• TonesInDeepFreeze
    3.8k
    what is actually quoted in the literature from Pierce seems to be about listing permutations of Boolean operators rather than showing truth.Banno

    What do you mean by "showing truth"? The paper shows a Peirce truth table with truth values.
  • TonesInDeepFreeze
    3.8k
    completeness and consistencyBanno

    completeness and soundness.

    consistency follows from soundness.
  • TonesInDeepFreeze
    3.8k


    Another exercise is proving the correctness of the everyday methods for addition, subtraction, multiplication and division. I have a book that shows some of it.
  • Banno
    25k
    The paper shows a Peirce matrix with truth values.TonesInDeepFreeze
    Sure, and that is where it seems to stop. Wittgenstein does the same thing in 5.101. Again, the novelty in the Tractatus is set out here:

    4.45 For n elementary propositions there are Ln possible groups of truth-conditions. The groups of truth-conditions that are obtainable from the truth-possibilities of a given number of elementary propositions can be arranged in a series. 
    4.46 Among the possible groups of truth-conditions there are two extreme cases. In one of these cases the proposition is true for all the truth-possibilities of the elementary propositions. We say that the truth-conditions are tautological. In the second case the proposition is false for all the truth-possibilities: the truth-conditions are contradictory . In the first case we call the proposition a tautology; in the second, a contradiction.

    Here Wittgenstein is pointing out that we can take any "grouping"* of elementary propositions and find if it is contradictory or tautological (or neither) by constructing its truth table.

    I don't see anything like this attributed to Peirce by Anellis. I'd be surprised if Peirce's logic would enable such a construct.

    Curious.

    A side line - I'm finding Google increasingly useless at searching for minutia of late. Any subtlety gets lost in irrelevancies.

    * - a wff.
  • Moliere
    4.7k
    - I'm finding Google increasingly useless at searching for minutia of late. Any subtlety gets lost in irrelevancies.Banno

    Same.
  • TonesInDeepFreeze
    3.8k
    Ludwig Wittgenstein is generally credited with inventing and popularizing the truth table in his Tractatus Logico-Philosophicus, which was completed in 1918 and published in 1921.[2] Such a system was also independently proposed in 1921 by Emil Leon Post.[3]
    — Wiki.
    Banno

    Yet, the Anellis paper says: "[T]he discovery by Zellweger of Peirce’s manuscript of 1902 does permit us to unequivocally declare with certitude that the earliest, the first recorded, verifiable, cogent, attributable and complete truth-table device in modern logic attaches to Peirce, rather than to Wittgenstein’s 1912 jottings and Eliot’s notes on Russell’s 1914 Harvard lectures."

    Anellis might not have adequately made the case for that assertion, as perhaps we would need to see Peirces's papers in more detail. But at least it must be allowed that Anellis may be correct. And, at least we see that Peirce was using truth tables.

    Then you moved to a different claim:

    What is original is that Witti points out how to use a truth table to determine tautology or contradiction.Banno

    But a truth table determines validity. And Peirce was using truth tables to determine validity. And 'valid' and 'tautology' are synonymous for sentential formulas. So Peirce was determining tautologousness.

    Then there's the question of "complete". But what is meant by 'complete' in this context? Does it mean observing that there is no finite bound on the number of letters in a truth table? Is it a given that Peirce didn't observe that and Wittgenstein did?

    The paper shows a Peirce matrix with truth values.
    — TonesInDeepFreeze
    Sure, and that is where it seems to stop.
    Banno

    No, he also showed the 16 truth tables for the 16 binary Boolean functions. And he did more work with truth tables. And we don't know the extent of his work without reading all that he wrote.
  • Banno
    25k
    Sure, all that can be agreed, and yet we still hold that Anellis has not carried his case.

    Is it a given that Peirce didn't observe that and Wittgenstein did?TonesInDeepFreeze
    But a truth table determines validity.TonesInDeepFreeze
    Does Anellis show explicitly that Peirce used a truth table in this way? I don't see that. In the diagram on p.61 he lists some values for three terms. In the diagram on p.62, he lists the possible values for binary connectives.
    Peirce’s object appears to have been to introduce matrices “partly as an aid in his classification of relations, and partly for the sake of illustrations or examples... — p.64
    ...not explicitly for determining the validity of any wff. Now in the absence of further evidence, it is reasonable to supose that Wittgenstein was the first to do this. What is absent is something showing that it had occurred to Peirce that the validity of a given wff can be shown by setting out it's truth table. Wittgenstein does set that out.

