2

I study maths, and I have found that a useful way of thinking about two propositions A and B being equivalent is to regard them as being two different ways of saying the same thing, or equivalently, interchangeable.

However, what I find puzzling from a philosophical perspective is why it may seem that one of the equivalent statements is much more apparent than the other. For example, it has been proven that Euclid's parallel postulate is equivalent to all of the following:

  1. two lines that are parallel to the same line are also parallel to each other
  2. there exists a triangle with internal angles summing to 180°
  3. every triangle in the plane has internal angles summing to 180°
  4. the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the legs of the triangle (Pythagoras's theorem)

I have listed them in the (subjective) order "most obvious" to "least obvious". In particular, the idea that (2) and (3) are equivalent is really quite puzzling. Intuitively, (2) seems to be a much less stringent condition than (3) - it is much more apparent, even though both provably say exactly the same thing!

My initial thought was that the fact that we perceive a difference indicates only that we understand some topics better than others i.e. that all equivalent statements ought to have the same apparency, but that is not true for us because we do not have enough knowledge. But I am not convinced that this fully explains (2)-(3).

Mauro ALLEGRANZA
  • 35,764
  • 3
  • 35
  • 79
legionwhale
  • 129
  • 4
  • 1
    Equivalent means "same truth value" in propositional logic and means inter-derivability (in a certain context) in a more broad context. But this does not explain all intuitive nuances of "equivalent" and thus we have [Intensional Logic](https://plato.stanford.edu/entries/logic-intensional/). – Mauro ALLEGRANZA Jul 07 '21 at 19:29
  • Because equivalence is a relative notion, it is relative to the background assumptions under which the propositions are considered, and which may involve conceptions not even present in the propositions themselves. In Euclid's case they are the remaining axioms of geometry. Equivalencies are not apparent to us for the same reason that logical consequences of a group of assumptions are not, the longer an inference chain that establishes a consequence the less we are able to "see" it. – Conifold Jul 07 '21 at 22:37
  • @Conifold In this case, that may be the issue. Unfortunately, I don't know which of Euclid's other axioms each proposition requires in order to be equivalent. However, I believe the problem can persist even for other sets of equivalent propositions founded on exactly the same axioms. Are you certain that it doesn't? – legionwhale Jul 08 '21 at 18:50
  • @MauroALLEGRANZA I read only as much as I could understand, but I disagreed strongly from the opening. It is correct to say that the morning(/evening) star is Venus. However, I do not think it is correct to say that Venus is the morning star. The morning star is specifically referring to "Venus is in the morning sky", not "Venus". Thus, I do not recognise this as an equivalence or equality, and therefore I do not have any problems such as "The morning star is the evening star". C.f. all dogs are tetrapods but obviously, not all tetrapods are dogs. Is this any different? I don't believe it is. – legionwhale Jul 08 '21 at 18:58
  • It is the complexity of inference chains that matters for "apparency", not whether background assumptions are the same. Different equivalencies may have proofs of different length and complexity even if they use the same axioms. See [What is the difference between depth and surface information?](https://philosophy.stackexchange.com/q/59391/9148) for one way to measure "apparency". – Conifold Jul 08 '21 at 23:07
  • @Conifold HI -- sorry to bother you, but I have explained this a bit better in a comment to Bumble's answer, if you wish to read that. I think that what I am referring to is slightly different. – legionwhale Jul 29 '21 at 00:09
  • The minimal length of inference chain from axioms is an objective number, it makes no difference what was proved first. And it can be different for equivalent statements. Even the two directions of the equivalence may have different lengths. – Conifold Jul 29 '21 at 01:40
  • @Conifold That's if you're presuming that the length of the inference chain is the only independent variable for apparency. Of course, it has a contribution, but I find it unlikely that, with our human biases, we would be as flawlessly rational as that belief would imply. – legionwhale Jul 29 '21 at 11:22
  • There are other random factors that differ from human to human, or even common intuitive shortcuts that allow bypassing some inference chains and not others. [Azzouni](https://as.tufts.edu/philosophy/sites/all/themes/asbase/assets/documents/azzouniStillASense.pdf) calls them "inference packages". This can be accounted for, to a degree, by setting up a proof system with compound inference rules that represent such leaps when calculating the length. – Conifold Jul 30 '21 at 04:48
  • Brings to mind Jerry Bona's famous quip: "The Axiom of Choice is obviously true, the Well–ordering theorem is obviously false; and who can tell about Zorn's Lemma?" The joke being that the three are logically equivalent. https://en.wikipedia.org/wiki/Jerry_L._Bona – user4894 Aug 06 '21 at 23:57
  • @user4894 This brings to mind what Bumble said in a comment under his answer :-) I still would disagree that the axiom of choice is obviously true. – legionwhale Aug 07 '21 at 15:25
  • @legionwhale Bumble misquoted the saying that I accurately stated, attributed, and documented. Regarding AC being true, a choice set is just a legislature. You have a country with 50 states and you pick a representative from each state. The US Senate is 2 applications of AC; the House is a choice set on the Congressional districts. If there was a country with infinitely many states, why couldn't they form a legislature? If nothing else, each state could choose a representative by lot. See https://en.wikipedia.org/wiki/Sortition – user4894 Aug 07 '21 at 20:31
  • @user4894 I forgive Bumble. Regarding AC, my opposition is not really the infinite collection of sets, but more the nature of the sets themselves, which may all be uncountably infinite in size. You cannot draw lots for an infinite set, and yet AC would imply there is a choice function for every non-empty subset of the real numbers. I am extremely unconvinced that such an assumption is well-founded. – legionwhale Aug 08 '21 at 15:54
  • @legionwhale Ok for the record I forgive Bumble too. I wasn't criticizing him, I was criticizing you for quoting him after I had supplied the correct quote :-) In any event, you can't have a uniform probability distribution on a *countable* set; but you CAN have one on an uncountable set. "Throwing darts at the real line," or flipping countably many fair coins to generate a random real in the unit interval. But again, why couldn't a country with infinitely many states have a legislature? That's undemocratic! – user4894 Aug 08 '21 at 20:44
  • @user4894 I'm sorry then :/ Regarding infinitely many states, in the analogy, that represents the infinitely many sets, no? As I said, that's not my problem—my problem is that each set itself may be infinite in size. Where can I find out more about uniform probability distributions on uncountable sets? I'm unconvinced. How do you rigourously describe throwing a dart at the real line? And if you can pick a random real number in [0,1] by flipping coins, why can't you do the same for an infinite countable set, such as the natural numbers (also using binary)? e.g. HHHTTTT... corresponding to 7. – legionwhale Aug 10 '21 at 13:13
  • @user4894 OK, after thinking a bit more, I realise that it doesn't work for the natural numbers because HHHHH... or indeed any outcome which is not always tails after some point does not correspond to any natural number. (Maybe these could be discarded? I'll think more.) But isn't the problem with uniform distributions on infinite sets that you get probability 0 everywhere? – legionwhale Aug 10 '21 at 13:34
  • @legionwhale One of the probability axioms is countable additivity. If you try to put a uniform distribution on, say, the natural numbers, either the probability of an individual number is positive, in which case the total probability is infinite; or it's zero, so that the total prob is zero. In no case can you make the total probability be 1. https://en.wikipedia.org/wiki/Probability_axioms So it's "easier" to pick a random representative from an uncountable set. The prob of an individual is zero, but that does not give a contradiction because additivity is only countable. – user4894 Aug 10 '21 at 19:11
  • @user4894 Huh. I've done the basics of probability, including countable additivity. In light of your information, it now seems to be an arbitrary delimitation. Looking it up, it seems it was chosen because it works. I'm still a bit curious about your picking a random number from [0,1]. I am aware of the *continuous* uniform distribution on [a,b], but I couldn't find more on picking a random number in practice with the method you described. Also, I hadn't noted it before, but I find it quite interesting, despite being trivial to prove, that the natural numbers can't biject with any power set. – legionwhale Aug 10 '21 at 20:33
  • @legionwhale Flip a countably infinite sequence of fair coins to generate an infinite bitstring and put a binary point in front of it to get a random real number in the unit interval. Of course you can't do this in real life, so perhaps that's what you mean. That the naturals can't biject with the powerset of the naturals is implied by Cantor's theorem. https://en.wikipedia.org/wiki/Cantor%27s_theorem – user4894 Aug 10 '21 at 21:05
  • @user4894 No, no, I understood that. I was just wanting to read more about where/when this method of selection is used (if it is used anywhere)? I am also well aware of Cantor's theorem. My remark is that the natural numbers do not biject with *any* power set. As I said, it's trivial to prove this—power sets of finite sets are finite and power sets of infinite sets are uncountable. It's just mildly interesting to note that they are never countable. Some fantastical thinking of mine has me wondering if we can do number theory with general powers of sets A^B—or at least some stuff with rings. – legionwhale Aug 10 '21 at 21:32

