• Do (A implies B) and (A implies notB) contradict each other?
    1. a → (b ∧ ~b)
    2. If b is true (b ∧ ~b) is false. If b is false (b ∧ ~b) is false, so (b ∧ ~b) is false.
    3.~a → ~(b ∧ ~b) - contraposition (1)
    4. ~a - modus ponens (2,3)     — Count Timothy von Icarus

    Modus Ponens is ρ→φ, ρ, ⊢ φ.

    Not seeing it in 4.
  • Do (A implies B) and (A implies notB) contradict each other?
    What I have consistently said is that reductio is not valid in the same way that a direct proof is. — Leontiskos

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

    While you are there, what does "FALSE" mean?
  • Do (A implies B) and (A implies notB) contradict each other?
    To me, RAA depends on modus tollens. — Lionino

    What does "depends on" do here?

    Modus Tollens: ρ→φ, ~φ ⊢ ~ρ
    RAA: ρ→(φ^~φ) ⊢ ~ρ

    Both are equally useable rules of inference.
  • Do (A implies B) and (A implies notB) contradict each other?
    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. — Leontiskos

    And yet your claim that Reductio is invalid is just wrong.
  • Do (A implies B) and (A implies notB) contradict each other?
    The grain of truth in @Leontiskos' position is that reductio arguments need to be used with care. If we have a bunch of assumptions that lead to a falsehood, we can throw out any and all of the assumptions.

    This has a place in many philosophical critiques. For example, it underpins the Duhem–Quine thesis, and greatly complicated Popper's fablsificationism.

    Not quite so in formal systems. In classical logic if assumptions a,b,c,d lead to a contradiction we can only conclude that their conjunct is false.

    But this allows is also to conclude certain implications. We might infer for example that a^b^c→~d. That is, if a, b, and c are true, it must be d that is false.

    Worth keeping in mind
  • Do (A implies B) and (A implies notB) contradict each other?
    ↪Lionino

    So here's the apparent problem:
    A, A→¬B∧B ⊢ ¬A
    A, A→¬B∧B ⊢ A
    It seems we can infer both A and ~A from the same thing. But that's because the two assumptions, A and A→¬B∧B, are inconsistent.
  • Do (A implies B) and (A implies notB) contradict each other?
    ↪Leontiskos
    I don't see where I require anything like that in that post. After all, it's your post.

    It might help if you explained what FALSE is.

    As it stands, I can't take much more of this gobble. It's clear that reductio is valid as used in classical logic.
  • Do (A implies B) and (A implies notB) contradict each other?
    truth incarnate... — Leontiskos
    :grin:

    Presumably that's TRUE? Or is it 'TRUE'?
  • Do (A implies B) and (A implies notB) contradict each other?
    ↪Lionino

    Well, first, ρ,μ ⊢φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ) is nonsense. I stuffed up. Am I allowed to edit it? :wink:

    It should be
    (ρ^μ) →φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ)
  • Do (A implies B) and (A implies notB) contradict each other?
    ↪Lionino
    :lol:

    I think you have here managed to set out Leo's confusion far more clearly than has Leo.

    Can you see the answer?
  • Do (A implies B) and (A implies notB) contradict each other?
    A→(B∧¬B)
    ¬(B∧¬B)
    ∴ ¬A     — Leontiskos
    If ¬(B∧¬B) is true, as it must be, then this is not a valid use of modus tollens.

    Again, as i pointed out previously, you are comparing two very different arguments. But further your case is not helped by your not setting out the inferential steps in each case. Indeed, your new first example is not well-formed in classical logic. Why is the second line a quote? Was that in error (if so, then I suggest you edit it...). And if you were to modify it so as to be well-formed, what would it look like, if not ¬(B∧¬B)?

    And how is "(B∧¬B) is false" and example of FALSE? Because of the quotes? What do they do?

    What you write here is just a muddle.
  • Do (A implies B) and (A implies notB) contradict each other?
    The post
    ↪Leontiskos
    does not show that I said I 'preferred the reductio to the modus tollens".

    And what I am after is a straight forward explanation of what "FALSE" is. Referring back to old posts that do not give a clear explanation anyway is in no way helpful.
  • Do (A implies B) and (A implies notB) contradict each other?
    you said you preferred the reductio to the modus tollens. — Leontiskos
    Rubbish.