    Now yes, it might have been that Peirce did understand this, but that is surmise.

    So if you like, I overstated the case with "Wittgenstein created truth tables", but maintain that their present use has more of Wittgenstein about it than Peirce. You are welcome to differ.

    I suggest it is time to move on.
  • Leontiskos
    3.1k
    1. A→(B∧¬B) assumption
    2. A assumption
    3. B∧¬B 1,2, conditional proof
    4. ~A 2, 3 reductio
    Banno

    1. A -> (B & ~B) {1}
    2. A {2}
    3. B & ~B {1, 2}
    4. ~A {1}
    TonesInDeepFreeze

    The reason these are not RAA is because there is no supposition taking place (and again, Tones' original attempt in this thread did not suffer from this problem). Banno and Tones will not understand RAA until they understand that the first step of the reductio portion of a proof (the "supposition" or "assumption") is different from a premise.

    For example:

    [ ∴ (P v ~P)
    1. __Suppose: ~(P v ~P)
    2. __∴ ~P {from 1}
    3. __∴ P {from 1}
    4. ∴ (P v ~P) {from 1; 2 contradicts 3}

    Rho is assumed and Mu is supposed, and if someone doesn't know the difference between an assumption/premise and a supposition then they won't understand a reductio.Leontiskos
  • Leontiskos
    3.1k
    An odd thing to say, since a contradiction will have "F" all the way down it's main operatorBanno

    The question here is the validity of a conclusion. See:

    Tones thinks that ¬(1) and ¬(2) both follow from (1, 2, 3).Leontiskos

    A truth table does not adjudicate between (1) and (2). It does not perform the and-elimination of the reductio for us. What Tones is doing is just arbitrarily ignoring inputs to the truth table:

    If you want to bring clarity you should explain what inference you used to draw (4). As it happens, truth tables don't adjudicate contradictions. I don't get to say:

    1. P→Q
    2. P
    3. ~Q
    4. ∴ Q {See truth table for 1, 2; avert eyes from 3 at all costs. I repeat: do not allow 3 a seat at the truth table!}

    (The fact that you think this sort of thing can be adjudicated by a truth table is proof that non-truth-functionality is in your blind spot.)
    Leontiskos

    To say ∴Q instead of ∴~P is to selectively consider the truth table for (1, 2), rather than the truth table for (1, 3). To think that a truth table settles the matter is to ignore the contradiction, which in this case is present in (1, 2, 3).
  • Banno
    25k

    Quoting only yourself might indicate a failure to address one's interlocutors. But I could make no sense of that last post anyway. Pointing out your errors has become too sad to continue. Especially if you are going to repeat things already refuted.
  • Leontiskos
    3.1k
    - At this point it is a very real question, whether you are even capable of reading at all.

    The simple version, for your benefit:

    Two premises and an inference:

    1. A -> (B & ~B) {1}
    2. A {2}
    3. B & ~B {1, 2}

    What can be drawn from these claims? <According to Tones> one can draw two different, contradictory conclusions (and this pertains to the misunderstanding of RAA). Further, when I ask Tones why he drew one conclusion rather than the other, he tells me to look at the truth table, which is the sort of nonsensical statement that I had thought only you were capable of, for earlier in the thread you had to stick your foot in your mouth any number of times over this same issue.
  • Banno
    25k
    Brilliant stuff. Seems we are done here, unless you have something substantive to say?

    Edit: Nice to see you have taken to quite radical edits of previous post.
  • Banno
    25k
    Do (A implies B) and (A implies notB) contradict each other?flannel jesus

    Just to sum up, here's the truth table:
    image.png
    If they did contradict each other, the third column would all be F's.

    So as things stand, 41% of folk got it wrong. Pretty sad.

