User blog comment:KthulhuHimself/I'm trying to (or could use some help) prove something interesting./@comment-1605058-20161121121812

Let $$S$$ be any statement and $$\neg S$$ its negation. Then we can't have both $$S\in P,\neg S\in P$$, since we've assumed $$P$$ is free of contradictions. By law of excluded middle, either $$S\not\in P$$, or $$S\in P$$ and then $$\neg S\not\in P$$. In either case, we have an example of $$A\not\in P$$.

However, this isn't Goedel's incompleteness, but the definition of consistency (=lack of contradictions). Goedel's incompleteness-like statement would state something like $$\forall P\exists A:A\not\in P\land\neg A\not\in P$$. Unfortunatelly, this is false, as there are consistent theories $$P$$ which, for every statement, prove either it or its negation. Goedel's incompleteness puts restrictions on what properties $$P$$ can satisfy (specifically, it can't have recursively enumerable set of axioms and interpret axioms of arithmetic at the same time).