What axioms do you believe in?

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    • #88713
      Anonymous
      Guest

      What axioms do you believe in?

    • #88714
      Anonymous
      Guest

      >believe
      I dont, I just look at the consequences

      • #88716
        Anonymous
        Guest

        If you believe anything at all, then you must believe some mathematical axioms. You only understand the world through ideas, which are almost identical (though not exactly) to sets.

        In other words, If you accept a certain species of animal exists, then you are already implicitly accepting the axiom of pairing, for example. Because the only way to accept it is to conceive animals as an idea in the bigger picture of ideas.

        • #88717
          Anonymous
          Guest

          sets dont exist in this universe so no

          • #88718
            Anonymous
            Guest

            yeah but ideas exist and thats what sets are referring to

            • #88719
              Anonymous
              Guest

              >ideas exist
              prove it

              • #88720
                Anonymous
                Guest

                Proof is an idea

                • #88721
                  Anonymous
                  Guest

                  Not if you write it down with chalk on a blackboard
                  Then it becomes reality

                  • #88723
                    Anonymous
                    Guest

                    yeah but the symbols are not ideas, symbols merely refer to ideas.

                    • #88737
                      Anonymous
                      Guest

                      Which are not real

              • #88722
                Anonymous
                Guest

                >prove it
                everything in the material world is only thought through an idea, that is why language and mathematics is identical with being. Therefore, ideas must be real to.

              • #88829
                Anonymous
                Guest

                You.
                I.

        • #88739
          Anonymous
          Guest

          >pretending all sets behave like finite sets

        • #88742
          Anonymous
          Guest

          I believe that the the axioms of euclidean geometry will produce good models of some real life situations and that the axioms of spherical geometry will produce good models of some other real life situations.

          I don’t believe in one set of axioms more than the other even though they have contradictory axioms.

          I don’t know what you mean by believing an axiom.

        • #88803
          Anonymous
          Guest

          Not him but I don’t believe anything at all. I’m paranoid that one day I’m going to punch 2+2 into the calculator and it will equal 3 or 0 or some shit. I can’t do arithmetic without punching into a calc 3 or 4 times and looking at it again and again, if I do an arithmetic problem without triplechecking I get increasing paranoia bording on hysteria.

      • #88753
        Anonymous
        Guest

        FPBP. The only rule in mathematics is consistency from A to B.

        • #88807
          Anonymous
          Guest

          exactly
          [math]2+2=5 implies 4+4=10[/math]

          • #88808
            Anonymous
            Guest

            2+2=5 => you’re a scrotebrained scrote

        • #88814
          Anonymous
          Guest

          >The only rule in mathematics is consistency from A to B.
          scrotebrain. That’s not even a rule.

    • #88715
      Anonymous
      Guest

      You mean as in postulations? Well, I’d argue I’m the ubermensch.

    • #88724
      Anonymous
      Guest
    • #88725
      Anonymous
      Guest

      that OP is a scrote

    • #88726
      Anonymous
      Guest

      Separation
      Union
      Powerset
      Collection
      Infinity

      Simple as.

      • #88730
        Anonymous
        Guest

        When are two sets equal.
        (Question not meant cynically)

        >Powerset
        Unnecessary.

        • #88732
          Anonymous
          Guest

          >Unnecessary.
          How so

          • #88735
            Anonymous
            Guest

            Well you can find more conservative claims that give you just as much.
            In math you usually don’t need to assume that big sets, even relatively small things as the reals, form a set (you can just deal with the class of all such numbers and express essentially all interesting theorems).

            On a more unrelated note, this quote is wild
            https://en.wikipedia.org/wiki/Elementary_function_arithmetic#Friedman’s_grand_conjecture

            Maybe for classical topology you actually need some of those strong axioms.

    • #88727
      Anonymous
      Guest

      Martin-Löf Type Theory + one impredicative universe + induction-recursion. What else?

    • #88728
      Anonymous
      Guest

      Lambda calculus is an underrated axiomatic system. It has fewer, simpler and more concrete rules than Set theory and properly defines what is valid formula/procedure/method. The theory is woke af upon the idea of substitution. Booleans,naturals,rationals,lists,sets,addition, multiplication and many more can be defined using Church encodings

    • #88729
      Anonymous
      Guest

      that reality isn’t a hallucination of my ego
      that empiricism and the scientific method are the only way to discover truths about the reality we live in.

