Talk:Intuitionistic logic

message: Very nice start to the article.  A bit POV, though, it's kind of obvious it's written by an advocate of intuitionism.  But still good start.
Ummm ... isn't it a bit narrow-based around the assumption of that old chestnut, intuitionism as one of the three main philosophies of mathematics? The Kolmogorov interpretation may have been the first step clear of that. Certainly the Brouwer not not not = not result needs a mention, too. Goedel used IL at one juncture, and after that, historically speaking, wasn't IL respectable in a technical sense? What happened then in relation to Markov was perhaps confusing.
Anyway, there are several strands to disentangle here: this is a bit ahistorical for me.
Charles Matthews 10:36, 15 Oct 2003 (UTC)
replies: text: Agree on the history issue.  For those into completeness, note that the Modal Logic "B" was intended to be a system matching Brouwer's intuition on "not". And that there are direct mappings between Intuitionist PC and the Modal Logic "S4"  Nahaj 01:11:29, 2005-09-08 (UTC) And Ian Hacking pointed out in an early paper that there are validity preserving mappings between Intuitionist PC and the Modal Logic "S3" (A weaker system than S4). Nahaj 13:50, 26 October 2005 (UTC)
thread_title:  comment by Charles Matthews 

message: Terms should not be equivocated.  Speakers end up suggesting that not isn't not, so classical logicians should use a term meaning the exact opposite: un. lysdexia 00:16, 13 Nov 2004 (UTC)
thread_title: my take

message: In the "law of the excluded middle" example under "Heyting algebra", R^2 is divvied up into a set A and its complement, the first comprising \{(x,y) : y>0\}. Now, it would seem to me that the complement of A would be \{(x,y) : y\le0\}. However, it's given as \{(x,y) : y<0\}, which is then used to show that the y=0 plane isn't included in the union of A and not-A. How can this be? grendel|khan 03:45, 2005 Jan 14 (UTC)
:It's really badly explained.  But the idea is that if you consider subsets of R^2, they form a Heyting algebra if you consider "or" and "and" to be set union and intersection, and (this is they key point) the complement of X to be the interior of R^2 - X. Then the complement of the open upper half-plane is the open lower half-plane.  The axioms of the Heyting algebra are satisfied in this case, but the law that X\vee\not X = R^2 doesn't hold.   -- Dominus 12:30, 14 Jan 2005 (UTC)
replies: text: I've expanded the explanation (which I think still needs some more work) and included detailed examples of both a valid and and invalid formula.  -- Dominus 16:19, 21 September 2005 (UTC)
thread_title:  I don't get it. How is this an example? 

message: I am doing an assignment at the moment and it is very important that I do well.  Can anybody give me a hand with it?  I can email the questions otherwise can discuss them one at time here.
Cheers,
:Does it say anywhere this is a wiki site, so obviously we help you cheat on your homework? Charles Matthews 07:16, 22 October 2005 (UTC)
replies: text: Requests to deal privately by email are pretty suspect, you know. Charles Matthews 21:43, 22 October 2005 (UTC)

text: Substitution is a rule of Intuitionist calculus (although not normally called an inference rule).  While Modus Ponens is the basic inference rule of most calculi, most have a number of "derived rules" such as adjunction and substitution of equivalents.  What counts as a rule needing to be given in your assignment depends on what your instructor has already given and what he has said.  Obviously we don't know this.  You could find therefore that an answer you get here (while in some sense "correct") is wrong for your assignment. Nahaj 21:03, 25 October 2005 (UTC)
replies: text: There is an alternative to this: instead of a number of explicit axioms together with a substitution rule, one gives axiom schemes that determine an infinite set of individual axiom instances.  Having a substitution rule is usual in modal logic, having axiom schemes seems most common elsewhere. --- Charles Stewart 21:22, 25 October 2005 (UTC)

text: Postscript: IIRC, Hilbert and Ackermann's Principles of Theoretical Logic uses axiom schemes. --- Charles Stewart 21:24, 25 October 2005 (UTC)
replies: text:  In the early days of  Modal Logic, Axiom Schemata were VERY common. (And also rules like Adjunction, that were later proved to be derived rules in most systems of interest.)  But by about the 1960's or so substitution had become more common in that field.  In automated theorem proving substitution is more common, although a lot of recent automated work has used Condensed Detachment, which can be looked at as a (constrained) substitution followed by Modus Ponens.   -- Nahaj 02:11, 30 October 2005 (UTC)
thread_title:  Help me please 

message: This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 04:08, 10 November 2007 (UTC)
thread_title:  WikiProject class rating

message: Intuitionistic logic, or constructivist logic, is the symbolic logic system originally developed by Arend Heyting to provide a formal basis for Brouwer's programme of intuitionism.
Verifiabilistic logic, is the symbol not yet developed by anyone to provide a formal basis for wikipedias WP:V policy. (Just kidding!) 88.159.74.89 (talk) 15:14, 3 August 2009 (UTC)
thread_title:  Verifiabilistic logic 

message: I am not very familiar with intuitionistic logic, however, there seems to be a problem with the explanation of the syntax. Notice that `=' is used to define `¬' but `=' is not on the basic list of connectives. Either way, as an explanation for negation, using an undefined complex connective or a primitive but unlisted connective is not a good way to approach basic issues in syntax. I am not sure how to resolve this worry but it should be resolved. —Preceding unsigned comment added by 206.241.2.45 (talk) 17:43, 10 November 2009 (UTC)
replies: text: You are misreading it. The "=" was merely denoting what is being defined by what, it is not part of the definition itself. It never ceases to fascinate me what things are people capable of misunderstanding. One would think that this one is completely obvious, since apart from its being a fairly standard notation, ¬ appeared on the left-hand side of the equality sign and couldn't possibly be part of definition of itself. — Emil J. 18:03, 10 November 2009 (UTC)
replies: text: I rearranged the words, in any case, but I generally agree. — Carl (CBM · talk) 18:09, 10 November 2009 (UTC)
replies: text: Ah, I missed the second occurrence. — Emil J. 18:15, 10 November 2009 (UTC)
thread_title:  `=' as Connective 

