one must be false. Picking one to be false without any other information is an arbitrary move. — Leontiskos

This isn't a proof of Modus tollens. — flannel jesus

I was not trying to say it was.

ρ→(φ^~φ) (premise)
~(φ^~φ) (law of non contradiction)
:. ~ρ (modus tollens) — flannel jesus

This is perhaps my favorite proof for the modus tollens thus far. The question is whether that second step justifies the modus tollens. Does the "law of non contradiction" in step two allow us to think of the contradiction as a simple kind of falsity, which requires no truth-assignment? And if so, does that thing (whatever it is), allow us to draw the modus tollens? These are the questions I have been asking for 12 pages.

See my posts <here> and <here> for some of the curious differences between (φ^~φ) and ¬(φ^~φ).

<I believe the reductio has failed and that the only strict way to draw ¬A is by using the modus tollens.>

• A→(B∧¬B)
• ∴ ¬A {modus tollens}

Now there is something special about (B∧¬B) which makes it seem plausible that we could successfully draw the modus tollens inference in this case, even though there is no other case where a one-premise modus tollens is possible. There is a way in which we can conceive (B∧¬B) such that the modus tollens goes through. What is this conception of (B∧¬B)? Let us call it ‘FALSE’, in order to be able to talk about it. FALSE is (B∧¬B) conceived in the manner which allows us to draw the modus tollens inference.

FALSE is a kind of emergent property of (B∧¬B) which no other conjunction of the form “(P∧Q)” possesses. It is unique, and its uniqueness is what ostensibly allows it to uniquely draw a one-premise modus tollens.

(B∧¬B) itself is also unique, insofar as it is both simple and compound. It is compound because it is a conjunction, and conjunctions are compound. It is simple because it can be seen to be false as a whole, without any variable assignment (hence the simplicity of FALSE). Both are necessary given the fact that a modus tollens requires the consequent as a whole to be false; and that there is no non-compound basis for something which is false without any variable assignment (and when we combine this with the fact that (B∧¬B) is a unique compound proposition,* we see why every other modus tollens requires a second premise).

Now a modus tollens requires that the consequent be false:

We can apply Aristotelian syllogistic to diagnose the way that the modus tollens is being applied in the enthymeme:

((A→(B∧¬B))
∴ ¬A

Viz.:

Any consequent which is false proves the antecedent
(B∧¬B) is a consequent which is false
∴ (B∧¬B) proves the antecedent

In this case the middle term is not univocal. It is analogical (i.e. it posses analogical equivocity). Therefore a metabasis is occurring. As I said earlier... — Leontiskos

The modus tollens inference will be valid if two conditions are met. First, it must be the case that (B∧¬B) can be legitimately interpreted as FALSE. Second, the modus tollens must be able to support FALSE as a consequent:**

Now one could argue for the analogical middle term, but the point is that in this case we are taking modus tollens into new territory. Modus tollens is based on the more restricted sense of 'false', and this alternative sense is a unfamiliar to modus tollens. This is a bit like putting ethanol fuel in your gasoline engine and hoping that it still runs.

Note that the (analogical) equivocity of 'false' flows into the inferential structure, and we could connote this with scare quotes. (B∧¬B) is "false" and therefore the conclusion is "implied." The argument is "valid." — Leontiskos

If both conditions hold then the argument cited at the beginning of this post is valid, and ¬A can be drawn by modus tollens. If at least one condition fails then the argument is not valid and ¬A cannot be drawn by modus tollens.

The more general point here is that the uniqueness of (B∧¬B) presents us with a difficulty: can it be used in an exceptional way within standard inferences, or not? Can it be treated as FALSE in order to produce strange inferences, such as one-premise modus tollens, or not? And is there a principled way to decide this question? Is there a way to know the ways it can be used and the ways it cannot be used? I don’t know, and my point throughout the thread is that I am wary of this whole approach, even though it seems intuitive to many.

* I take it that formally equivalent propositions such as ¬(B∨¬B) do not count as separate bearers of FALSE.

** I realize I am mucking up the original definition of FALSE a bit here, but that is hard to avoid. The point is only that the interpretation must be both supportable, and it must suffice for the inference.