    • #88733
      Anonymous
      Guest

      Axiom of infinity is absolutely scrotebrained and the fact that it doesn’t bother most people in the field makes me physically sick.
      The rest is fine.

      • #88736
        Anonymous
        Guest

        >Axiom of infinity is absolutely scrotebrained and the fact that it doesn’t bother most people in the field makes me physically sick.
        See pic related.

        • #88741
          Anonymous
          Guest

          roasties are braindead and always follow the trend

          cantor created crappy maths

    • #88734
      Anonymous
      Guest

      Only the ones required for Lain Point Theory.

    • #88738
      El Arcón
      Guest

      A real number is a cut in the real number line.

    • #88740
      Anonymous
      Guest

      RCA_0 is all you need. In fact, it’s all you want.
      (Possible sans LEM. I don’t know if there’s any serious repercussion from dropping LEM in RCA_0. Likely not.)

      https://en.wikipedia.org/wiki/Reverse_mathematics#The_base_system_RCA0

      • #88750
        Anonymous
        Guest

        The reverse math page btw. has a list of what classical arithmetic with only Sigma_1 induction and second order Sigma_1cup Pi_1 comprehension (computable sets) give you, pic related.

        Admittedly it’s not all of math, but it’s still hella strong for a theory that isn’t a set theory (and the models of e.g. the reals aren’t super pretty in either)

    • #88743
      Anonymous
      Guest

      You know, I don’t really know what axioms I believe in. But I can tell you what I don’t believe in. The natural numbers. You see, I don’t believe in the axiom of induction… Captcha: YT00D.

      • #88745
        Anonymous
        Guest

        1 is a natural number
        If n is a natural number then n+1 is a natural number

        If a statement is true for every natural number n then the same statement is true for 1 and n+1.(axiom of induction)

        I believe that this axiom is valid by proof of contradiction.

        If a statement is true for any natural number n but not for n+1 we would have a contradiction since n+1 is any natural number as well. It’s like saying that if a statement is true for any natural n it is also false for any natural k. It doesn’t make sense for induction not to work

        • #88747
          Anonymous
          Guest

          >If a statement is true for every natural number n then the same statement is true for 1 and n+1.(axiom of induction)
          That’s not Induction.
          In Induction, the universally quantified claim for the unary predicate is the conclusion (consequent), not the assumption (antecedent).

          You know, I don’t really know what axioms I believe in. But I can tell you what I don’t believe in. The natural numbers. You see, I don’t believe in the axiom of induction… Captcha: YT00D.

          Induction is rejected in the work of predicative arithmetic of
          https://en.wikipedia.org/wiki/Edward_Nelson
          (The reasoning being that PA "defines" natural numbers by it’s axioms, and so, roughly, his case is that then those axioms shouldn’t have "for all numbers" in them already – lest you’re, in a very nitpicky way – circular)

          That said, I’m

          RCA_0 is all you need. In fact, it’s all you want.
          (Possible sans LEM. I don’t know if there’s any serious repercussion from dropping LEM in RCA_0. Likely not.)

          https://en.wikipedia.org/wiki/Reverse_mathematics#The_base_system_RCA0

          and I personally think that at least for [math]Sigma_1[/math]-formulas in the arithmetical hierarchy, Induction should be fine.
          Those are the predicates which only have at most one unbound quantifier, namely 1 [math]exists[/math] upfront. E.g. the subset
          [math]{nin{mathbb N}mid exists k. (n=2cdot k)}[/math]

          is [math]Sigma_1[/math]-definable. For any n, know that you can at least (iff it’s actually even) computationally certify membership via giving a k. This example in fact collapses to [math]Sigma_0[/math] if you accept the mod functions as a total function symbol, since then the set is extensionally the same as
          [math]{nin{mathbb N}mid (n % 2) = 0}[/math]

          The [math]Sigma_1capPi_1[/math] formulas are exactly those which a Turing machine can compute. Since those certify both membership and non-membership.
          https://en.wikipedia.org/wiki/Certificate_(complexity)
          And all the [math]Sigma_1[/math] (certifying membership) ones are at least semi-decidable.
          Given the structure of induction, I think it’s fine to adopt induction for those formulas with one existential quantifier.