    Here's an actual contradiction, for comparison:
    image.png
  • Leontiskos
    3.1k
    So as things stand, 41% of folk got it wrong. Pretty sad.Banno

    That rare combination of hubris and senility. Gotta love it.

    I would suggest reading Lionino's first post on page 1, but that would require reading. I see that in the last page or two you managed to misread all sorts of things re: Peirce and Wittgenstein.
  • Banno
    25k
    Well, thank you for your erudite and productive contributions. Without you, this thread would have been much abridged.
  • Leontiskos
    3.1k
    - Without you and Tones the thread would have been filled with good-faith argumentation, and that's a sobering fact. Can you at least answer a simple question, or is that too much?

    Is this statement true or false:

    Tones thinks that ¬(1) and ¬(2) both follow from (1, 2, 3).Leontiskos
  • Leontiskos
    3.1k
    Wrong. By the definition of 'valid' in context of classical logic, they are valid.TonesInDeepFreeze

    According to what definition are both proofs valid?

    Or if you like, when I asked what rule of inference allows you to draw (4), you simply said, "RAA." What do you suppose the inference rule "RAA" means? "RAA" is no answer at all, and appealing to the mere name begs the question at hand.

    On my reading reductio ad absurdum ("reduction to absurdity") is the idea that a supposition can be rejected if it leads to an absurdity. What is necessary for a reductio is <an isolated supposition> which can then be "reduced" to an absurdity. Without a supposition there can be no reductio, for premises are equal one to another. If two premises contradict then our system is inconsistent. A reductio is not what happens when premises contradict.
  • Lionino
    2.7k
    So as things stand, 41% of folk got it wrong.Banno

    It was a misclick from my part. :confused:
  • TonesInDeepFreeze
    3.8k
    all that can be agreed, and yet we still hold that Anellis has not carried his case.Banno

    As far as I can tell, Anellis outlines the case but does not make it fully. I have not claimed that Peirce developed the notion of truth tables sufficiently to deserve being called 'the inventor of the truth table regarding its use for testing validity of formulas with any finite number of letters'. Only that it is arguably the case that he did per Anellis's conclusion that is based on grounds adduced by Anellis but that seem to me to require more evidence from Peirce's writings. So I wouldn't so strongly conclude that Peirce did not develop the notion of truth tables sufficiently to deserve being called ''the inventor of the truth table regarding its use for testing validity of formulas with any finite number of letters'.

    TonesInDeepFreeze
    But a truth table determines validity.
    — TonesInDeepFreeze
    Does Anellis show explicitly that Peirce used a truth table in this way?
    Banno

    He does say:

    "Richard Zach [...] reminds us that “Peirce, Wittgenstein,
    and Post are commonly credited with the truth-table method of determining
    propositional validity.”

    and

    "Lane [...] [tells us] that, “For many years, commentators have
    recognized that Peirce anticipated the truth-table method for deciding whether
    a wff is a tautology.”"

    Again, I'd need to see the writings.

    Now in the absence of further evidence, it is reasonable to supose that Wittgenstein was the first to do this.Banno

    And reasonable to be more cautious pending finding out more about Peirce's writings.
  • Lionino
    2.7k
    Nothing special about truth tables. For all we know some monk in China in 1030AD or Iranian in the Muslim Golden Age used some but we don't know because nobody here speaks Mandarin or Farsi.

    Calculus was invented by two people simultaneously. The Zermelo-Russell paradox was found by three different people independently. I think that putting 0s and 1s on a grid is much less special than those two.

    Our usage of them however does come from Wittgenstein.
  • TonesInDeepFreeze
    3.8k
    1. A -> (B & ~B) {1}
    2. A {2}
    3. B & ~B {1, 2}
    4. ~A {1}
    — TonesInDeepFreeze

    The reason these are not RAA is because there is no supposition taking place
    Leontiskos

    The proof is RAA since it fulfills the definition of RAA. I've shown that several times already.

    /

    Tones thinks that ¬(1) and ¬(2) both follow from (1, 2, 3).Leontiskos

    My reply was and is:

    1. A -> (B & ~B) {1}
    2. A {2}
    3. B & ~B {1, 2}
    4. ~A {1}

    1. A -> (B & ~B) {1}
    2. A {2}
    3. B & ~B {1, 2}
    4. ~(A -> (B & ~B)) {2}

    Look at the proofs exactly. They show that ~(A -> (B & ~B)) follows from (2), and ~A follows from (1).

