On this page https://en.wikipedia.org/wiki/Principle_of_bivalence :
Intuitionistic logic is a two-valued logic but the law of excluded middle does not hold.
On this page https://en.wikipedia.org/wiki/Truth_value :
In intuitionistic logic, and more generally, constructive mathematics, statements are assigned a truth value only if they can be given a constructive proof. It starts with a set of axioms, and a statement is true if one can build a proof of the statement from those axioms. A statement is false if one can deduce a contradiction from it. This leaves open the possibility of statements that have not yet been assigned a truth value. Unproven statements in intuitionistic logic are not given an intermediate truth value (as is sometimes mistakenly asserted). Indeed, one can prove that they have no third truth value, a result dating back to Glivenko in 1928.[3]
Instead, statements simply remain of unknown truth value, until they are either proven or disproven.
Could you explain to me what does it mean ? Is intuitionist logic two-valued and there is a subtlety, or is the first quote wrong as asserted by the second ?
Best Answer
First, give us a definition of what it means for a logic to be $n$-valued; then, we'll be able to tell whether intuitionistic logic is 2-valued or not.
Since there is no standard, widely-accepted notion or definition of "$n$-valued logic" at all, discussion regarding whether a given logic is $n$-valued or not tend to be very confused at best. The discussion on Wikipedia is no exception in this regard.
One could define many different notions of $n$-valuedness. Below, I consider four possible such definitions.
In algebraic logic, one has a notion of $n$-valued semantics. This is not the definition we have in mind when we say that classical logic is 2-valued, since, while classical propositional logic has a complete 2-valued semantics, it also has a complete 4-valued semantics, where the four possible truth values are the sets $\emptyset, \{1\}, \{2\}, \{1,2\}$, conjunction is interpreted as intersection, and the negation of $x$ is interpreted as the relative complement $\{1,2\} \setminus x$. Saying that classical logic is therefore a 4-valued logic would be weird and confusing, to say the least. On the other hand, intuitionistic logic cannot have a complete $n$-valued semantics for any natural number $n$.
We could also try to pin down the notion of $n$-valuedness internally, without committing to a particular semantics. E.g. one could say that a subsystem of classical logic has no more than $n$ truth values if $$\bigvee_{0 \leq i \leq n} \bigvee_{0 \leq j < i} (A_i \leftrightarrow A_j)$$ is a tautology, i.e. out of every $n$ formulas, at least two are logically equivalent. Then, as expected, classical logic has no more than 2 truth values since $(A_1 \leftrightarrow A_0) \vee (A_2 \leftrightarrow A_0) \vee (A_2 \leftrightarrow A_1)$ is a classical tautology, and is not $1$-valued since $A_1 \leftrightarrow A_0$ is not a classical tautology. However, using the disjunction property, one can again show that intuitionistic logic is not $n$-valued in this sense for any $n \in \mathbb{N}$.
The Wikipedia page links a Stanford Encyclopedia of Philosophy article, which implicitly uses another ad-hoc definition of truth value. The basic story is as follows: we can say that a logic "admits $n$ truth values" if we can find $n$ formulas $f_1(P),\dots,f_n(P)$ in one free propositional variable $P$, so that we can prove $f_i(P) \rightarrow \neg f_j(P)$ for all $i \neq j$, and for each $f_n$ we can find some formula $X_n$ for which $f_n(X_n)$ has a proof in our logic. In this sense classical logic admits at least 2 truth values: we can take $f_1(P) = P$, $X_1 = P \vee \neg P$, $f_2(P) = \neg P$ and $X_2 = P \wedge \neg P$. Using a result from Glivenko's 1928 paper, one can show that it's not possible to extend this set $f_1,f_2$ with any third $f_3$ and $X_3$, even in intuitionistic logic: no "third truth value" is ever taken. Using later results of Rieger and Nishimura, one can prove a much stronger statement: for any 3 formulas $A_1,A_2,A_3$ in one free propositional variable $P$, at least one of $A_i \rightarrow A_j$ with $i \neq j$ is always an intuitionistic tautology. In this sense, intuitionistic logic admits only two truth values, as does classical logic.
There are, of course, systems that people traditionally call systems of many-valued logic, such as Łukasiewicz's 3-valued logic, or Belnap's four-valued logic. But these are just traditional names, and do not fall under any general definition of $n$-valued logic. In particular, intuitionistic logic is not usually included at all in the study of "many-valued logic".
Again, none of these definitions are standard or widely used across logic. This should answer your main question: there is a subtlety, namely that since "$n$-valuedness of a logic" doesn't actually have a standard definition, whether intuitionistic logic is two-valued or not will depend on the definition you choose to adopt.