# Propositional logic — syntactical completeness

Lets consider propositional logic. We say a proof system for propositional logic is syntactically (negation) complete if for every $\alpha$, either $\alpha$ or $\neg \alpha$ are provable within the system, i.e., $\Sigma \vdash \alpha$ or $\Sigma \vdash \neg \alpha$. (I assume standard definitions for $\Sigma$ and $\vdash$).

It seems to be me that no sound proof system for propositional logic could prove $p$ or $\neg p$ from empty set of sentences, where $p$ is a propositional atom. So, propositional logic is not syntactically complete.

Now, it is to my understanding that this is kind of completeness Godel was considering in his Incompleteness Theorem for FOL. Since FOL (with abuse of terminology) subsumes propositional logic, it trivially holds that FOL is not syntactically complete either. This seems to simple, so I am wondering what am I doing wrong? Or his proof focused on Peano arithmetic specifically?

• I think you're mixing up propositional and first-order logic. The incompleteness theorem applies to first-order logic, not propositional logic. – templatetypedef Aug 23 '14 at 18:13
• I have stated that his theorem applies to FOL. My confusion stemmed from the wrong assumption of what constitutes a FOL theory, in particular the language part of it. – bellpeace Aug 23 '14 at 18:30

You are not doing anything "wrong", you just missed the point of the incompleteness theorem, a bit.

What you claim (correctly) is that a proof system with no axioms is incomplete. You also say that this is not very impressive...

What Godel showed is something much stronger: consider the theory of Peano arithmetic, then there does not exist a complete axiomatization for it.

How does this relate to your post? Well, in your suggestion, even when you consider an empty set of axioms, there are still some true sentences (the tautologies), and there are axiom systems that capture exactly those sentences.

• Simple follow-up question. Does one usually prove (in)completeness w.r.t. a subset of the language also? As far as I know, PA axioms do not tell anything about propositional atoms, so incompleteness would then trivially hold by my argument. – bellpeace Aug 23 '14 at 14:37
• I don't really understand the question. Please elaborate, preferably as a new post. – Shaull Aug 23 '14 at 15:50
• I'll try to do it here, as it is probably just my simple misunderstanding. Consider the theory of PA, and consider any propositional atom $p$. Then, again you cannot prove $p$ nor $\neg p$, hence incompleteness. We don't consider propositional atoms in PA? Basically, are we just focusing on a subset of FOL language in this case? – bellpeace Aug 23 '14 at 16:18
• Well, PA is a first-order theory. There, propositional atoms are captured by 0-ary relations. In PA, there are no 0-ary relations, so you can't even say $p$. This makes this problem nonexistent. – Shaull Aug 23 '14 at 16:24
• Ok, this is what I had in mind when saying a subset of the FOL language. Thanks! – bellpeace Aug 23 '14 at 18:31