    Leo, what is "FALSE"?
  • Do (A implies B) and (A implies notB) contradict each other?
    So I ask again: How is it that something and its negation can both [function as the second premise of a modus tollens]? — Leontiskos

    It doesn't. Explained last time you made this claim...

    Oh, and what is "FALSE"?
  • Do (A implies B) and (A implies notB) contradict each other?
    I think Leontiskos is talking about choosing between the conjuncts, while Banno is correctly stating that reduction ad absurdum is formally valid. — Lionino

    I'm glad you followed this.
  • Do (A implies B) and (A implies notB) contradict each other?
    ↪Leontiskos
    , what is "FALSE"?

    It's your term.

    This conversation is increasingly inane. Again, I seem to have reduced you to reciting gobbledygook.
  • Do (A implies B) and (A implies notB) contradict each other?
    I literally said it was an interpretation, not a translation. — Leontiskos
    Ok, Presenting a statement that someone has not made is not presenting an interpretation. :roll:


    This does not contradict what I have been saying. — Leontiskos
    Quite so. So what? It remains that RAA is a valid inference in classical propositional logic.
  • Do (A implies B) and (A implies notB) contradict each other?
    What is the supposed difference between "false" and "FALSE"?
  • Do (A implies B) and (A implies notB) contradict each other?
    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
    Neither do I. This distinction between false and FALSE is not my doing. It seems to be another case of Leontiskos confabulating arguments on the part of those who disagree with him.
    That was my interpretation of Banno, not Banno himself. — Leontiskos
    Presenting a statement that someone has not made is not presenting a translation.
    I think your charges of "misrepresentation" are all bosh — Leontiskos
    I agree with Tones that you habitually misrepresent positions that are counter to your own, here and elsewhere.
    _________________

    Has everyone agreed by this point that ↪Banno's truth table does not fully capture what a reductio is? — Leontiskos
    I'll agree with that. It is incomplete. As Tones pointed out RAA is an inference rule, not a sequent within classical propositional logic. The inference allows one to infer ~ρ given a proof of (μ ^ ~μ) with ρ as assumption, a form displayed in the truth table.
    The easiest way to see this is to note that a reductio ad absurdum is not formally valid — Leontiskos
    This is rubbish. Given a proof of B and ~B from A as assumption, we may derive ~A as conclusion. This is the form of reductio inferences and is quite valid.


    I think that what has Leo worried is the notion that in an informal reductio with multiple assumptions, we have to have grounds for choosing which assumption we deny. So for example if we have assumption A and assumption B and assumption C, and from these we infer some contradiction, we then have the option of rejecting any or all of the assumptions, and a choice to make.

    This is not the case in formal uses of reductio.

    Given ρ,μ ⊢φ^~φ, we can write that ρ→~μ or we can write that μ→~ρ. (Tree proof 1; Tree proof 2)

    Leo seems to think that choosing between ρ→~μ and μ→~ρ somehow involves an act of will that is outside formal logic. He concludes that somehow reductio is invalid. His is a mistaken view. Either inference, ρ→~μ or μ→~ρ, is valid.

    Indeed, the "problem" is not with reduction, but with and-elimination. And-elimination has this form
    ρ^μ ⊢ρ, or ρ^μ ⊢μ. We can choose which inference to use, but both are quite valid.

    We can write RAA as inferring an and-sentence, a conjunct:

    ρ,μ ⊢φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ)

    (ρ^μ) →φ^~φ⊢ (ρ→~μ) ^ (μ→~ρ)
    (fixed error)

    ...and see that the choice is not in the reductio but in choosing between the conjuncts.

    Leo is quite wrong to assert that Reductio Ad Absurdum is invalid.


    _________________
    For the folks following along at home, the greek letters allow us to write about the sentences of classical propositional logic. We can substitute for any greek letter, consistently, a well formed formula from that logic. "⊢" is read as "infer". or "we can write" So we can set out modus ponens as

    ρ,ρ→μ ⊢μ

    Read this as "given rho and rho implies mu, infer mu". Substitute any WFF from classical logic into this form, consistently, and you will have a valid inference.

    More often folk will use capital letters instead of greek, but here I thought it useful to seperate these out from the use of capital letters in the OP
  • An Argument for Christianity from Prayer-Induced Experiences
    All of your premises are wrong except for number 1. — Lionino

    And even if they were right, the conclusion does not obviously follow - indeed, it is very unclear what the structure of the argument is.

