TITLE: Don't Gödel's completeness and incompleteness theorems contradict each other?
QUESTION [7 upvotes]: Gödel's completeness theorem: Given a set of axioms, if we cannot derive a contradiction, then  the system of axioms must be consistent.
Gödel's incompleteness theorem:'Given any consistent, computable set of axioms, there's a true statement about the integers that can never be proved from those axioms'.
Since the incompleteness theorem assumes that the given set of axioms is consistent, then, by using the completeness theorem shouldn't all statements derived from this set of axioms be true thereby rendering the claim that 'there exists a true statement that cannot be proved' false?

REPLY [6 votes]: To add to Asaf's useful answer.

"Gödel's completeness theorem: Given a set of axioms, if we cannot derive a contradiction, then the system of axioms must be consistent." No, that would trivialize the result, since consistency is defined as failing to entail a contradiction! The completeness theorem says that, if a bunch of sentences is consistent in the sense of not entailing a contradiction [in a standard system of first-order logic] then it has a model.
"Gödel's incompleteness theorem:'Given any consistent, computable set of axioms, there's a true statement about the integers that can never be proved from those axioms'." Better, 'Given any consistent, computable set of axioms, there's a statement $\varphi$ in the language of arithmetic such that the axioms prove neither $\varphi$ nor $\neg\varphi$' Since, given an interpretation, one of $\varphi$ and $\neg\varphi$ will be true, your statement follows. But the fundamental result doesn't involve the notion of truth but is purely syntactic.
With those clarifications, we can see better why there is no tension between them. The completeness theorem tells us that, given a set of first-order axioms in a language that includes the language of arithmetic, there is an interpretation which makes all those axioms [and their consequences] true. But nothing at all follows from that about whether the axioms will suffice to prove one of $\varphi$ and $\neg\varphi$ for any arithmetical sentence $\varphi$. It only shows that there is a model for all the sentences which it can prove.