(@Lionino, @Banno, @Count Timothy von Icarus)

(I am probably going to need to begin moving away from this thread.)

We don't know whether P is true or not. We know S is true. S being true and P being false leads to a contradiction. Therefore we have ascertained that P is true. No assumption is needed or allowed. — Lionino

As I see it, the problem is that this is a misunderstanding of a reductio. A zero-premise reductio makes no sense, and a one-premise reductio misunderstands what a reductio is doing. As pointed out, a reductio is meant to show the inconsistency of some assumption given a set of premises (i.e. more than 1!). My response was here.

When you say "we know S is true" you are stipulating. What does it really mean to "suppose" that P is true? The supposition move is not a purely formal move. The LEM says that everything is either true or false, not that some things are supposedly true. A reductio is doing something over and above a formal move. A system with two inconsistent assumptions results in the dichotomy between the two assumptions, and there is no formal difference between an assumption and a supposition.

Here's another way to put it:

We can say:

• A→(B∧¬B) {Assumption}
• A {Supposition}
• ∴ ¬A {reductio ad absurdum}

But we could equally say:

• A {Assumption}
• (A→(B∧¬B)) {Supposition}
• ∴ ¬(A→(B∧¬B)) {reductio ad absurdum}

This is different from the principle of explosion, but it results in the same formal indifference between the two conclusions, given the four assumptions (two of which are called 'suppositions').

When you run a reductio with only one premise you are basically acting out the same principle as, "One man's modus ponens is another's modus tollens." The second "argument" is no more silly than the first. We just think the first is less silly because the premise of the first argument seems to contain more complexity, and therefore it seems to mimic Tones' system of premises. But it arguably does not contain more complexity (and even if it did, it is only a single premise). It is just a simple conditional with a contradiction in the consequent. My following post tries to draw out the way that a contradiction is complex/compound only in a curious and unique way.

This is the formal conclusion, before the and-elimination step of the reductio takes hold:

1. A→(B∧¬B) {Assumption}
2. A {Assumption}
3. ∴ (¬A ∨ ¬(A→(B∧¬B)))

...which is the same as, "∴ (A ∨ (A→(B∧¬B)))" We are merely picking an assumption to be true or false, for no reason.

Whether we call (1) a supposition or (2) a supposition is arbitrary, and purely stipulative. There is no formal reason to draw one of the disjuncts of (3) rather than another.

But the issue is that we already know S is true. — Lionino

The issue is that we don't know S is true any more than we know that ¬P is true. All we really know is that S and ¬P are inconsistent. Given their inconsistency, one must be false. Picking one to be false without any other information is an arbitrary move.

I don't understand.
(S∧¬P)→(B∧¬B)
S
∴ P is supposed to be the definition of RAA according to where I got it from. — Lionino

What's at all wrong with this?:

(S∧¬P)→(B∧¬B)
¬P
∴ ¬S

We're going in circles. Time to go <meta>.

This is the RAA, innit? :smile: — Lionino

I was already convinced that RAA is insufficient. That as you say:

So the question is: how do we choose between either? Isn't it by modus tollens? — Lionino

RAA will not prove ¬A.

(S∧¬P)→(B∧¬B)
S
∴ P — Lionino

The logic of the RAA proves (¬S v P), and the RAA choses one or the other.

The truth-functionalist is likely to object to me, “But your claims are not verifiable within classical logic!” Yes, that is much the point. When we talk about metabasis eis allo genos, or contradiction per se, or reductio ad absurdum, we are always engaged in some variety of metalogical discourse.

...

How can we start inching towards the difference between ‘false’ and ‘FALSE’? First I should say that the “proposition” (b∧¬b) can be either. It can be interpreted as false or as FALSE each time we touch it with our mind. What this means is that terms like (b∧¬b) or ‘false’ are metalogically equivocal or ambiguous given the question we are considering... — Leontiskos

The problem as I see it is that those who will not move into an analysis of the language are trying to solve a metalogical problem with the logic itself, and this cannot be done. We must move into metalogical discourse, and because of this I would propose analyzing the nature of (b∧¬b) and the attendant inferences using English rather than (redundant) logical translations. The logical translations involving that term seem by this point to be clearly underdetermined.