    SO I supose one question is, can such an argument be constructed?
  • Do (A implies B) and (A implies notB) contradict each other?
    It depends upon the values given to the variables. — creativesoul

    Hello, creative. How are the fish hooks?

    It exactly does not depend on the values given to the variables. That's kinda the point of using variables - you get to put different things in and get the same result.

    So a+b = b+a regardless of what number you stick in to the formula, and a^(a→b)⊢b regardless of what statement you put in, too. Or so it is supposed to go.
  • Do (A implies B) and (A implies notB) contradict each other?
    ..we end up with ¬(a→(b∧¬b)), and this can't be read as "It is not the case that a implies a contradiction" — Lionino
    Why not? I'm not seeing the issue here.
  • Do (A implies B) and (A implies notB) contradict each other?
    So, since ¬◇(a→(b∧¬b)) would be read by many as "It is not possible that A implies a contradiction", is that the same thing as "It is necessary that not-A implies a contradiction"? — Lionino

    □¬(a→(b∧¬b)) would be "It is necessarily not the case that A implies a contradiction"
  • Do (A implies B) and (A implies notB) contradict each other?
    ↪Lionino


    Isn't it something like that "if it is not possible that A implies a contradiction, then A is necessarily true"?

    Or "If in no possible world A implies a contradiction, then A is true in every possible world"?
  • Do (A implies B) and (A implies notB) contradict each other?
    ↪flannel jesus
    Yep.

    ↪Lionino
    I gather you worked through this? Nice.

    ↪flannel jesus
    Leo does that sort of thing - claims you have said something you haven't, if it suits his purposes.
  • Do (A implies B) and (A implies notB) contradict each other?
    (It would seem that you are wrong in claiming that classical logic treats contradictions as false. — Leontiskos
    Again, no.

    Pasted-Graphic-1.png

    F's all the way down.
  • Do (A implies B) and (A implies notB) contradict each other?
    Here's another outright error.
    A reductio is not truth-functional. — Leontiskos

    Given a proof of B and ~B from A as assumption, we may derive ~A as conclusion — Lemmon

    Or if you prefer: φ→(ψ^~ψ)⊢~φ

    Or if you think it is only truth-functional if it fits in a truth-table,
    Pasted-Graphic.jpg

    At some point one has to realise that Leo has such an odd notion of logic that he is unreachable.
  • Do (A implies B) and (A implies notB) contradict each other?
    I have already responded to these charges. — Leontiskos
    Maybe not as much as you think.

    At this point you either have an argument for "∴¬A" or you don't. Do you have one? If not, why are you still saying that ¬A is implied? — Leontiskos
    I'm not seeing a salient point here. Pretty demonstrably, you have made a series of claims that have been shown to be in error.

    At this stage it is unclear what your general point concerning "metalogic" might be, beyond an "esoteric mystery".
  • Do (A implies B) and (A implies notB) contradict each other?
    ...but they have no way of knowing when their logic machine is working and when it is not. — Leontiskos

    "Machine", singular. So back to my point, that
    Each of these systems sets out different ways of dealing with truth values. How the truth value of a contradiction is treated depends on which of these systems is in play. — Banno
    and so
    Asking, as you do, how to treat the truth value of a contradiction apart from the system that sets out how a truth value is to be dealt with makes little sense. — Banno

    What I hope to have done over the last page is to show that you are mixing logics, resulting in your own confusion. You do not succeed in showing that "...something and its negation can both be false" in classical propositional logic.
  • Do (A implies B) and (A implies notB) contradict each other?
    A reductio is as much a proof in classical propositional logic as is modus tollens. — Banno
  • Do (A implies B) and (A implies notB) contradict each other?
    ↪Leontiskos
    I seem to have reduced you to reciting gobbledygook. My apologies.
  • Do (A implies B) and (A implies notB) contradict each other?
    ↪Leontiskos
    What?

    The reductio shows that A→(B∧¬B)⊢~A. As
    ↪hypericin
    pointed out.

    It could equally be used to show that A⊢~A→(B∧¬B); but that was not the issue you raised.

