Three atomic forms expression both in disjunctive and in conjunctive form?

Question

we know that A v B is in both conjunctive and in disjunctive normal form.

we also know that A ^ B is in both conjunctive and in disjunctive normal form.

Does it follow from this, that A v B v C is in CNF and in DNF

and that A ^ B ^ C is in CNF and in DNF?

And could we continue this infinitely?

(we can put the parentheses there if we want and as we want)

Thanks!

Answer 1

Actually, I don't know that A v B is in conjunctive normal form.

Here's a definition of a clause adapted from Merrie Bergmann's An Introduction to Many-Valued and Fuzzy Logic p. 20:

1. A literal (a letter or negation of a letter) is a clause.
2. If P and Q are clauses, then (P∨Q) is a clause.

Definition of conjunctive normal form.

1. Every clause is in conjunctive normal form.
2. If P and Q are in conjunctive normal form, then (P∧Q) is in conjunctive normal form.

Here's a definition of a phrase from p. 18 of the same book.

1. A literal is a phrase.
2. If P and Q are phrases, so is (P$\land$Q).

And disjunctive normal form:

1. Every phrase is in disjunctive normal form.
2. If P and Q are in disjunctive normal form, so is (P$\lor$Q).

Now, A, B, and C are all phrases, as well as clauses by condition 1. of both definitions above. Thus, A, B, and C are in conjunctive normal form and disjunctive normal form, and (A$\land$B) and (A$\lor$B) are in disjunctive normal form, and conjunctive normal form. Now, C is a clause and thus in conjunctive normal form by the above. So, since C and (A$\land$B) are in conjunctive normal form, so is [(A$\land$B)$\land$C]. Similarly, one can show that [(A$\lor$B)$\lor$C] is in disjunctive normal form.

If we subscript literals, A$_1$, ..., A$_n$, then by an inductive argument we can show that ((A$_1$$\land$A$_2$)$\land$...A$_n$) is in both conjunctive normal form and disjunctive normal form... at least once the parenthesizitation becomes clear.