• Pippen
    80
    @Michael: We are dealing just in the scope of classical (two-valued) logic and there a statement like "This statement is false" is just without a truth-value as far as I know.
  • Michael
    14k
    We are dealing just in the scope of classical (two-valued) logic and there a statement like "This statement is false" is just without a truth-value as far as I know.Pippen

    So not having a truth value is the third option. If we have the conjunction p ∧ q and if p is false and q doesn't have a truth value then the conjunction as a whole is false.
  • Michael
    14k
    Although actually there's also Bochvar's internal three-valued logic which has a different truth table to Łukasiewicz's, and has a conjunction of this kind not having a truth value (e.g. propositions like "thiggledy piggledy and grass is green" are meaningless), which is consistent with your view.

    I suppose which to choose is just a matter of preference.
  • Pippen
    80
    So not having a truth value is the third option. If we have the conjunction p ∧ q and if p is false and q doesn't have a truth value then the conjunction as a whole is false.Michael

    Not in classical logic. Not having a truth value is not a third option there. I think 2. of my proof is dubious, maybe nagase will check that out.
  • Michael
    14k
    Not in classical logic. Not having a truth value is not a third option there. I think 2. of my proof is dubious, maybe nagase will check that out.Pippen

    But your argument rests on your own truth table in which there are three options. You're saying that if p is false (or true) and if q doesn't have a truth value then p ∧ q doesn't have a truth value.

    If there wasn't a third option then q ("this sentence is false") must be either true or false.

    If it helps, don't think of them as truth values but as predicates. You can have "true" as a predicate, "false" as a predicate, or "neither true nor false" as a predicate.

    So the question is on what predicate a conjunction has if one of its operators has "false" as its predicate and the other has "neither true nor false" as its predicate. According to Łukasiewicz such a conjunction has "false" as its predicate and according to Bochvar such a conjunction has "neither true nor false" as its predicate. Unfortunately I don't know enough to determine how one goes about choosing one truth table over another. Maybe it's simply axiomatic. But as you say, perhaps @Nagase has some insight into the matter.
  • Michael
    14k
    However, the above might not even be relevant to the particular issue at hand. It could be that your argument conflates. To explain this, consider the statement "this statement is false and grass is red". There are two different ways to interpret this, depending on what we consider to be the referent of "this statement":

    1. a) "this statement is false" is false and b) grass is red
    2. a) "this statement is false and grass is red" is false and b) grass is red.

    We might say that if "this statement is false" is false then "this statement is false" is true, giving us our paradox . But can we say that if "this statement is false and grass is red" is false then "this statement is false and grass is red" is true, giving us another paradox? I don't think we can. So given that grass isn't red, both 2a and 2b are false, and so the conjunction of 2a and 2b is false.

    I also think that 2 is the correct interpretation of the statement "this statement is false and grass is red".

    Now replace "grass is red" with "every other statement is false".
  • Srap Tasmaner
    4.6k

    Let S be the set of all statements.
    Let z be the string "If x ∈ S, then x is false."
    Assume z ∈ S.
    If z is true, then z is false.
    If z is false, then it is false that if x ∈ S, then x is false.
    Therefore if z ∈ S, z is false, and it is false that if x ∈ S, then x is false.

    Let z* be the string "z* is false and if x ∈ S/z*, then x is false."
    Assume z* ∈ S.
    Let S/z* be the complement of z* in S.
    If z* is true, then z* is false.
    If z* is false, then either z* is true or it is false that if x ∈ S/z*, then x is false.
    Therefore if z* ∈ S, z* is false, and it is false that if x ∈ S/z*, then x is false.
  • Pegasys
    4


    It's true. If "all statements" implies that the statement "all statements are false" is a part of the set of all statements, then "false" is identical. But this does not mean that the statement is false, if the definition of identity is merely reflexive. If however, "identity" is transitive, then this would imply a higher resolution in the proof, implying that more statements are added to the block of proof. And since these statements are elements of the set of all statements, then it follows that some statements which are not on the right side of the equation, are not in the game, but they remain elements of the set of all statements. It follows that the "belief" you mention induces meta-perception, intelligent observation, but this does not imply omni. It only means that self-referentials implicitly define the truth of the statement if "limited truth" is observed by climbing the meta-ladder disregarding the left-hand side, if symmetry is not defined. but how-could it be symmetrical if "identity" is implicit and "false" is not? So it must be true since the interpretation of "all statements" is "false" and "false" is an non-valid intepretation on a lower level but on the level in question it is a "non-valid interpretation" reflecting some symmetrical property of belief dynamics, that is true, and hence a part of the negation of your statement "all statements are false". This implies the importance of the "some" quantifier.

