• TonesInDeepFreeze
    3.6k
    1. A→(B∧¬B) assumption
    2. A assumption
    3. B∧¬B 1,2, conditional proof
    4. ~A 2, 3 reductio
    — Banno

    As I noted earlier in response to Tones' reductio, a reductio is an indirect proof which is not valid in the same way that direct proofs are.
    Leontiskos

    This caveat will cover my posts in general: People may have different contexts with different meanings of terms, different fundamental conceptions, different philosophical positions, and especially different formal systems that may diverge from classical logic, contract it, or extend it. If a context is not made explicit, then it is reasonable to respond in the context of classical logic, which I will do. And doing so is especially apt when responding to criticisms or examination of classical logic.

    Reductio ad absurdum is proven valid as it is proven that inferences using it are truth preserving, just as with the other rules of classical sentential logic. An argument is valid if and only if there are no models in which the premises are all true and the conclusion is false. Reductio ad absurdum is proven to provide only valid arguments.

    You can see this by examining your conclusion. In your conclusion you rejected assumption (2) instead of assumption (1). Why did you do that? In fact it was mere whim on your part, and that is the weakness of a reductio.Leontiskos

    The argument may be better written:

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

    (2) A ... assumption toward a contradiction

    (3) B & ~B ... 1, 2 modus ponens

    (4) ~ A ... 2, 3 RAA with 2 discharged

    There we see that the premise A -> (B & ~B) proves ~A, while the temporary assumption A was discharged.

    And it is valid, as every model in which "A -> (B & ~B)" is true is a model in which "A" is true.

    I am attributing the modus tollens to you because you are the one arguing for ¬A. If you are not using modus tollens to draw ¬A then how are you doing it? By reductio?Leontiskos

    Modus tollens and reductio are two sides of the same coin.

    A Hilbert system may have modus tollens (or to have classical, the non-intuitionistic version) in axiom form, and a natural deduction system may use RAA (and, to have classical, the non-intuitionistic version) in rule form.

    To prove a negation, there have to be axioms or rules to do it. Modus tollens and RAA are in a sense versions of each other and they both do the job.
  • TonesInDeepFreeze
    3.6k
    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

    Tones replied that that is not true for all contradictions but for some interpretations.
    Lionino

    That's not what I said.

    If I recall correctly, you said that "A -> (B & ~B)"* may be translated as "A implies a contradiction". (*Or it might have been a related formula; not crucial since my point pertains to all such examples.)

    That is not the case as follows:

    (1) The sentence has a sub-sentence that is a contradiction, but the sentence itself does not mention the notion of 'contradiction'.

    (2) To say "a contradiction" is to implicitly quantify: "There exists a contradiction such that A implies it". And that quantifies over sentences. If we unpack, we get: "There exists a sentence Q such that Q is a contradiction and A implies it".

    A translation of "A -> (B & ~B)" is:

    If A, then both B and it is not the case that B.

    and not

    "A implies a contradiction".

    (3) "B & ~B" is a particular contradiction, not just "a contradiction". Even though all contradictions are equivalent, a translation should not throw away the particular sentences that happened to be mentioned.

    (4) If we have that A implies B & ~B, then of course, we correctly say "A implies a contradiction". But that is a statement about A, not part of a translation.

    "If A implies B & ~B, then A implies a contradiction" is true, but it is a statement about the sentences, not a translation of them.
  • TonesInDeepFreeze
    3.6k
    If A is false, then A implies anything.flannel jesus

    We need to be careful to recognize that "A is false" is not simpliciter. Rather, "A is false in model M" (with sentential logic, a model may be represented as a row in a truth table).

    You can check the truth-table on implication: A -> B is always true if A is false.flannel jesus

    But I would add (and this was my point to @Leonino) it is not the case that

    "Winston Churchill was French" implies every sentence, simpliciter.