- I'm quite serious. See my edit to that post, which may help you.

Straight RAA does not require the "and elimination". It's an additional step when there are multiple assumptions. — Banno

I have never seen a reductio that does not have multiple assumptions.

Edit: this is what I think a one-premise reductio would look like:

A→ABSURD
∴ "A cannot be affirmed"

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Introducing ABSURD in the way I did above destroys the LEM of classical logic. — Leontiskos

While you are there, what does "FALSE" mean? — Banno

If you don't want to read the posts where I quite sincerely tried to get at this, we could just say that FALSE is what is necessary to get the modus tollens to run with only one premise. It is the sense of the contradiction required for the valid inference.

It is what is supposed to answer this question:

how do you prove that you may derive ~ρ from ρ→(φ^~φ)? — Lionino

I consider it an open question as to whether this question is answerable.

So, what is a "direct proof"? I gather you think using MT is direct, but RAA isn't? WHat's the distinction here? — Banno

Modus tollens requires no "and-elimination" step. Is that a good way to put it in your language?

But couldn't we just assume B here and get ~A just the same? "If B then ~A," seems to work fine here because the conjunct is still going to come up false. — Count Timothy von Icarus

You're basically preaching to the choir. <This> is the third time I presented that idea. But a proof that requires an additional assumption is different from one that does not.

Edit: Sorry, misunderstood - I guess I'm wondering how a proof by exhaustive cases fits in. That's what you're introducing, and it's a good introduction.

And then we can do the same thing assuming ~B. That covers all our options assuming LEM. — Count Timothy von Icarus

I am wondering if allowing contradictions in this way fiddles with the LEM, but this is just conjecture.

I guess I'm not seeing the trouble here. I can see the trouble with proofs by contradiction in mathematics that prove things for which no constructive proof exists. That makes sense because, on some philosophies of mathematics, an entity doesn't exist until the constructive proof does (and perhaps it can't exist). — Count Timothy von Icarus

The trouble is simply that (b∧¬b) has been consistently creating unexpected behavior in this thread, but I find your approach interesting. Do continue.

And yet your claim that Reductio is invalid is just wrong. — Banno

What I have consistently said is that reductio is not valid in the same way that a direct proof is. Perhaps I slipped at some point and called it invalid. In any case, I don't see this as a big mistake. As I said earlier, what is entailed is the disjunction, not some subset of the disjuncts. That is, in this case if we are to avoid the contradiction as a reductio requires that we do, then we must reject either (A→¬B∧B) or else A.

The problem is that this A, A→¬B∧B ⊢ ¬A was given as reductio ad absurdum. But this entails anything. — Lionino

To my mind the explosion only occurs if you don't reject either of the two premises. If you reject either of the two premises via reductio, explosion is avoided, no? When we reject (A→¬B∧B) and accept A, explosion no longer follows. The same is true if we accept (A→¬B∧B) and reject A.

What if we reject (1) instead? Then A is made true, but it does not imply (B∧¬B). — Leontiskos

This is the path that @Banno and @TonesInDeepFreeze have chosen:

• (a→(b∧¬b)) → ¬a

They have two possible routes which could be used to reach their destination: Mt. Caradhras or the Mines of Moria. Gimli suggest that they take the Mines, but Banno knows that "the dwarves dug too deep in their greed, awakening horrors in the depths." So they try the pass of Caradhras, but it turns out to be unworkable, smothered by snow and storm. Do they dare tempt the Mines of Moria? Viewers must wait and see... :grin:

• Mt. Caradhras = reductio ad absurdum
• Mines of Moria = A modus tollens with only one premise

Note: For those considering the mines, two posts may be especially useful: first, second.

The grain of truth in Leontiskos' position is that reductio arguments need to be used with care. — Banno

But my fuller position is that any inference utilizing strange senses of would-be familiar logical concepts must be used with care. I am not opposed to the Mines of Moria, but I don't think people are taking enough care in traversing them.