# What is the different semantic of existential and universal qualifier between classical and intuitionistic logic?

Take the semantic definition of existential and universal qualifier outlined here

Is that valid for classical or intuitionistic logic? Or is it a general definition?

Take the following sentence:

"It exists a man that, if he owns an hat, then all mans own an hat"

Take $C(x)$ is the predicate:
"Man $x$ owns an hat"

than the sentence can be formally written as follow:

$\exists x(C(x) \rightarrow \forall xC(x))$

How can that sentence be semantically interpreted using the initial definition of existential and universal qualifier?

• Are you asking about mathematical structures, like models of (classical) first order logic or about the philosophical/conceptual differences? In the former case you should look up realizability models, Kripke models or topological models. In the latter case, the Brower Heyting Kolmogorov interpretation of intuitionistic logic. – aws Jul 22 '13 at 12:46
• Where have you looked? What particular type of definition are you looking for? – Carl Mummert Jul 22 '13 at 12:47
• @CarlMummert: I've looked on Wikipedia ( First order logic ) since now it is my only available source. – Robbo Jul 22 '13 at 12:59
• @aws: By now might be sufficient the conceptual difference. – Robbo Jul 22 '13 at 13:03

The semantic definitions you linked to in the first line of the question would be acceptable intuitionistically as well as classically, but an intuitionist would take "there exists an evaluation $\mu'$" to mean that one can actually exhibit an appropriate $\mu'$. In other words, these definitions are intuitionistically acceptable in the context of an intuitionistic meta-theory.

From a philosophical viewpoint, one usually regards intuitionistic logic as differing from classical logic mainly in the interpretations of $\lor$ and $\exists$. Intuitionism requires, for the assertion of $\phi\lor\psi$ or of $\exists x\,\phi(x)$, that one can (at least in principle) assert one of the disjuncts (in the case of $\lor$) or one specific instance (in the case of $\exists$). Because intuitionistic reasoning respects these requirements, proofs in intuitionistic systems often have far clearer computational content than proofs in similar classical systems.

I think there's quite a nice explanation in the SEP article on constructive mathematics, in particular the section on the constructive interpretation of logic. It's best not to look at existential and universal quantifiers in isolation, but look at the interpretation as a whole, especially since disjunction and implication are also very important.

Since there are already good explanations around, I'll just give a quick example of when existential quantification is different in classical and intuitionistic logic:

Let $P$ be any proposition. For instance, take it to be some unproved conjecture eg the twin prime conjecture. In classical logic we can easily show that there is some number $n$ such that either $n$ is $0$ and $P$ is false or $n$ is $1$ and $P$ is true, by applying an instance of excluded middle $P \vee \neg P$. In intuitionistic logic you can't do this in general because you need to supply an actual witness for $n$. In particular you need to know whether $n$ is $0$ or $1$, which you can't do without either proving or disproving $P$.

• Taking P is the twin prime conjecture, what does $n$ stand for? – Robbo Jul 22 '13 at 15:45
• In classical logic $n$ is defined to be $1$ if there are infinitely many twin primes or $0$ if there are not infinitely many. So it is probably equal to $1$. – aws Jul 22 '13 at 16:14
• There's actually another classic example that you might find much clearer: there exist $a$ and $b$ irrational such that $a^b$ is rational. It's in the introduction of the article I linked to (plato.stanford.edu/entries/mathematics-constructive/#1). – aws Jul 22 '13 at 16:18

One example is that in intuitionistic logic the usual De Morgan's laws do not hold (that is, only three out of four implications are true, e.g. see here), in particular

$$\neg(p \land q)\quad \text{ does not imply }\quad \neg p \lor \neg q.$$

Similarly

$$\neg \forall x. P(x) \quad \text{ does not imply } \quad \exists x. \neg P(x).$$

I hope it helps ;-)

Edit: Removed some false claims.

• Sorry for such a late reply. I don't see how to prove ¬(¬p ∨ ¬q) ⇒ p ∧ q. It looks as if one would need to manufacture proofs of p and q out of thin air (i.e by contradiction). I can prove ¬p ∧ ¬q ⇔ ¬(q ∨ p), though. I can also prove (∀x. ¬(P x)) ⇔ ¬(∃x. P x). – James Wood May 27 '15 at 19:00
• @JamesWood You can use Heyting algebras to show it is wrong (like here). It should have been $$\neg (p \land q) \iff \neg\neg(\neg p \lor \neg q),$$ I have no idea what made me write such nonsense, sorry for that. – dtldarek May 28 '15 at 7:58