    Rather, it is only in interpretations in which "Winston Churchill is French" is false is it the case that for every sentence "P", ""Winston Churchill is French" implies P" is true.
  • TonesInDeepFreeze
    3.6k
    His ready-made approach doesn't answer the questions that are being askedLeontiskos

    It would be rare that any one poster can answer all questions. I have given corrections, explanations about the logic that is being critiqued, answers to specific questions, and perspective on the subject. And describing my approach as "ready made" is empty polemics.

    A person may have all kinds of independent, self-fashioned, not ready-made ideas. And one may critique "ready-made" classical logic, but when a critique misconstrues or misrepesents what it is critiquing, then it is quite appropriate to point out the errors and even explain why they are errors. Moreover, it is appropriate to point it out when what is critiqued is subjected to criteria not pertinent. Of course, classical logic doesn't accord with much of everyday discourse. And a pencil doesn't accord with the colors in the world. That doesn't entail that, for certain purposes, a pencil is not preferable to a box of crayons. Black and white photography doesn't accord with the colors in the world. That doesn't entail that black and white photography is wrong. An x-ray doesn't accord with the way we see things in everyday life. That doesn't make an x-ray wrong. And it is apt to explain how x-rays work, what they represent, and their purpose, just as it is apt to explain how classical logic works, what it represents and its purpose, especially in context of ill-premised remarks about it.
  • TonesInDeepFreeze
    3.6k
    They will be required to examine the logic machine itself instead of just assuming that it is working.Leontiskos

    Logicians and philosophers of mathematics examine logics with intense scrutiny and may be interested in all kinds of formal and philosophical alternatives. And critiques of classical logic should be welcomed; indeed they have been vital in the history of logic and mathematics. But when critiques are ill-premised and describe the logic in inaccurate ways, then, of course, how can that be dealt with other than indeed explaining how the "machine" does operate?

    But in what sense is the "machine" working?

    Well, it does not work to express much of everyday discourse. Is there anyone who has denied that? And it does not accord with certain philosophies and alternative formal systems. And it does not provide answers to all the philosophical questions. But it does work as an axiomatization for mathematics for the sciences (indeed, the P vs NP problem is commonly described as the most important mathematical problems for practical implications for purposes of business, with a one million dollar prize for its solution). It does work as the basis for the computers we're using when we post. It does work in the sense that it gives rigorous definitions and unambiguous formulations of mathematics and for other fields of study. It does work in the sense that, in principle, we can most objectively (that is algorithmically) check that any given purported proof is indeed a proof in the system. It does work in the sense that the soundness and completeness of the logic is rigorously proven. And it does work in the sense that it provides beauty to its students and experts, and such that it induces great creativity and insight and questions that themselves lead to not only flourishing of mathematical inquiry but discoveries applicable in other fields. And it works in the sense that it provides the starting point and the basic rubrics for the invention of alternatives to it.
  • TonesInDeepFreeze
    3.6k
    A reductio requires special background conditions. In this case it would require the background condition that (1) is more plausible than (2).Leontiskos

    That is a bad misconception. It's not how it works.

    Proof is from a set of premises. A natural deduction proof lists premises. A reductio premise however is discharged at the end of the proof. It is not an ordinary premise.

    Let G be a set of premises and a sentence P is not a member of G. And we want to show that G proves ~P. Then we may use any of the members of G in our argument. But, along with members of G, we also may suppose P to derive a contradiction, thus to show that G proves ~P.

    "plausibility" is not invoked.
  • TonesInDeepFreeze
    3.6k
    To add to my remarks about the number of posts and length of posts. A reply to a short post or to a small portion of a post can be long because of the number, extent and fundamental nature of the errors or even one error in the short post. And beyond that, anyone should be welcome to expatiate as much as they like. At least for me, that's at the heart of an open forum: People get to share their ideas, beliefs, knowledge, and disagreements without being limited by arbitrary restrictions. Again, I think about how petty, hypocritical, and close minded the poster is when he posts so very much yet begrudges someone else from answering with posts in a row, especially given such obvious conditions as catching up after being away for even a few days or posting in hours when others are not.
  • TonesInDeepFreeze
    3.6k


    Modus tollens is an inference rule or axiom, depending on the system. That's syntactical.