          Here’s a listing for schemata of that axiom

          https://en.wikipedia.org/wiki/Induction,_bounding_and_least_number_principles

          • #88748
            Anonymous
            Guest

            >(The reasoning being that PA "defines" natural numbers by it’s axioms, and so, roughly, his case is that then those axioms shouldn’t have "for all numbers" in them already – lest you’re, in a very nitpicky way – circular)
            That doesn’t make sense. PA is not the definition of the natural numbers, we already know by Godel and others that PA is not enough to define them. The actual definition is that
            – 0 is a natural number
            – if x is a natural number, so is SX
            – only natural numbers are those resulting from a repeated application of the above two points
            It doesn’t refer to PA at all. In light of this, induction is valid for all formulas, not just quantifier-free ones.
            >hat said, I’m

            RCA_0 is all you need. In fact, it’s all you want.
            (Possible sans LEM. I don’t know if there’s any serious repercussion from dropping LEM in RCA_0. Likely not.)

            https://en.wikipedia.org/wiki/Reverse_mathematics#The_base_system_RCA0 and I personally think that at least for Σ1-formulas in the arithmetical hierarchy, Induction should be fine.
            Why wouldn’t it be fine for all formulas? And you mean both Sigma_1 and Pi_1 formulas correct?

            • #88749
              Anonymous
              Guest

              First off, I’m not defending Nelsons view.

              As for your argument, you say
              >only natural numbers are those resulting from a repeated application of the above two points

              But if you drop induction IND from PA, i.e. look at Robinson arithmetic (which actually adds some stength of IND back), then that theory already has models with "infinite natural numbers at the end". It’s shortly described at the bottom here
              https://ncatlab.org/nlab/show/Robinson+arithmetic

              The statement you wrote down can, as far as formal theory concerned, only be held if you also consider induction.

              >Why wouldn’t it be fine for all formulas?
              Just seems like too strong an assumption and extra assumptions can always be stated explicitly (or, in a text, you just state that now you switch to a stronger theory.)

              >And you mean both Sigma_1 and Pi_1 formulas correct?
              I just meant to require Sigma_1 induction, but I think depending on the rest of the context you get both of those sides for induction in particular

              • #88751
                Anonymous
                Guest

                >The statement you wrote down can, as far as formal theory concerned, only be held if you also consider induction.
                My point is that it’s a separate definition independent of induction. I’m not referring to any formal system, not PA, not ZFC, not Q not anything else to define the natural numbers. After you do so, you should see why induction for ALL formulas is true, and that there’s no circularity.
                Of course
                >only natural numbers are those resulting from a repeated application of the above two points
                cannot be formulated in first order logic, but I’m not trying to. Definitions don’t all have to be formulated in a strict formal system.
                >Just seems like too strong an assumption and extra assumptions can always be stated explicitly (or, in a text, you just state that now you switch to a stronger theory.)
                Too strong? Too strong for what? What matters is that it’s true, so you use it to make your job easier. Unless you also want to consider nonstandard models where induction for more complicated formulas fails, but nobody cares about that.
                >I just meant to require Sigma_1 induction, but I think depending on the rest of the context you get both of those sides for induction in particular
                Is there any reason to prefer Sigma_1 formulas to Pi_1? To me they seem two sides of the same coin.

                • #88752
                  Anonymous
                  Guest

                  >Definitions don’t all have to be formulated in a strict formal system.
                  You do you, I’m not gonna argue you should be more concise with your background if you don’t like

                  >Too strong? Too strong for what?
                  Too strong as in including subsets of N which we don’t even know are decidable. Again, I’m not pushing this conservatism on anyone, so I’m not gonna argue more for that

                  >To me they seem two sides of the same coin.
                  Yeah pretty much. I think in complexity theory the Sigma side is the more "positive" phrasing, namely unbounded search without giving a time bound, but with certainty that till you certainly find something for the positive cases. (E.g. you can dovetail algorithms and computably generate the subset of N defined as all the numbers for which the "3x+1" collatz function applications eventually end in 1. Those searches will give us all such numbers even if it may be the case that there’s counter-examples, or uncomputable n’s.)

                  I think historically Sigma_1 and the induction principle for it is just where computability theorists and complexity theorists see the theory diverge into the less computable realm / oracles / set theory like comprehension.

                  As opposed to the pic in

                  The reverse math page btw. has a list of what classical arithmetic with only Sigma_1 induction and second order Sigma_1cup Pi_1 comprehension (computable sets) give you, pic related.