message: PRED-1: (\forall x \ \phi (x)) \to \phi (t), if no free occurrence of x in \phi  is bound by a quantifier quantifying a variable occurring in the term t
Am I the only one for whom the condition does not even parse? I see how a variable that is free in one formula can be bound to a quantifier in another formula, but how can an occurance of one variable in a formula be bound in a term? Carsten Milkau (talk) 11:31, 27 September 2010 (UTC)
replies: text: It was apparently trying to say "if no free occurrence of x in \phi  is within the scope of a quantifier quantifying a variable occurring in the term t". I've replaced it with a more usual description.—Emil J. 12:38, 27 September 2010 (UTC)
replies: text: if no occurrence of any variable in t becomes bound in \phi(t) This is a lot better, thank you very much! — Carsten Milkau (talk) 14:58, 27 September 2010 (UTC)
thread_title:  Quantifier eleminiation 

message: there's something I am not convinced in this: 
 (\forall x \ \phi (x)) \to \phi (t),  if the term t is free for substitution for the variable x in \phi (i.e., if no occurrence of any variable in t becomes bound in  \phi(t)
Let's consider the sentence: 
  \forall x \ (\forall y \ (x,y > 0 \to x+y>0))  
which is of the form  \forall x \ \phi(x).  From my understanding of the definition, we cannot use the quantifier elimination axiom in this case to infer the sentence:
 \phi(y) = \forall y \ (y>0 \to y+y>0) 
that is \phi in which x is substituted with y, because y becomes bounded in \phi(y). However, there seems to be nothing wrong with this inference to me...
Maybe I didn't understand well the definition. I was thinking whether we could explain this better, maybe through some example.  
Enomisbomber (talk) 15:36, 18 February 2021 (UTC)
thread_title:  Again on the quantifier elimination 

message: In reference to the section on the non-interdefinability of the operators, I built a proof of the reverse direction of one of the identities that the article says is only provable in the forward direction. The identity in question is \neg (\phi \to \psi) \leftarrow \neg (\neg \phi \vee \psi). I can't do it in the combinator-style axiomatization presented (because I got confused trying to eliminate the abstractions) but the proof term in the equivalent abstraction-style axiomatization is: \lambda k ^ {\neg (\neg \phi \vee \psi) }. \lambda f^ {\phi \to \psi }. k (\iota_1 (\lambda a^\phi. k(\iota_2 (f a)))). Am I going insane here or is the article just mistaken? 209.51.184.11 (talk) 18:18, 9 October 2013 (UTC)
replies: text: Intuitionistic provability of this implication follows immediately from Glivenko’s theorem. I’ll fix that.—Emil J. 11:12, 10 October 2013 (UTC)

text: On second thought, I don’t see the point of listed this separately in the first place. Any of the other identities becomes intuitionistically provable as well if one puts negation on both sides. There’s nothing remarkable about this fact, and it does not really express interdefinability of connectives.—Emil J. 11:17, 10 October 2013 (UTC)
thread_title:  One more classical identity is provable intuitionistically 

message: Constructive logic redirects to this article, which begs the question of why it doesn't mention the constructive perspective on the semantics (BHK interpretation or the like)? Like many other mathematicians, my main impression of intuitionistic logic used to be that it seemed to be a lot of sophistery about arbitrarily declaring some proof techniques as "unclean", but with little reason to the matter beyond "L. E. J. Brouwer said so," and the present state of the article doesn't do much to counter that impression.
* "Intuitionists don't like the Law of Excluded Middle? Well, since I happen to be taught to use a Hilbert-style deduction system which doesn't even have that as an explicit inference rule or axiom, this means nothing to me."
* "Intuitionistic logic gets its semantics from Heyting algebras? Well, since these seem to be defined specifically to match the intuitionistic logic (rather than the other way around), then why should I care? Why not rather do fuzzy logic, if the classical one seems too traditional?"
The first thing that gave me any hint that there really could be something here was precisely the constructive perspective: we exclude those proof techniques because they are not valid in the constructive interpretation. Therefore the article should, at least as motivation, present some variant of this interpretation. 130.243.68.105 (talk) 17:14, 26 November 2019 (UTC)
thread_title:  Constructive semantics 

message: Hi. 
So I've read that in classical logic, you can basically define the binary logical operators with truth tables. (This may be an oversimplification, let me know if so)
Assuming that is the case, can you do something similar with intuitionistic truth operators? That is, does each intuitionistic binary truth operator accept two arguments, each from a particular set of (the intuitionistic equivalent of) truth values, and output a particular value from the same set?
Thanks for reading!
JonathanHopeThisIsUnique (talk) 23:56, 1 February 2020 (UTC)
thread_title:  Intuitionistic Truth Operators 

message: This page's introduction says 
> In the semantics of classical logic, propositional formulae are assigned truth values from the two-element set  ("true" and "false" respectively), regardless of whether we have direct evidence for either case. This is referred to as the 'law of excluded middle', because it excludes the possibility of any truth value besides 'true' or 'false'.
However, law of excluded middle is defined as (P or ~P) is true, not as there are only two truth values. Which page is correct? Wqwt (talk) 23:55, 19 September 2022 (UTC)
thread_title:  Definition of "law of excluded middle" in introduction