    The notion of falsity is semantical.

    Syntax and semantics work in synch in classical logic. But to use modus tollens in and of itself does not require mentioning falsity.
  • TonesInDeepFreeze
    3.6k
    in your reductio you do not treat the contradiction as falseLeontiskos

    The inference rules don't opine as to falsity. Rather, syntactically, when a contradiction results from a conditional premise, then the rules allows ending the proof with the negation of the conditional premise and discharging the conditional premise. (Same for subproofs within a proof.)

    Truth and falsehood are handled in the semantics. And we show that the syntax and semantics are in accord as:

    For any set of sentences G and any sentence P:

    P is provable from G if and only if there is no model in which all the members of G are true and P is false.

    Intuitively, a 'model' represents a possible set of circumstances.

    In other words, if P is provable from G, there is no set of circumstances in which all the sentences in G are true but P is false.

    And we prove that rigorously.
  • TonesInDeepFreeze
    3.6k
    One must think about the difference between a reductio ad absurdum and a direct proofLeontiskos

    To prove a negation, we must have a rule to use to do that. And any alternative (that adheres to soundness and provides for the completeness of the calculus) to modus tollens or RAA would be equivalent with them in the sense that the resulting systems would provide the same inferences as each other.
  • TonesInDeepFreeze
    3.6k
    I am simply misunderstanding what "→(B(x)∧¬B(x)" means, it can't be just "any contradiction", as Tones has pointed.Lionino

    What I say is that "P implies a contradiction" is not a translation of "P -> (Q & ~Q).
  • TonesInDeepFreeze
    3.6k
    why can we read a→(b∧¬b) as "a implies a contradiction"Lionino

    I would not accept that reading, for the reasons I mentioned several posts ago.

    Most briefly: Yes, if P implies Q & ~Q, then P implies a contradiction. But that is a remark about "P -> (Q & ~Q)" and not a translation of it.
  • TonesInDeepFreeze
    3.6k
    A reductio is not truth-functional.Leontiskos

    RAA is an inference rule.

    A sentence Phi is truth functional if and only if the truth or falsity of the sentence is a function of the assignment of truth or falsity to the atomic sentences occurring in Phi.

    I don't know what you mean by an inference rule being truth functional.

    But an inference rule may be truth preserving, as RAA is:

    RAA is among the natural deduction rules. And regarding those rules:

    If the rules provide that P is provable from a set of premises G, then any model in which the sentences in G is true is a model in which P is true.
  • TonesInDeepFreeze
    3.6k
    In general, the consistency of an axiomatic system isn't provable in an absolute sense due to Godel's second incompleteness theoremsime

    What does "absolute sense" mean?

    Godel-Rosser is that system of a certain kind don't prove their own consistency. That doesn't entail that there are not other systems proven to be consistent by secure means.
  • TonesInDeepFreeze
    3.6k
    such that it can prove its own consistency. Then in this case, a proof of ¬¬a metalogically implies that ¬a isn't provable, i.e that a does not imply a contradiction.sime

    The system S could be inconsistent, in which case, if "S is consistent" is expressible in the language of S, then S proves "S is consistent" even though S is inconsistent.