                  Admittedly it’s not all of math, but it’s still hella strong for a theory that isn’t a set theory (and the models of e.g. the reals aren’t super pretty in either)

                  , pic related is what you get if you adopt some weak choice ("for every infinitely big tree there exists an infinite path").
                  Clearly this has some important mathy theorems, so it’s not like you should stop at RCA0. But you can be explicit about your assumptions beyond it too.

                  Or you just adopt ZFC, I don’t care, do what you want. I’m off for today.

        • #88802
          Anonymous
          Guest

          >I believe that this axiom is valid by proof of contradiction.
          >This axiom
          >Valid by proof

          • #88813
            Anonymous
            Guest

            I could have worded it better, I agree. I see it more like a logical consequence rather than a rule

    • #88744
      Anonymous
      Guest

      OP is always a scrote.

    • #88754
      Anonymous
      Guest

      Maybe said differently, if you adopt induction only for [math]Sigma_1[/math]-predicates [math]exists k. P(n, k)[/math], where [math]P[/math] only contains computable operations,
      then via induction you enable yourself to prove statements
      [math]forall n. exists k. P(n, k)[/math]
      for which you know that indeed you can, for each k, computably certify why [math]P(n, k_n)[/math] holds.

      If you adopt induction for predicates in n with longer chains of quantifiers, then the axiom itself is what proves them.

      • #88758
        Anonymous
        Guest

        Are there any arithmetic statements that you can prove using induction only for Sigma_1 and Pi_1 predicates but cannot prove using induction only for Sigma_1 predicates?

        • #88759
          Anonymous
          Guest

          No, over Robinson arithmetic (PA with induction removed but the [math]Sigma_1[/math] claim added back that each nonzero number has a predecessor), the induction schemas on level n are equivalent, i.e.
          [math]Qvdash {mathsf {ISigma_n}} iff {mathsf {IPi_n}}[/math]

          which are the names for Q+the n level induction.

          Of course, this is something "I knew" but in reality only knew from Mathoverflow and the bottom of this Wikipedia page
          https://en.wikipedia.org/wiki/Induction,_bounding_and_least_number_principles#Relations_between_the_principles

          So I dug up the reference and downloaded this 1978 Springer book with the paper, pic related. (The proofs are quite concise there, but basically clearly seem to imply that you can switch out the outer quantifiers and do the translation explicitly)

          • #88761
            Anonymous
            Guest

            What is a in the proof? Free variable or actual numeral? If free variable i dont get why we suppose not theta(a), rather than not for all a, theta(a) or there exists a s.t. theta(a).

            • #88762
              Anonymous
              Guest

              Again, I find those proofs far to concise if you’re not in the field, and so hard to read.

              That said, I think what he tries to say is the following:

              Sidenote: Induction reads
              [math]Big(theta(0) land forall(x:N). big(theta(x)totheta(x+1)big)Big)to forall (y:N). theta(y)[/math]

              Then the first paragraph, with a bunch of verbosity filled in by me, would read:

              "Let [math]theta[/math] be a unary predicate over the numbers of either Sigma or Pi type. Suppose
              [math]theta(0) land forall(x:N). big( theta(x) to theta(x+1) big) [/math]
              but
              [math]negforall (y:N). theta(y)[/math],
              i.e. but
              [math]exists (z:N).neg theta(z)[/math],
              i.e.
              [math]neg theta(a)[/math]
              for some particular [math]a:N[/math],
              i.e. assume the consequent in Induction is wrong despite the antecedent being validated.
              Then note that we can also form those two predicates in y for which Induction fails as well
              (…)"

              • #88768
                Anonymous
                Guest

                The thing that confuses me is that we’re working at the level of formal logic, and not theta(a) seems to imply that we have an actual witness for not for all x, theta(x), but I see no reason for this to be the case. At the level of formal logic, there’s no reason to expect to have a witness just from the statement not for all x, theta(x).
                Perhaps he meant to say [math]exists a neg theta(a)[/math], in which case the statement [math]exists w (a = y+w land neg theta(w) ) [/math] really is
                [math]exists w exists a(neg theta(a) land a = y+w land neg theta(w) )[/math], which then through coding three variables at once can be transformed into a [math] Sigma_n[/math] statement as required. For the other statement I guess we want to use [math]forall a[/math] and add a condition which takes a to be the smallest one to single it out.
                Perhaps for you it’s trivial but for someone like me with not much experience it’s far from trivial.
                Do you think that’s what we meant?
                And then induction fails because inductive hypotheses (2 of them) are true and the forall statement is false. And it’s false by taking y= any of the a’s satisfying that property because then w is forced to be zero and it would assert [math]neg theta(0)[/math] which we know is false. Did I get that right?

