# Conjunction of Clauses and Well-Formed Formulas

Here is a theorem in my notes:

If $$\phi$$ is any wff such that $$\neg \phi$$ is not a tautology, then $$\phi$$ is tautologically equivalent to a conjunction of clauses.

My question is that...can this theorem hold if $$\phi$$ is not a tautology and so $$\neg \phi$$ is tautologically equivalent to a conjunction of clauses?

• How would you write a tautology as a conjunction of clauses? Does your definition allow $x \vee \neg x$ as a clause? – Fabio Somenzi Oct 15 '18 at 5:41
• Apply the theorem to $\lnot \phi$ : "If $\lnot \phi$ is any wff such that $¬¬ \phi$ (i.e. $\phi$) is not a tautology, then $\lnot \phi$ is tautologically equivalent to a conjunction of clauses." – Mauro ALLEGRANZA Oct 15 '18 at 12:51
• @MauroALLEGRANZA - Sorry, I wrote the answer while you were writing your comment. If you rewrite your comment as an answer, I can delete my answer. – Taroccoesbrocco Oct 15 '18 at 13:03

If $$\varphi$$ is any wff such that $$\varphi$$ is not a tautology, then $$\lnot \varphi$$ is tautologically equivalent to a conjunction of clauses.
is a corollary of the theorem stated in your question. Indeed, if $$\varphi$$ is any wff such that $$\varphi$$ is not a tautology, then $$\lnot \varphi$$ is a wff and $$\lnot \lnot \varphi$$ (which is equivalent to $$\varphi$$) is not a tautology. According to the theorem stated in your question (applied to $$\lnot \varphi$$), $$\lnot \varphi$$ is tautologically equivalent to a conjunction of clauses.
• In the theorem from the notes, the "such that $\neg \phi$ is not a tautology" part is redundant. There are plenty of ways to write $\phi$ as a conjunction of clauses if $\neg \phi$ is a tautology. If clauses like $x \vee \neg x$ are not allowed, then if $\phi$ is a tautology (as opposed to $\neg \phi$) then $\phi$ has no equivalent conjunction of clauses. – Fabio Somenzi Oct 15 '18 at 21:35