    I don't know what role A is suggested to have. Some formula A does not prove a contradiction, I guess. I don't know how that is supposed to bear upon the consistency of a system with other axioms. ~A and ~~A. Don't know why choosing that pair rather then A an ~A, or maybe this has to do with intuitionism.
  • Lionino
    2.7k
    "If A implies B & ~B, then A implies a contradiction" is true, but it is a statement about the sentences, not a translation of them.TonesInDeepFreeze

    Yes, granted. I used the word "translation" wrong in basically all of my posts. I meant "is a true statement about..." instead.
    Now, the conclusion that I arrived at is that "A does not imply a contradiction" is not an accurate statement about ¬(A→(B and ¬B)), it would be a true statement about (A→¬(B and ¬B)) instead. When it comes to ¬(A→(B and ¬B)), as it is the same as (¬A→(B and ¬B)), "not-A implies a contradiction" is a true statement about it.
  • Leontiskos
    2.8k


    Regarding reductio ad absurdum, last night I was having a dream. I was walking a trail I know well and I noticed that the topography was inaccurate. I am usually semi-lucid when I dream, and so I decided to try to change the topography to make it more like it is in real life. As soon as I did this I noticed that this is similar to a reductio. In both cases an additional, uncharacteristic level of will emerges.

    The reductio is an uncharacteristically teleological move for truth-functional logic:

    (3) A ... toward a contradictionTonesInDeepFreeze

    The one who performs the reductio sees an opportunity to produce a contradiction and then decides to pursue it in order to achieve the inference desired (which inference is, again, a metabasis).

    The Medievals would have called the truth-functionality of classical logic something which pertains to the intellect (as opposed to the will). It is supposed to be purely formal, purely intellectual, and in no way willful. As you and I know, this is not entirely true since any human act involves the will, and therefore in any logic the teleological end of the acts at hand must involve the will. Nevertheless, a reductio involves the will over and above the way that direct inferences involve the will. The reductio attempts to leverage a contingency about the problem at hand in order to wield the contradiction and draw a conclusion. Hence the difference between a supposition and a mere assumption is that the supposition acknowledges the teleological motive of the will in a way that an assumption does not (Tones called his move a supposition whereas Banno called the same move an assumption).

    So we could say that the metabasis eis allo genos and the essence of a reductio ad absurdum is found not only in the unique inference that concludes a reductio, but also in its starting point: the supposition. This is what separates the supposition from the other assumptions, even though this difference is mental and not formal. And if we pay very close attention we will see that the formal conclusion is different from the teleological conclusion. The formal conclusion is that the system which includes the supposition is inconsistent. The teleological conclusion is that we should reject the supposition rather than a different premise. There is a miniature inference from the formal conclusion to the teleological conclusion, and this tends to be ignored by most students of classical logic. Put differently, the reductio strictly speaking only tells us that something cannot be supposed. It is a second step to say that that which cannot be supposed is in fact false.

    Note too that if someone is a strict univocalist with respect to inference then the reductio and all metabasis is disallowed. A reductio is an inference in a sense that is analogous to the way that, say, a modus tollens is an inference. If -inference- cannot function analogically, the reductio cannot succeed. These are fun little wrinkles in the purported truth-functionality of classical logic. Of course some might in fact disallow reductio and prefer a stricter logically system, but this system will be less powerful vis-a-vis achieving natural inferences.
  • Leontiskos
    2.8k
    Sorry - falling behind in this thread.

    ↪Leontiskos I solved my main problem just right above.Lionino

    I don't know if you saw my edit, which may now be redundant:

    Does this support my claim that what is at stake is something other than a material conditional? The negation does not distribute to a material conditional in the way you are now distributing it.Leontiskos

    But for the record I do accept this as a valid rhetorical move. However when it comes to propositional logic, from
    P1: A
    P2: A→contradict
    The conclusion can be whatever we want, from explosion
    Lionino

    Interesting, thanks for digging into this. Actually thanks for digging into all of the stuff you have dug into in this thread. It has helped me piggyback onto a lot of other ideas.

    I understand that you'd think that B∧¬B should be able to be replaced by any proposition P, but that is not the case.

    Example:
    (A∧(B∧¬B))↔(B∧¬B) is valid
    But (A∧C)↔C is invalid.
    Lionino

    Very good. This is another instance of the wrinkle that is created when the contradiction is allowed.