                • #88770
                  Anonymous
                  Guest

                  ¬∀(n:N). θ(n)
                  to
                  ∃(n:N). ¬θ(n)
                  is classically, a stronger form of the Markov rule
                  Using this to instantiate an
                  a, with ¬θ(a)
                  is a natural deduction rule too.

                  https://en.wikipedia.org/wiki/Markov%27s_principle#In_Intuitionistic_logic
                  https://en.wikipedia.org/wiki/Existential_instantiation

                  I don’t want to go through the pains of interpreting a super short 70’s typewritten book, I’d want to read a more common text – I think those papers and books aren’t so much concerned with foundations than with complexity theory

                  • #88771
                    Anonymous
                    Guest

                    >Using this to instantiate an
                    >a, with ¬θ(a)
                    >is a natural deduction rule too.
                    What is it called? Do you have a link to the rule? As I understand, here a is neither a free nor bound variable, but some other kind of variable?

                    • #88772
                      Anonymous
                      Guest

                      I literally linked it, wtf

                      • #88773
                        Anonymous
                        Guest

                        Oh, didnt notice the second link, thanks

                      • #88774
                        Anonymous
                        Guest

                        Oh, didnt notice the second link, thanks

                        So without this rule of inference and using Markov rule can you also form a proof along the lines I indicated?

                      • #88775
                        Anonymous
                        Guest

                        As I said a couple of times, I’m not really interested in taking it apart. Interpret that as "I donno."

                        I’ve never seen a theory without that rule of inference, though ∃. It’s even more viable if you don’t have LEM.
                        I’d think you need a Markov rule, which imho one should adopt for the lower levels of the hierarchy. (Sidenote: Brower’s school of constructivism can’t accept Markov’s rule for consistency reasons, but Markov’s school of course does.)

                      • #88776
                        Anonymous
                        Guest

                        I’m mainly interested in classical logic here.

                      • #88777
                        Anonymous
                        Guest

                        Under Curry Howard, if you’ve proven
                        [math]exists u. P(u)[/math],
                        then this literally means you got a pair [math](u, p)[/math], that typechecks as
                        [math](a, p_a) colon exists u. P(u)[/math],
                        where p witnesses [math]P(a)[/math], and so typechecks as
                        [math]p_a colon P(a)[/math]

                        So the left projection of any proof of [math]exists u. P(u)[/math] literally is the [math]a[/math] where’s looking for and that’s what the rule more abstractly allows you to do.
                        tl;dr if you have proven an existential statement constructively, then you got that witnessing a anyway, so the rule is a sort of real life tautology in that context.

                        Of course classically the rule is somewhat of a meme, as ZFC will e.g. proof the existence of choice functions that provably can’t be defined.
                        It proves
                        [math]exists f. text{f is a choice function for this and that}[math]
                        but actually there’s no term g that does the job.
                        The rule is okay still, I suppose, insofar as that "term a" is only used during a proof and doesn’t have to linger around after.

                      • #88784
                        Anonymous
                        Guest

                        Not if you interpret [math]exists[/math] as the squashed variant of the existential type, or the one in Prop if you were in CIC. In general, this type does not enjoy choice.

                        If you extend your theory with classical logic in propositions it then makes perfect sense because there are no such choice functions in general but it’s fine to merely accept their existence.

                      • #88788
                        Anonymous
                        Guest

                        >I’ve never seen a theory without that rule of inference, though ∃.
                        You’ve never heard of LK and LJ?

                      • #88789
                        Anonymous
                        Guest

                        I wouldn’t know by heart which rules those fulfill

    • #88755
      Anonymous
      Guest

      for each n,

    • #88756
      Anonymous
      Guest

      >Assume a proper class of Woodin cardinals exists

      • #88757
        Anonymous
        Guest

        for what purpose?

      • #88769
        Anonymous
        Guest

        Can you turn down the bass?

    • #88760
      Anonymous
      Guest

      The one that tells us we’re bonked and can’t really know everything anyways.

    • #88763
      Anonymous
      Guest

      The axiom of choice, but ONLY the axiom of choice.

      • #88767
        Anonymous
        Guest

        woke af

      • #88782
        Anonymous
        Guest

        For me, it’s global choice.