    Human nature has a tendency to contradict in order to control environment. Logic is the web in that its elements are of a dual nature, but they only are dual through the lens the spider uses. If an element becomes conscious of itself, then it can only do so by attaching meaning to the situation, to the set it belongs to, like using Gödel numbering for example. Then this element needs to evoke contradiction in order for its statements to make sense, but only to itself, the rest may challenge the element or not. It's a game for machines and rises the question if machines are conscious or not, amongst other things, dependend on the individual, or machine, or element, dependent on the sociology involved.
  • Owen
    24
    Ep(p).

    (All statements are false) is false, and it is equivalent to
    (Some statements are true).

    1. (All p)(~p) -> ~q.
    2. (All p)(~p) -> ~(~q).
    3. (All p)(~p) -> q.
    4. (All p)(~p) -> (~q & q).
    5. ~(All p)(~p).
    6. (Some p)(p).

    Also,
    (All statement are true) is false, and it is equivalent to
    (Some statements are false).
  • Pippen
    80
    Here is what I found out so far:

    The logicians formulate "All (S)tatements are (F)alse" as follows: All x: (Sx -> Fx). If you do that you can indeed prove that this is just plain false since if it's true it's a contradiction and so by RAA it's its negation that is consistent.

    But why do I have to formulate the statement like above? Why can't I just formulate: All x: (Sx & Fx). This statement is not false, it is not well formed since it entails the liars paradox.

    Both versions of the upper statement say roughly the same - that every x in the set of statements is false - but their form is different and so are their results. So who is right? Or why am I wrong?
  • Srap Tasmaner
    4.6k
    The logicians formulate "All (S)tatements are (F)alse" as follows: All x: (Sx -> Fx). If you do that you can indeed prove that this is just plain false since if it's true it's a contradiction and so by RAA it's its negation that is consistent.

    But why do I have to formulate the statement like above? Why can't I just formulate: All x: (Sx & Fx). This statement is not false, it is not well formed since it entails the liars paradox.

    Both versions of the upper statement say roughly the same - that every x in the set of statements is false - but their form is different and so are their results. So who is right? Or why am I wrong?
    Pippen

    Because and don't say the same thing.
  • BlueBanana
    873
    (All statements are false) is false, and it is equivalent to
    (Some statements are true).
    Owen

    A small error there: ((All statements are false) is false) is equivalent to (some statements are true). (All statements are false) is equivalent to (there are no statements that are true).

    And this is, of course, only considering the classical logic with only two truth values, which we aren't considering here: more correct statement would be ((All statements are false) is false) is equivalent to (some statements are either true or have no truth value). This actually proves OP wrong, because (this statement is false) implies (there is a statement with no truth value), which of course implies (there is a statement that is either true or has no truth value), which then implies ((All statements are false) is false).
  • Pippen
    80
    Because ∀x(Sx→Fx) and ∀x(Sx∧Fx) don't say the same thing.Srap Tasmaner

    Ok, but why is my formulation of "All statements are false" as ∀x(Sx∧Fx) impossible and only ∀x(Sx→Fx) the correct formulation? Is there any reason? Because, again, both formulations lead to different results for the proposition. In case of ∀x(Sx→Fx) it's false, in case of ∀x(Sx∧Fx) it's not well formed.
  • Srap Tasmaner
    4.6k

    One says, if x is a statement, then it is false. There is a longish tradition of so interpreting universal statements (Russell and Ramsey both for slightly different reasons), and thus denying them existential import.

    The other version says, everything is a statement and everything is false. Is that what you want to claim?
  • Pippen
    80
    I assume a set of all statements as the domain. Then indeed ∀x(Sx∧Fx) means: everything is a statement and everything is false. But that's exactly what "all statements are false" says, isn't it? Why can't I formulate it that way? My guess is there is no real reason, it's just an agreement that we formalize universal statements as implications (and use universal sets as domains).

    But the consequences are huge, because if my case is formally possible then ∀x(Sx∧Fx) fails to be true or false in the one and only case of applying it to itself. It'd be false because it says so, it'd be true because it's false and says so. Wouldn't that be enough to make ∀x(Sx∧Fx) a not-wff-formula?
  • Srap Tasmaner
    4.6k

    Read this again. I did it that way specifically to leave room for the conclusion that the claim must not be a statement (i.e., not truth-apt), and that would be the result for the Liar. (You could think of that as concluding "If there's a set of all truth-apt strings, the Liar isn't in it.")