    ...so I think now it is a bit more clear why ¬(a→(b∧¬b)) is True only when A is True, the second member is always False and the or-operator returns True when at least one variable is True.Lionino

    Yes, but as @Count Timothy von Icarus and I have noted, it seems simpler to say that (¬(p→q)→p). The antecedent of a negated material conditional is always true, and this goes back to my point in the edit you may have missed above.

    (a→b) ↔ (¬a∨b)
    ¬(a→b) ↔ ¬(¬a∨b)
    However (a∨b) and ¬(¬a∨b) aren't the same
    So ¬(a→b) and (a∨b) aren't the same

    (a→(b∧¬b)) ↔ (¬a∨(b∧¬b))
    ¬(a→(b∧¬b)) ↔ ¬(¬a∨(b∧¬b))
    (¬a→(b∧¬b)) ↔ ¬(¬a∨(b∧¬b))
    Since ¬(¬a∨(b∧¬b)) is the same as (a∨(b∧¬b))
    (¬a→(b∧¬b)) ↔ (a∨(b∧¬b))
    Lionino

    Good. This back to 's point about ¬¬a.
  • TonesInDeepFreeze
    3.6k


    Thank you for recognizing my point.

    /

    For any A, we have these possibilities:

    (1) There is a contradiction Q such that A implies Q.

    (2) There is a contradiction Q such that A does not imply Q.

    (3) For all contradictions Q, A implies Q.

    (4) For all contradictions Q, A does not imply Q.

    If there is a contradiction that A implies, then A implies all contradictions.

    If there is a contradiction that A does not imply, then there are no contradictions that A implies.

    So we could state equivalences among (1), (2), (3), (4).

    /

    Theorems:

    (5) (A -> (B & ~B)) <-> ~A

    (6) ~(A -> (B & ~B)) <-> A

    (7) (~A -> (B & ~B)) <-> A

    (8) ~(~A -> (B & ~B)) <-> ~A

    From (A -> (B & ~B)) we infer that A implies all contradictions.

    From ~(A -> (B & ~B)) we infer that A implies no contradictions.

    From (~A -> (B & ~B)) we infer that A implies no contradictions.

    From ~(~A -> (B & ~B)) we infer that A implies all contradictions.
  • TonesInDeepFreeze
    3.6k
    The one who performs the reductio sees an opportunity to produce a contradiction and then decides to pursue it in order to achieve the inference desired (which inference is, again, a metabasis).Leontiskos

    There's no metabasis (change).

    Again, to show a derivation that a set of premises G proves a sentence ~P, we may use any of the members of G as premises in the derivation, then enter P as a conditional assumption, then derive a contradiction and infer ~P with the conditional assumption P discharged.

    There is no switching or "metabasis". Rather, at the very start, we state our premises and stick with them. The conditional assumption P is not one of our premises, but rather it is conditional assumption that is then discharged.

    You would do well to look at a specific natural deduction system to see the exact way its rules are formulated. And also to look at a proof of the deduction theorem that is the basis of natural deduction.
  • Leontiskos
    2.8k
    (3) "B & ~B" is a particular contradiction, not just "a contradiction". Even though all contradictions are equivalent, a translation should not throw away the particular sentences that happened to be mentioned.TonesInDeepFreeze

    This is part of the difficulty. If (b∧¬b) is a particular contradiction, then what is a non-particular contradiction? That is what you must ask yourself. When I said things like:

    Thus when Banno says that a contradiction (b∧¬b) is false, does he mean that it is false or that it is FALSE?Leontiskos

    ...or when @Lionino distinguished proposition-qua-variable from proposition-qua-truth-value, we were both pointing to this same valence where a material symbol (b∧¬b) has two legitimately different mental conceptions associated with it. In your language we would say that it can be conceived as a particular contradiction or a non-particular contradiction (non-particular being, in my terms, "falsity incarnate," or FALSE, or ABSURD, and in Lionino's earlier phrasing, contradiction-proposition-qua-truth-value, which truth value is necessarily false as opposed to contingently false).

