7

In many contexts we distinguish between proofs using AC and proofs which do not use AC. (To phrase this somewhat differently: If there is a proof without AC, this proof is usually preferred.)

I would like to know whether there are names commonly used for these two types of proofs.

In my mother tongue I have heard the names which could be translated to English as effective-proof (for the proof avoiding AC) and non-effective proof or ineffective proof (for the proofs employing AC).

When I searched for these term on internet 1, 2, 3 and in Google Books 1, 2, 3, I found some hits. But not enough to be persuaded that these two terms are widely used.

So I would like to ask

  • Can some of the names for the proofs with/without AC which I mentioned above considered standard?
  • Are other names for such proofs commonly used?
Martin Sleziak
  • 53,687
  • 1
    Without AC: 'constructive'. With AC: 'non-constructive' if it's actually referred to. –  Feb 20 '15 at 08:53
  • 1
    I had the feeling that the phrase constructive proof is often used in different meaning. Wikipedia article I linked mentions famous example of with $\sqrt2^{\sqrt2}$ and $(\sqrt2^{\sqrt2})^{\sqrt 2}$. – Martin Sleziak Feb 20 '15 at 08:59
  • Or the proof of existence of transcendental number based on cardinality argument, although this proof can be made constructive if we are more careful. – Martin Sleziak Feb 20 '15 at 09:00
  • 4
    I've not run into a pair of terms that could reliably be interpreted that way. Constructive doesn't work, because it frequently excludes much more than just $\mathsf{AC}$, and neither does effective. (I'm curious about your first sentence, though: the only context in which I find any need to make the distinction is that of specifically investigating consequences of $\mathsf{AC}$ and its variants. In any other context I consider its use utterly unremarkable.) – Brian M. Scott Feb 20 '15 at 09:37
  • 3
    Same as Brian. I don't know a "correct" term. I usually just write "choice free" or "in ZF" or "the axiom of choice is not needed". – Asaf Karagila Feb 20 '15 at 09:45
  • @BrianM.Scott What I meant was that at least some authors often indicate whether the proof uses AC or not. But I will admit that I have mostly seen this in texts related to set theory. But basically for any results from any area it seems reasonable to ask whether AC was used. (I have heard some such questions asked, for example, in talks about topics from algebra and real analysis.) In any case, I rephrased the first paragraph a bit. – Martin Sleziak Feb 20 '15 at 09:46
  • 1
    @mistermarko "Constructive proof" is different concept with the "proof without AC". König's lemma is provable in ZF without AC, but it is considered to nonconstructive. – Hanul Jeon Feb 21 '15 at 02:50
  • 2
    @tetori: You need to be careful when saying that König's lemma is provable without choice. It is true if one adds the assumption that the tree is countable, but if one phrases it in a general way "Every tree of height $\omega$, with finite levels has an infinite branch" then the axiom of choice is needed. – Asaf Karagila Feb 21 '15 at 10:12
  • 2
    I quite like the terms "choiceless" for proofs that avoid AC (which suggests the complementary "choiceful" for proofs using AC) – Neil Barton Feb 25 '15 at 00:57
  • 2
    Why put a bounty? As mentioned in the comments there is no standard term. – Vivaan Daga Mar 26 '22 at 07:07
  • 2
    I vote for "Choicefuln't" – ℋolo Mar 26 '22 at 11:20
  • @ℋolo: I think "mostly not-non-constructive proofs" is even better. – Asaf Karagila Mar 26 '22 at 14:30
  • @AsafKaragila unfortunately I believe that "mostly constructive proofs" is a bit of a confusing name. – ℋolo Mar 26 '22 at 16:36
  • "Why put a bounty? As mentioned in the comments there is no standard term." But that was 6 years ago.. and clearly an update/refresh was appropriate here; btw +1 Martin – Verónica Rmz. Apr 02 '22 at 00:56

0 Answers0