    There was no reason to conclude either version of the claim is not truth-apt; both versions must be false.
  • Srap Tasmaner
    4.6k
    Besides which, there are loads of reasons to think this claim is simply false. You'd have to have a language without the logical constants.

    Even if you changed it to "All atomic sentences of L are false", then the question is: under what interpretation? Systematic falsehood under one interpretation is systematic truth under another.
  • Michael
    14k
    But the consequences are hugePippen

    They're not. Whether you want to say "all statements are false" is false or "all statements are false" is neither true nor false, it is still the case that "all statements are false" isn't true and "at least one statement is true" is true.
  • Pippen
    80
    They're not. Whether you want to say "all statements are false" is false or "all statements are false" is neither true nor false, it is still the case that "all statements are false" isn't true and "at least one statement is true" is true.Michael

    Yeah, but that statement wouldn't be false either! And it's a huge difference if a statement is false or not false (no matter if it's true or not besides that).

    In summary, my problem is why nobody interprets the statement "All sentences are false" as "All sentences are false and this very sentence is false", because in this version the whole thing wouldn't be true or false. I just don't see an error in infering one from the other.
  • Michael
    14k
    And it's a huge difference if a statement is false or not false (no matter if it's true or not besides that).Pippen

    Not really. What difference does it make if we consider the sentence "grass is red and wub a lub a dub dub" to be false or neither true nor false?

    Either way, it isn't true. It doesn't describe some fact about the world.
  • Pippen
    80
    @Michael: It may not be true, but it wouldn't be false either. It would hang in the middle between the two values.
  • Srap Tasmaner
    4.6k
    In summary, my problem is why nobody interprets the statement "All sentences are false" as "All sentences are false and this very sentence is false", because in this version the whole thing wouldn't be true or false. I just don't see an error in infering one from the other.Pippen

    You can make the inference if you like, but the second version is also false, as I showed a month ago. I'll do it again:

    Let S be "S is false and all statements are false".

    Now we try assigning truth values to S to see if it is possible for S to be true or false. We may find that it must be assigned the value "true" in all models of English (that it is a tautology), that it must be assigned the value "false" in all models (that it is a contradiction), or that it can be true in some models and false in others, like most statements, or that it cannot be assigned a truth value in any model (like the Liar).

    1. Assume S is true.
    2. If S is true, then both conjuncts are true.
    3. If both conjuncts are true, then the first is true, so it is is true that S is false.

    This is contradiction, because at the moment we are assuming S is true. So there are no models of English in which S is true. But perhaps S is not false either, as you claim.

    4. Assume S is false.
    5. If S is false, then at least one of the conjuncts is false.

    We now try each conjunct in turn. First conjunct:

    6. Assume it is false that S is false.
    7. If it is false that S is false, then S is true.

    But we assumed in (4) that S is false, so this is a contradiction; thus there are no models of English in which S is false and its first conjunct is false. Second conjunct:

    7. Assume it is false that all statements are false.

    Here at last we have a possible truth-value assignment that doesn't immediately produce a contradiction. Thus all models of English must assign truth values this way: S is false (our assumption 4) and it is the second conjunct that is false, so it is also false that all statements are false. This is the only possible way to assign truth values without the model contradicting itself.

    Your view I think is something like this: the usual way of determining whether a conjunction is a tautology, a contradiction, contingently true or false, or just not truth-apt at all, is to assign all possible truth-value combinations to the conjuncts and use a little truth table to see how the conjunction comes out. An arbitrary P & Q is

    T & T : T
    T & F : F
    F & T : F
    F & F : F

    so it could be either true or false. On the other hand, P & ~P goes like this:

    T & F : F
    F & T : F

    It's always false. So you're thinking that since you have, in essence, "[the Liar] & P" as your conjunction, we'll be unable to construct a truth table because the first conjunct is not truth-apt. True.

    But conjunction is a short circuit with respect to falsehood, as disjunction is with respect to truth. If we can establish that one of a pair of conjuncts is false, we are never forced to evaluate the other conjunct to know that the conjunction as a whole is false. In my version, I never do assign a truth value to the first conjunct, but it doesn't matter. In essence we end up treating it as a pseudo-conjunction: blah-blah-blah P, and P happens to be a contradiction.