    • #88765
      Anonymous
      Guest

      Praxeological axioms.

      • #88781
        Anonymous
        Guest

        Woke af

    • #88766
      Anonymous
      Guest

      Axiom 1: Inconsistency is allowed wherever I deem necessary.

      • #88817
        Anonymous
        Guest

        Woke af schizo tyrant

        • #88818
          Anonymous
          Guest

          This meme or any of its variations have never been funny. You’re a low IQ scrote and I hate you.

          • #88820
            Anonymous
            Guest

            You’re the scrote scrote for assuming it was meant to be funny. Please suffocate.

            • #88821
              Anonymous
              Guest

              What is it meant to do if not be funny?

              • #88822
                Anonymous
                Guest

                It’s schizo logic for the schizo. Go die now.

    • #88778
      Anonymous
      Guest

      [math]exists f. text{f is a choice function for this and that}[/math]

    • #88779
      Anonymous
      Guest

      https://i.imgur.com/nuaAbvb.gif

      For me, its the field axioms. Its all I need for my work.

      • #88787
        Anonymous
        Guest

        Your work must be very boring then. There’s very little you can do with just the field axioms.

    • #88780
      Anonymous
      Guest

      My math, my choice.

    • #88785
      Anonymous
      Guest

      Falso, the patrician choice.
      https://inutile.club/estatis/falso/

    • #88790
      Anonymous
      Guest

      You cannot trust axioms. Any axiom is true in the theory for which it is an axiom.
      But it is likely to be wrong for other theories.
      Example: the theory for spiders says as an axiom that each animal has eight legs.
      But an ostrich has two legs, which is fewer than eight. hence the axiom is wrong and
      it cannot be trusted.

      • #88791
        Anonymous
        Guest

        >You cannot trust axioms. Any axiom is true in the theory for which it is an axiom.
        The second sentence exactly says that it’s not a matter of "trust", but choice

        >Example
        Convoluted way of saying that "theory" and "axioms" always go together and axiom sets don’t mix among each other.

        • #88792
          Anonymous
          Guest

          A sane person of course agrees with you.
          But how else explain the concept of axioms to Wildberger and his followers?

          • #88793
            Anonymous
            Guest

            He’s an old old man, he gets a free pass from me.

          • #88800
            Anonymous
            Guest

            you think wildberger doesn’t understand the concept of axioms?

            • #88801
              Anonymous
              Guest

              Stop replying to scrotebrains.

          • #88824
            Anonymous
            Guest

            >Wildberger and his followers
            They’re all just LARPers anon, no one is stupid enough to reject the notion of real numbers.

    • #88795
      Anonymous
      Guest

      What does ZFC taste like?

    • #88797
      Anonymous
      Guest

      ZFC + Con(ZFC) + Con(ZFC + Con(ZFC)) + Con(ZFC + Con(ZFC) + Con(ZFC + Con(ZFC))) +…

      to believe otherwise is sheer lunacy

      • #88799
        Anonymous
        Guest

        What about Con(ZFC + Con(ZFC) + Con(ZFC + Con(ZFC)) + Con(ZFC + Con(ZFC) + Con(ZFC + Con(ZFC))) +…)?

        • #88816
          Anonymous
          Guest

          You still can’t prove it’s consistent, you’d need to add transfinite towers of consistency. And then starts the problem of the kind of ordinals you believe in.

          • #88827
            Anonymous
            Guest

            >You still can’t prove it’s consistent
            Make it an axiom.

    • #88798
      Anonymous
      Guest

      Why do they always say that NBG is bad?

    • #88806
      Anonymous
      Guest

      All of them.

    • #88811
      Anonymous
      Guest

      I believe that a line is an interval between two points that can be extended indefinitely in either direction.

      • #88812
        Anonymous
        Guest

        go on…

    • #88819
      Anonymous
      Guest

      Transitive Closure OR Countable Choice, but not both

      • #88823
        Anonymous
        Guest

        >OR
        >but not both
        That’s called XOR.

      • #88828
        Anonymous
        Guest

        Why?

        Also, isn’t countable choice vastly milder than transitive closure for all sets?

    • #88825
      Anonymous
      Guest

      BUMP

    • #88826
      Anonymous
      Guest

      a point on a line is a point on a line.
      so far in X years of this shit, no one has managed to express a dissenting opinion that makes any sense

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