    Of course, (1) and (2) together are inconsistent, so both ~(A -> (B & ~B)) and ~A follow from (1) with (2).

    Meanwhile, it's not needed to mention (3) since it comes merely by inference from (1) and (2).

    /

    A truth table does not adjudicate between (1) and (2). It does not perform the and-elimination of the reductio for us. What Tones is doing is just arbitrarily ignoring inputs to the truth table:Leontiskos

    My reply to that was and is:

    It goes without saying that there is no rule of inference that forces one rather than the other.
    — Leontiskos

    Yes, in the exact sense that there is no inference rule that dictates what set G must be.
    TonesInDeepFreeze

    And, regarding a different example (modus ponens) given by the poster, but applicable to the current examples:

    There are two separate things: the deduction system, (such as natural deduction) which is syntactical, and truth evaluation (such as truth tables), which is semantical.

    But we have the soundness and completeness theorem that states that a formula P is provable from a set of formulas G in the deduction system if and only if there is no row in the truth table such that all the members of G are true and P is false.

    The proof would be this:

    1. P -> Q {1}
    2. P {2)
    3. Q {1, 2}

    The rule applied there is modus ponens:

    From P and P -> Q, infer Q and charge it with all lines charged to P and to P -> Q.

    The truth table would be:

    P true, Q true ... P -> Q true

    P true, Q false ... P -> Q false

    P false, Q true ... P -> Q true

    P false, Q false ... P -> true

    There are two rows in which both P and P -> Q are true, and Q is true in both of those rows. There is no row in which both P and P -> Q are true but Q is false, so modus ponens is valid.

    But if you want to include ~Q:

    "P→Q
    P
    ~Q
    ∴ Q {See truth table for 1, 2; avert eyes from 3 at all costs. I repeat: do not allow 3 a seat at the truth table!}"
    — Leontiskos

    There's no "avert eyes", "don't allow 3 a seat"

    1. P -> Q {1}
    2. P {2}
    3. ~Q {3}
    4. Q {1, 2}

    or if we are required to use (3):

    1. P -> Q {1}
    2. P {2}
    3. ~Q {3}
    4. Q {1, 2}
    5. Q & ~Q {1, 2, 3}
    3. Q {1, 2, 3}

    Truth table:

    P true, Q true ... P -> Q true ... ~Q false

    P true, Q false ... P -> Q false ... ~Q true

    P false, Q true ... P -> Q true ... ~Q false

    P false, Q false ... P -> Q true ... ~Q true

    There are no rows in which P -> Q, P, and ~Q are all true and Q is false. The argument from {P -> Q, P, ~Q} to Q is valid. ~Q is in the truth table.[/quote]

    /

    when I ask Tones why he drew one conclusion rather than the other, he tells me to look at the truth tableLeontiskos

    Actually, I've given the poster a full explanation. I'll give it again, since he is unwilling to read what he even responds to:

    One conclusion is from one set and the other conclusion from another set.

    ~A is syntactically inferred from the set {A -> (B & ~B)} per RAA.

    ~(A -> (B & ~B)) is syntactically inferred from the set {A} pre RAA.

    Those are two different inferences, and both of them are valid.

    To see that they are valid, we can look at the truth tables.

    Every row in the truth table in which the member of {A -> (B & ~B)} is true is a row in which "~A" is true.

    Every row in the truth table in which the member of {A} is true is a row in which "~(A -> (B & ~B))" is true.

    /

    Wrong. By the definition of 'valid' in context of classical logic, they are valid.
    — TonesInDeepFreeze

    According to what definition are both proofs valid?
    Leontiskos

    By the definition I posted in this thread probably at least three times. Again:

    An inference from a set of formulas G to a formula P is valid
    if and only if
    every interpretation in which all the members of G are true is an interpretation in which P is true.

    For sentential logic, that is equivalent with:

    An inference from a set of formulas G to a formula P is valid
    if and only if
    Every row in the truth table in which all the formulas in G are true is row in which P is true.