1 Answers1

1

Your question is related to the question: how can a deductively valid argument tell you something that you don't already know?

Human beings are not logically omniscient, i.e. we do not know the logical consequence of everything we know. One of the purposes of a good proof is to take a potentially non-obvious logical consequence relation and express it as a sequence of simple and obvious steps. In the case of logical equivalence, there is a two-way logical consequence relation. If A entails B and also B entails A, then A and B are logically equivalent. But as with any logical consequence relation, the fact that A and B entail each other does not mean that it is obvious that they do.

To say of two propositions that they "say the same thing" suggests something stronger, perhaps something like they convey the same information or the same meaning to you. This is a more finely-grained distinction than logical equivalence. Meanings relate to intensions as well as extensions. Also, propositions that are logically equivalent are not always interchangeable. It is possible even for propositions that share the same intension to fail to preserve truth when substituted. This is sometimes referred to as hyperintensionality.

So, rather than thinking of logically equivalent propositions as "saying the same thing" or as being "interchangeable", it is better to think of them in terms of a pair of consequence relations.

Bumble
  • 20,757
  • 2
  • 27
  • 65
  • I think I myself was slightly confused by my choice of Euclid's postulates, since they are axioms rather than results. But I disagree that this is about logical omniscience. I agree that a result proved from axioms may be less apparent than the axioms -- I presume that apparency depends on the length of the inference chain. However, with two equivalent statements (say A and B), we have an inference chain from both sides. Therefore it is not clear why we should regard A as less apparent than B. Because we proved A first? Aliens who proved B first (finding it more apparent) would disagree. – legionwhale Jul 28 '21 at 23:56
  • 1
    I suppose you could fall back on saying that one of a pair of logically equivalent propositions just seems more reasonable in some intuitive way because it makes a better fit with our mental models. I remember a mathematician joking that in set theory, the well-ordering theorem is obviously false, the trichotomy is obviously true, and he couldn't decide whether the axiom of choice is true or not. – Bumble Jul 29 '21 at 00:58
  • Yes, I suppose so. When I asked this, I was wondering why it might be that certain propositions fit better with our mental models, especially in a pure field like maths, but I guess that none of us could really answer definitively. I like the joke. The axiom of choice seems pretty shaky to me, but I should probably study set theory before I make a judgement. – legionwhale Jul 29 '21 at 11:43