Conjunctive Normal Form Is a statement of the form $\phi \vee \psi \vee \xi$ considered to be in its conjuntive normal form (CNF), given that $\phi \vee \psi$ is considered to be in CNF? 
Example: While converting $\phi \wedge \psi \rightarrow \xi$ to its CNF, we get $\neg(\phi \wedge \psi) \vee \xi$ which gives $(\neg \phi) \vee \neg(\psi) \vee \xi$. Is this statement in its CNF?
 A: Here's a definition of a clause adapted from Merrie Bergmann's An Introduction to Many-Valued and Fuzzy Logic p. 20:


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*A literal (a letter or negation of a letter) is a clause.

*If P and Q are clauses, then (P $\lor$ Q) is a clause.


Definition of conjunctive normal form.


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*Every clause is in conjunctive normal form.

*If P and Q are in conjunctive normal form, then (P$\land$Q) is in conjunctive normal form.


So, even though ϕ∨ψ∨ξ is not in conjunctive normal form (note the parentheses), 
((ϕ∨ψ)∨ξ) is in conjunctive normal form and
(ϕ∨(ψ∨ξ)) is also in conjunctive normal form.
Demonstration:
Suppose that ϕ, ψ, and ξ are literals.  Since ϕ, and ψ are literals, and literals are clauses, by definition of a clause and detachment, (ϕ∨ψ) is a clause.  Since ξ is a literal, ξ is a clause.  Thus, by definition of a clause and detachment, ((ϕ∨ψ)∨ξ) is a clause.  Since every clause is in conjunctive normal, ((ϕ∨ψ)∨ξ) is in conjunctive normal form.
One can similarly show that ((ϕ∨ψ)∨ξ) is a clause by building it up from its literals using part 2. of the above definition of a clause, and invoking the definition of conjunctive normal form to infer that ((ϕ∨ψ)∨ξ) is a clause.