    Hey, Leontiskos, read!
  • Leontiskos
    3.1k
    By the definition I posted in this thread probably at least three times. Again:

    An inference from a set of formulas G to a formula P is valid
    if and only if
    every interpretation in which all the members of G are true is an interpretation in which P is true.

    For sentential logic, that is equivalent with:

    An inference from a set of formulas G to a formula P is valid
    if and only if
    Every row in the truth table in which all the formulas in G are true is row in which P is true.
    TonesInDeepFreeze

    So then looking at either example:

    1. A -> (B & ~B) {1}
    2. A {2}
    3. B & ~B {1, 2}
    4. ~A {1}

    1. A -> (B & ~B) {1}
    2. A {2}
    3. B & ~B {1, 2}
    4. ~(A -> (B & ~B)) {2}
    TonesInDeepFreeze

    Or:

    1. P -> Q {1}
    2. P {2}
    3. ~Q {3}
    4. Q {1, 2}
    TonesInDeepFreeze

    Either way, your claim is not fulfilled. In the first two arguments (4) does not follow from the truth of (1) and (2), and in the third argument (4) does not follow from the truth of (1), (2), and (3). As I've noted multiple times, your appeal to the "truth table" involves an arbitrary selection of certain premises and an arbitrary exclusion of others.

    For example, in the third example the "truth table" would be one where the first three premises are true, and no such truth table exists. Given that it doesn't even exist, it surely is not going to help us in the way you claim that it will:

    To say ∴Q instead of ∴~P is to selectively consider the truth table for (1, 2), rather than the truth table for (1, 3). To think that a truth table settles the matter is to ignore the contradiction, which in this case is present in (1, 2, 3).Leontiskos
  • TonesInDeepFreeze
    3.8k


    I'm replying to a bot programmed to not understand anything about this subject, not even to understand the inference rules of sentential logic nor how to read a truth table, and to skip recognition of replies already given.

    Again:

    Every row in which "A -> (B & ~B)" is true is a row in which "~A" is true.

    Every row in which "A" is true is a row in which "~(A -> (B & ~B))" is true.

    Every row in which both "P -> Q" and "P" are true is a row in which "Q" is true.

    And the third example is not vitiated by the fact that (1), (2), (3) together are inconsistent:

    (a) The inference is from (1) and (2) to (4). The fact that (3) is inconsistent with (1) and (2) does not entail that (1) and (2) do not imply (4).

    (b) I previous showed a truth table that does include a column for ~Q. And again, in that truth table:

    Every row in which both "P -> Q" and "P" are true is a row in which "Q" is true.

    Moreover, indeed, there are no rows in which "P -> Q", "P" and "~Q" are all true, so, vacuously, every row in which "P -> Q", "P" and "~Q" are all true is a row in which "Q" is true.

    Moreover, "Every row in which all the Xs are true is a row in which Y is true" is equivalent with "The is no row in which all the Xs are true and Y is false". In this case, there is no row in which "P -> Q", "P" and "~Q" are true and "Q" is false.

    Leontiskos, if you're going to critique basic beginning logic, then know what you're critiquing!
  • TonesInDeepFreeze
    3.8k
    Referring to another poster, Leontiskos wrote:

    That rare combination of hubris and senility. Gotta love it.Leontiskos

    Don't gotta love the crude, disgusting ageism there, no matter what the other poster's age is.
  • Banno
    25k
    Leon has lost much of his credibility in this thread. You have been remarkably patient and persistent.
  • Leontiskos
    3.1k
    Leon has lost much of his credibility in this thread. You have been remarkably patient and persistent.Banno

    From the moment Tones entered the thread there have been complaints about the way he comports himself. His infantile <third-person> nonsense was the most recent chapter in this book.
  • TonesInDeepFreeze
    3.8k
    Go back and see.Leontiskos

    Indeed! Look at the actual posts to see who "comported themself" how.
  • Banno
    25k
    there have been complaints about the way he comports himselfLeontiskos
    Were there any that were not from you?

    Seems that you cannot coherently address what he says, so you attack the man.
  • TonesInDeepFreeze
    3.8k


    Actually, there've been other first insulters in this thread.
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