    Edit: correction
  • Do (A implies B) and (A implies notB) contradict each other?
    a reductio is an indirect proof which is not valid in the same way that direct proofs are. — Leontiskos
    A reductio is as much a proof in classical propositional logic as is modus tollens.
    In your conclusion you reject (2) instead of (1). Why do you do that? — Leontiskos
    Simply because I matched your example, which has
    A→(B∧¬B)
    ∴ ¬A     — Leontiskos
    and not ~A⊢A→(B∧¬B).


    Again, don't blame me for your problems.

    edit: corrected A⊢A→(B∧¬B)/~A⊢A→(B∧¬B)
  • Do (A implies B) and (A implies notB) contradict each other?
    This answer proves that you do not understand the questions that are being asked. If one wants to understand what is being discussed here they will be required to set aside their ready-made answers. — Leontiskos
    Ah, so it's an esoteric mystery. :wink:

    Nowhere in that post do you affirm (B∧¬B).
    — Banno
    I never said I did. Read again what you responded to. "     — Leontiskos
    The consequent is (B∧¬B)
    The negation of the consequent is ~(B∧¬B)
    Affirming the negation of the consequent is ⊢~(B∧¬B)
    if you don't affirm the negation of the consequent, you affirm (B∧¬B).

    Nowhere do you do this. Nowhere in these examples is it the case that "...something and its negation can both be false". That is, you do not show that somehow classical propositional logic affirms both
    ~(B∧¬B) and (B∧¬B).

    Indeed, while your second example is a case of modus tollens, this is not clear for the first.

    (The second is
    1. A→(B∧¬B) assumption
    2. ¬(B∧¬B) assumption
    3. ¬A 1,2 modus tollens)

    Modus Tollens tells us that "Given ψ→ω, together with ~ω, we can infer ~ψ". In the first example you do not have ~ω. It might as well be a Reductio, although even there it is incomplete. It should be something like:

    1. A→(B∧¬B) assumption
    2. A assumption
    3. B∧¬B 1,2, conditional proof
    4. ~A 2, 3 reductio

    ans so A→(B∧¬B)⊢~A

    _________________
    And yes, A→FALSE is not well-formed in classic propositional logic. So if your first example is to be understood as using MTT, it is not an example from classic propositional logic. Again, that is not something I have supposed, and you misattribute it.

    Which takes us back to what I pointed out earlier - you are mixing various logical systems. The equivocation here is on your part. Don't put the blame for your poor notation on to me.

    (Edit: Actually, Open Logic builds propositional logic from, amongst other things, ⊥. (Definition 7.1). And v(⊥) = F - the valuation of ⊥ is "false" - in Definition 7.15. In this sort of build, φ → ⊥ could be well formed. @TonesInDeepFreeze might be able to clarify.)
  • Do (A implies B) and (A implies notB) contradict each other?
    I've been ignoring Tones... — Leontiskos
    Your loss.

    I think the thread shows that this is not true. — Leontiskos
    Then the thread is in erorr. (p ^ ~p) is false in classical propositional logic.

    The problem here is that your answer lacks specificity — Leontiskos
    Not at all. A contradiction in first order predicate logic is an expression of the form (φ ^ ~φ). It is not an expression of the form ~φ. The lack of specificity here is your attempt to make use of a notion of contradiction that is not found in classical propositional logic.

    How is it that something and its negation can both be false? — Leontiskos
    Whether or not we affirm the negation of the consequent... — Leontiskos
    Nowhere in that post do you affirm (B∧¬B).
  • Do (A implies B) and (A implies notB) contradict each other?
    So far as I can see, it was you who proffered
    the notion of contradiction in its entirety — Leontiskos
    I'm puzzling over what this might be.

    pray tell how a contradiction is to be dealt with in classical propositional logic? — Leontiskos
    As has been explained at length, in classical propositional logic contradictions are false.

    ...you seem to be implying that, according to the logic, one person is right and one person is wrong when they disagree about whether a given instance of (b∧¬b) should be treated as a proposition/variable or as a simple truth value. — Leontiskos
    Another example of your practice of misattributing stuff to your interlocutors - as you did with
    ↪TonesInDeepFreeze
    . What I said is that the disagreement here is as to which system is in play. Hence there is no absolute answer as to which view is "right".
  • Do (A implies B) and (A implies notB) contradict each other?
    A contradiction is a contradiction. It is neither true nor false. It is the basis for both truth and falsity. — Leontiskos
    This seems to be the source of your difficulties.

