Is $A \vee B$ in its Conjunctive Normal Form?

Question

Since a conjunctive normal form consists of a conjuction of disjunctions, why is, say, $A \vee B$ in the conjunctive normal form?

Answer 1

For this we need to have a look at the definitions of conjunctive normal form, where recalling the definitions of finite disjunctions and conjunctions may be useful.

From van Dalen Logic and Structure (2008), p.25:

Definition 1.3.7 (Finite conjunctions/disjunctions):

\begin{cases} \bigwedge_{i \leq 0} \varphi_i = \varphi_0\\ \bigwedge_{i \leq n+1} \varphi_i = \bigwedge_{i \leq n} \varphi_i \wedge \varphi_{n+1}\\ \end{cases}

\begin{cases} \bigvee_{i \leq 0} \varphi_i = \varphi_0\\ \bigvee_{i \leq n+1} \varphi_i = \bigvee_{i \leq n} \varphi_i \vee \varphi_{n+1}\\ \end{cases}

Definition 1.3.8 (Conjunctive Normal Form): If$$\varphi =\bigwedge_{i \leq n}\bigvee_{j \leq m} \varphi_{ij}$$ where $ϕ_{ij}$ is atomic or the negation of an atom, then $\varphi$ is a conjunctive normal form.

Now we can see that what you observed is a particular case of when $n=0$: $$\varphi = \bigwedge_{i \leq 0}\bigvee_{j \leq m} \varphi_{ij} \Rightarrow \bigvee_{j \leq m} \varphi_{0j}$$ this follows from definition 1.3.7 above. Now let $m=1$: $$\varphi = \bigwedge_{i \leq 0}\bigvee_{j \leq 1} \varphi_{ij} \Rightarrow \bigvee_{j \leq 1} \varphi_{0j} \Rightarrow \bigvee_{j \leq 0} \varphi_{0j} \vee \varphi_{01} \Rightarrow \varphi_{00} \vee \varphi_{01}$$ Therefore $\varphi_{00} \vee \varphi_{01}$ is in conjunctive normal form (note that we can show that a similar result on conjunction holds for the disjunctive normal form).