Diagonal lemma

For every one-variable arithmetical formula $\theta (x)$, a sufficiently strong theory of arithmetic has a sentence $\gamma$ satisfying $\gamma \leftrightarrow \theta (┌\gamma ┐)$.

Every consistent recursively axiomatized extension of elementary arithmetic is incomplete. If it were complete, its theorem set would be decidable; representing that decision procedure and diagonalizing against it yields a contradiction.