    As has been explained, in classical logic a contradiction is false. Dialetheism considers what must be the case if some contradictions are considered to be true.The various paraconsistent logics consider what must be the case if A, ~A ⊨ B; that is, if contradictions are not explosive.

    All this to say that there are various ways to treat truth values, each with its own outcomes. (the list is not meant to be exhaustive - there are other options)

    Each of these systems sets out different ways of dealing with truth values. How the truth value of a contradiction is treated depends on which of these systems is in play.

    Asking, as you do, how to treat the truth value of a contradiction apart from the system that sets out how a truth value is to be dealt with makes little sense. It does not make much sense to speak of "the notion of contradiction in its entirety".
  • Do (A implies B) and (A implies notB) contradict each other?
    "The car is green" and "The car is red" is not a contradiction. But if we add the premise: "If the car is red then the car is not green," then the three statements together are inconsistent. That's for classical logic and for symbolic rendering for classical logic too. — TonesInDeepFreeze
    Yep. Worth noting that parsing this correctly shows that the original was incomplete - implied nothing.

    More generally, parsing natural languages in formal languages, while not definitive, does occasionally provide such clarification. That's kinda why we do it.

    Also worth noting that (A → B) ^ (A → ¬B), while not a contradiction, does imply one, given A:

    (A → B) ∧ (A → ¬B)→(A→(B∧¬B))

    So in answer to the OP
    Do (A implies B) and (A implies notB) contradict each other? — flannel jesus
    Taking "implies" as material implication, they are not contradictory but show that A implies a contradiction.

    I'd like to see what formation rules you come up with. — TonesInDeepFreeze
    I had the same thought when I read that. It's wellformed. It is also invalid: A∧¬A

    This thread is bringing out some rather odd attitudes towards the relation between logic and natural languages.
  • A Case for Moral Anti-realism
    ...but are refusing to make sense of them... — Michael

    An obligation is simply something you ought to do. Your inability to make sense of obligation is not our problem. Eventually this reduces to a personal psychological issue.

    But I wonder how widespread this inability is, and what place it plays in odd political ideals.

    Edit: I was unable to make anything of this:

    So the proper comparison would be:

    1. You were given an order
    2. Do this

    I have no problem with (1). Is this all "you ought do this" means?     — Michael
  • A Case for Moral Anti-realism
    1. You ought do this
    2. Do this
    The first appears to be a truth-apt proposition, whereas the second isn’t. But beyond this appearance I cannot make sense of a meaningful difference between them. The use of the term “ought” seems to do nothing more than make a command seem like a truth-apt proposition.     — Michael

    You conclude that there are no such thing as obligations.

    Compare:
    1. You were asked to give an answer to what we get when we add six and five.
    2. What is six and five?
    The first appears to be a truth-apt proposition, whereas the second isn’t. Beyond this appearance is there a meaningful difference between them? Will you say that the use of the term "asked" seems to do nothing more then make a question seem like a truth-apt proposition?

    Do you also conclude that there are no such things as answers?

    I think not. Answers are brought about by asking questions, just as obligations are bought about by (amongst other things) commands and promises.

    Or this:
    1. She greeted you
    2. "Hello"

    The use of the term "greeted" seems to do nothing more than make "Hello" seem like a truth-apt proposition?

    Will you conclude that there is no such thing as a greeting?

    We bring answers and greetings into existence; they are things we do with words, and a part of our social life. As are obligations.

    Anscombe argued against the moral "ought" found in ethics, but was very clear that there was a place for "non -emphatic ought" apart from a moral sense:
    I will end by describing the advantages of using the word "ought" in a non-emphatic fashion, and not in a special "moral" sense; of discarding the term "wrong" in a "moral" sense, and using such notions as 'unjust'. — MMP, p.13
    There follows a passionate defence of the justice. Your girlfriend did you an injustice when she reneged on the promise she made. It was an injustice because she undertook an obligation to you, which she did not fulfil. One ought fulfil one's obligations, since that is what an obligation is.

    To my eye, this and my last post answer your objection.

    Heading back a few days and a few pages, this was all in answer to your attempted defence of
    There is nothing that exists beyond the act. — AmadeusD
    I will maintain that questions, greetings and obligations are examples of things that exist "beyond the act", along with property, currency, marriage, incorporation, institutionalisation, legality... and a few other things.
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