I'm wondering about the possibility of circumventing the problem of incompleteness posed by Roger Penrose in his book "Shadows of the Mind".
It occurred to me (and, Googling has revealed to me, others) that you can purchase a formal system's completeness as the price of consistency, if you adopt a paraconsistent logic and a "falsificationist" approach by which your aim is to find consistent subsets of all your enumerable propositions. This seems to me to be exactly what human beings are doing when they "intuit" the truth of the Gödelian sentence. I don't yet know enough about logic to fully develop the idea, however.
In particular, it seems to me that for a falsificationist approach to work, you need to have it such that $(\Gamma \models \perp) \rightarrow (\Gamma \vdash \perp)$ in order to guarantee (or at least render highly probable) that you will eventually discover an inconsistency in your system, if it exists.
It seems that for any system with an effective procedure for determining if $\Gamma \models \phi$ for some $\phi$, the requirement is met since for a procedure to be "effective" it (caveat: seems to me) that it would have to be equivalent to some kind of syntactic manipulation of the formulae in $\Gamma$.
In addition to a straightforward "yes", "no", "you're asking the wrong question", etc. style of answer I'd also highly appreciate links to the literature concerning work on the above problem specifically.
(EDIT) I've found another question which bears rather directly on this one, here: Gödel's Completeness Theorem and logical consequence
This is the content of Godel's completeness theorem. $T\models\varphi$ if and only if $T\vdash\varphi$, at least when we are discussing first-order logic. Note, by the way, that since there is no structure which satisfies $\bot$, so writing $T\models\bot$ means by definition that $T$ has no models.
Interestingly, this is something which hinges on the axiom of choice. Namely, the proof of the completeness theorem uses the axiom of choice. To see how the implication you're interested in uses the axiom of choice, consider the case there exists a set $A$ which cannot be linearly ordered (something which strongly contradicts the axiom of choice).
Let $\mathcal L_A$ is the language with constants $\{c_a\mid a\in A\}$ and a binary relation symbol $<$. Now take $T$ to be the theory stating that $c_a\neq c_b$ for $a\neq b$ and $<$ is a linear ordering of the universe. This theory is finitely satisfiable, so it cannot prove $\bot$; but it has no models since from a model of $T$ we obtain a linear ordering of $A$. (Note that the language here is uncountable, since $A$ is uncountable. If $\cal L$ is a countable language in first-order logic, then the completeness theorem for $\cal L$-theories is in fact "choice free". So for things like arithmetics we don't use the axiom of choice.)
Finally, let me point out that incompleteness here in the sense of Godelian sentences and Godel-Rosser sentences, have nothing to do with the completeness of the underlying logic, or the completeness/consistency of a theory. Those are different type of completeness (and lack thereof), and they should not be confused.