Propositional calculus and intuitionist logic

Truth table semantics is particular to classical propositional logic. So if the author were being specific, they would have added this qualifier. (However it’s probably a good rule of thumb that if you just see “propositional calculus,” in most contexts they are referring to classical.)

While “propositional” is a descriptor that mainly refers to the language (in particular, the absence of non logical symbols other than sentence letters), as soon as we start talking about proof systems or semantics, we have a particular logic in mind: classical, intuistionistic, minimal, modal, etc.

In classical propositional logic, a semantic interpretation can be more generally provided by a Boolean algebra, which is a set $X$ with constants $\top, \bot$, unary operation $\lnot$, and binary operations $\wedge, \vee$ satisfying certain axioms. The prototypical Boolean algebra is $\{ T, F \}$ with the obvious interpretations.

In intuitionistic propositional logic, the corresponding construction is a Heyting algebra, which is a set $X$ with constants $\top, \bot$, and binary operations $\wedge, \vee, \rightarrow$ satisfying certain axioms. Given a Heyting algebra, we can then often define $\lnot x := (x \rightarrow \bot)$; however, in a general Heyting algebra, some of the Boolean algebra axioms or derived identities such as $x \vee \lnot x = \top$, $\lnot(\lnot x) = x$, and $\lnot(x \wedge y) = (\lnot x) \vee (\lnot y)$ no longer hold.

An interesting example of a Heyting algebra which is not a Boolean algebra is if $X$ is the set of open subsets of $\mathbb{R}$, with $\top := \mathbb{R}$, $\bot := \emptyset$, $U \wedge V := U \cap V$, $U \vee V := U \cup V$, and $U \rightarrow V := \operatorname{int}((\mathbb{R} \setminus U) \cup V)$. In this Heyting algebra, for example if we set $U := \mathbb{R} \setminus \{ 0 \}$, then $\lnot U = (U \rightarrow \emptyset) = \operatorname{int}(\{ 0 \}) = \emptyset$, so $U \vee \lnot U = \mathbb{R} \setminus \{ 0 \} \neq \top$, and $\lnot (\lnot U) = \mathbb{R} \ne U$. (Of course, there's nothing particular special about $\mathbb{R}$ in this example: the same construction will work for the open subsets of any topological space.)

In case you're not comfortable with topology or real analysis, there's also a finite example that's often used: $X := \{ 0, \frac{1}{2}, 1 \}$; $\top := 1$; $\bot := 0$; $x \vee y := \max(x,y)$; $x \wedge y := \min(x,y)$; and $x \rightarrow y$ is given by a table. In this Heyting algebra, $\lnot(\frac{1}{2}) = (\frac{1}{2} \rightarrow 0) = 0$, so $\frac{1}{2} \vee \lnot(\frac{1}{2}) = \frac{1}{2} \ne \top$.