    The analogical equivocity of the particular metabasis under consideration incorporates both particularity and non-particularity simultaneously. This equivocity is precisely what the reductio runs on, for when the reductio is begun the implicit contradiction is considered in particular terms, but by the end of the reductio it has been isolated and re-conceived in its non-particular sense. Or, at the beginning of a reductio when we only suspect a contradiction, that contradiction lives within the system in a particular way and not in a non-particular way. Once it is "ferreted out" it becomes non-particular, a contradiction qua contradiction, and this non-particular sense is what is required for the special reductio inference.
  • TonesInDeepFreeze
    3.6k
    (Tones called his move a supposition whereas Banno called the same move an assumption).Leontiskos

    Again: We derive that a set of premises G implies a sentence ~P by using any of the members of G as lines, then entering P on a line, then deriving a contradiction, then inferring ~P with the line for P discharged.

    The rule is the same whether we call the line entry of P an 'assumption', 'conditional assumption', 'assumption toward a contradiction', 'assumption to be discharged', 'supposition', 'conditional supposition', 'supposition toward a contradiction', 'supposition to be discharged', 'conditional premise', 'premise toward a contradiction', or 'premise to be discharged'.

    Those are merely ways of referring to the line; they handles we use; the logical basis of the rule does not depend and is not affected by what handle we happen to use to describe the line entry.
  • TonesInDeepFreeze
    3.6k
    what is a non-particular contradiction?Leontiskos

    'non-particular' is your word. It's up to you to say what you mean by it.

    There are particular contradictions and we can generalize about them. One such generalization is that all contradictions are equivalent.
  • Leontiskos
    2.8k
    'non-particular' is your word. It's up to you to say what you mean by it.TonesInDeepFreeze

    If you know what you mean by 'particular', then surely you know what you mean by 'non-particular'? If you can identify a particular contradiction, surely you can identify a contradiction that is not particular?

    Perhaps now you are beginning to see the point?
  • TonesInDeepFreeze
    3.6k


    There are particular apples and we can generalize about them. There is no apple that is not a particular apple. But we do say things like "If x is an apple, then x has a core". That is not claiming that there is an apple that is not a particular apple, but rather we can make generalizations about apples.

    Perhaps now you are beginning to see the point?Leontiskos

    Perhaps now you're beginning to see the point that a poster has no fault in posting several posts, some of them fairly long, in reply to several posts, some of them fairly long. Moreover that it is not unreasonable to post a long post in reply to a short one. And that it is a lot better not to lie about a poster by claiming he said things he did not say and indeed as he said the opposite of what he said. And that it is ridiculous to fault a posters for sometimes needing to delete posts, such as when needing to get rid of an otiose posting box left from a post that was started deferred to another post or not even posted.

    /

    Back to RAA: You've not shown any fault in RAA, but rather your fault in not understanding it.
  • TonesInDeepFreeze
    3.6k
    does he mean that it is false or that it is FALSE?Leontiskos

    Banno may speak for himself, but I don't know what difference in reference you mean by spelling 'false' without caps and with all caps.

    Nothing is "reconceived" in natural deduction. It seems you don't know how a natural deduction system is formulated.

    In your language we would say that it can be conceived as a particular contradiction or a non-particular contradiction (non-particular being, in my terms, "falsity incarnate," or FALSE, or ABSURD, and in Lionino's earlier phrasing, contradiction-proposition-qua-truth-value, which truth value is necessarily false as opposed to contingently false).Leontiskos

    What in the world? That is not my "language"; it's a bunch of your own verbiage.

