I'm stuck on some fallacious argument. A theory $T=(\Delta, \Sigma)$ is a pair -- of axioms $\Delta$ with a signature $\Sigma$. And I have the following "proof" that every theory is complete (that is $\Delta \vdash p$ or $\Delta \vdash \neg p$).
Let $T$ be an arbitrary theory and $p\in \Sigma$ be arbitrary. Suppose that $\Delta \vdash \neg p$ then we're done; otherwise $\Delta \not \vdash \neg p$. By the Completeness Theorem it follows that $\Delta \not \models \neg p$. But this means that there exists an interpretation $v:\Sigma \to \{0,1\}$ such that $v(q)=1$ for all $q\in \Delta$ and $v(\neg p)=0$. But as part of the definition of "interpretation" we have $v(\neg p)=0$ if and only if $v(p)=1$. Then $\Delta \models p$ and so by the Completeness Theorem we get $\Delta \vdash p$. Therefore, $T$ is a complete theory. Therefore, every theory is complete. .... but this is obviously false.
It seems the definition of "interpretation" begs the question ..... that is to say, in our definition already assumes that either $p$ or its negation is semantically true.
It is not correct to conclude that $\Delta\models p$. To conclude that, you would need to know that $v(p)=1$ for every interpretation $v$ such that $v(q)=1$ for all $q\in\Delta$. You only know there exists such an interpretation for which $v(p)=1$. Maybe for some such interpretations $v(p)=1$, and for others $v(p)=0$.