    So we could skip most of the steps above and go that way instead. We show that "All statements are false" can only be false, which you agree to, and then short-circuit any conjunction it appears in.

    I claim this is reasonable because whether to count some string of English words as a statement at all, as truth-apt, is up to us. There is no formal system on offer here to tell us whether some string is or isn't well-formed. We can rule out anything we like, but we can only rule in strings we can successfully assign a truth value to. We can't rule in the Liar. But we can rule in "All statements are false" by calling it always false, and we can rule in "This statement is false and all statements are false" by calling it always false. We're not compelled to, but we can.

    ADDED: I would claim further that this approach is reasonable precisely on the grounds that we would ordinarily expect "All statements are false" and "This statement is false and all (other) statements are false" to be equivalent.

    Your version has a truth-apt statement being equivalent to the Liar. If we can avoid that, we ought.

    AND STILL MORE: Note that short-circuiting is consistent: "Dinosaurs are extinct" is contingently true; there are possible models in which it is false. But "Dinosaurs are extinct and all statements are false" is always false. "It's raining or it's not, and all statements are false" is always false, even though "It's raining or it's not" is always true.
  • Jake Tarragon
    341
    In formal logic directly self referential statements like the one you showed do not exist and can not be defined.Meta

    Surely there's a "New Logic" somewhere that incorporates them? An extension of formal logic ...
  • Meta
    185

    I think a better expression instead of self- referential would be self-containing maybe?
    Like the symbol of the meta language denoting a sentence can not be a part of that sentence.

    You must be right but these new logics must not have the same axioms as our classical one and must be weaker in at lest one aspect. (Could be stronger in other aspects.)
  • Pippen
    80
    So you're thinking that since you have, in essence, "[the Liar] & P" as your conjunction, we'll be unable to construct a truth table because the first conjunct is not truth-apt. True.Srap Tasmaner

    Yes, that's my point.

    Now, a conjunction of the form "[liar] & P" is neither a conjunction nor a proposition at all, because a conjunction is formally defined as "p & q" and p,q need to be propositions with truth values. That's just predetermined in the language of predicate and propositional logic. So "[liar] & p" can't be true and can't be false, it's just syntactically not a wff formula if I am right. I see your point and you are quite right from a practical point of view, but formally it's not possible I think.
  • Srap Tasmaner
    4.6k

    If your point is only that this sentence can't be represented within classical logic, then duh.

    I still say the English sentence in question can justifiably and consistently be considered false.

    Logic is a tool, one of my faves, but it does not sit in judgment of natural language, which may very well be just as formal only vastly more complex and powerful.
  • Pippen
    80
    If your point is only that this sentence can't be represented within classical logic, then duh.Srap Tasmaner

    That's my point. The problem is this: To prove that (S) "All sentences are false" is not a wff, I have to assume that S = S' with (S') "All sentences are false and this sentence is false". Then I can prove that S' is not a wff and since S = S' then S can't bei either, but how can a not well formed formula like S' be equal/equivalent to anything at all? It's not working. This is the reason why I wanted to formalize "All sentences are false" right away as a conjunction "All sentences are false and this sentence is false" and wondered if it would be possible and if not why.
  • Srap Tasmaner
    4.6k

    You can't do this in classical logic because you would need the predicates "… is a statement" and "… is false". You can't have either of those. Classical logic is swell, but it gets some of its swellness from being carefully circumscribed.
  • Meta
    185
    Sticking with naive logic. You say that the statement in the OP cant be false because of the liar paradox.
    The liar paradox says that the statement A for which A <-> not A can't have a truth value.
    Your statement has the form A <-> not A & B
    This sentence is not contradictiory in itself because A can be false. There is no need to add a new truth value. However it is not prohibited. Just totally unnecessary.

    Edit: Saying A is a paradox because it has similarities with the liar paradox is dogmatic and irrational imo and totally misses the point of what makes a paradox paradoxical.

    Edit2: Let's say X is the statement: Every statement is either true false or a paradox.
    Now X has similarities with the liar paradox and based on your logic it is a paradox. But I would say X is trivially true.
  • Pippen
    80
    @srap: Why can't I have "is a statement" and "is false" in predicate logic?

    @Meta: Your formulation in propositional logic is too simple, you can't express there what "All statements are false" want to say. In propostional logic, as a matter of fact, the liar paradox is just plain false since it says there: p <-> ~p.
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