    I can't make sense of almost all of the rest of your post and similar posts. Probably, I'll focus on the parts that are most blatantly false or ill-premised about the logic you're talking about.
  • Leontiskos
    2.8k
    Has everyone agreed by this point that 's truth table does not fully capture what a reductio is? (See bottom of post for truth table)

    ((a→(b∧¬b)) ↔ ¬a) is truth-functionally valid, but the implication in the first half of the biconditional is not the same implication that is used in a reductio ad absurdum.

    The easiest way to see this is to note that a reductio ad absurdum is not formally valid, and we can see this by noting that the reductio in 's post (now highly edited) does not prove his conclusion:

    As I noted earlier in response to Tones' reductio, a reductio is an indirect proof which is not valid in the same way that direct proofs are. You can see this by examining your conclusion. In your conclusion you rejected assumption (2) instead of assumption (1). Why did you do that? In fact it was mere whim on your part, and that is the weakness of a reductio.*

    * A reductio requires special background conditions. In this case it would require the background condition that (1) is more plausible than (2).
    Leontiskos

    That the conclusion of a reductio is not formally provable is the first hint that the implication in Banno's formula is not the implication of reductio ad absurdum. If it were then a reductio would be formally provable.

    This is related to Lionino's point about the associativity of the → operator in the case of a contradiction:

    I think I finally solved my own problem. When translating it to natural language, I was misplacing the associativity of the → operator in this case.
    So ¬(A → (B∧ ¬B)) is the same as (¬A) → (B∧ ¬B), which may be read as "Not-A implies a contradiction", it can't read as "A does not imply a contradiction". We would have to say something like A ¬→ (B∧ ¬B), which most checkers will reject as improper formatting, so we just say A → ¬(B∧ ¬B), which can be read as "A implies not-a-contradiction", more naturally as "A does not imply a contradiction".
    Lionino

    And it is also related to Tone's point

    "If A implies B & ~B, then A implies a contradiction" is true, but it is a statement about the sentences, not a translation of them.TonesInDeepFreeze

    If "A implies a contradiction" were a translation of the sentences, then it would mean ¬A. That is, it would be formally equivalent to ¬A. This is what Banno thinks his truth table has shown, and such is in line with 's claim that "(p ^ ~p) is false in classical propositional logic," as if we could formally translate a contradiction as "false" (whatever that is supposed to mean). Banno's response to Tones should therefore be, " <If A implies B & ~B, then A implies a contradiction> is true, and it is a translation of the sentences, not a statement about them."


    Or if you think it is only truth-functional if it fits in a truth-table:Banno

    Pasted-Graphic.jpg
  • Leontiskos
    2.8k
    Banno may speak for himself, but I don't know what difference in reference you mean by spelling 'false' without caps and with all caps.TonesInDeepFreeze

    That was my interpretation of Banno, not Banno himself. See the post just above, posted a few seconds ago, for more detail on Banno's view. For instance:

    ...and such is in line with ↪Banno's claim that "(p ^ ~p) is false in classical propositional logic," as if we could formally translate a contradiction as "false" (whatever that is supposed to mean).Leontiskos

    My point in response to you is that by limiting yourself to truth-functional logic you don't know what you mean by a "particular contradiction." Unless you think you can use the word "particular" without having any idea what it would mean for something to be non-particular?
  • TonesInDeepFreeze
    3.6k
    If "A implies a contradiction" were a translation of the sentencesLeontiskos

    It's not a translation of the sentences discussed. That point was recognized by @Lionios who originally claimed it to be a translation.
  • Leontiskos
    2.8k
    It's not a translation of the sentences discussed.TonesInDeepFreeze

    You said:

    "If A implies B & ~B, then A implies a contradiction" is true, but it is a statement about the sentences, not a translation of them.TonesInDeepFreeze

    In saying this are you saying, among other things, that these two claims are not equivalent?

    1. "If A implies B & ~B, then A implies a contradiction"
    2. (a→(b∧¬b))→¬a

    (My claim here is that (1) represents a reductio whereas (2) does not, even though thinks his truth table has shown that (2) translates a reductio.)
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