In a logic system, a set $\Sigma$ of formulas is said to be inconsistent if $\Sigma \vdash \bot$, and consistent otherwise.
Does it mean that $\Sigma$ is consistent if and only if $\Sigma \vdash \top$? Thanks.
[Math] A consistent set of formulas
logic
Related Solutions
As a footnote to Asaf's answer, I suspect that what the OP was after by way of an "intuitive definition" was in fact the idea that
(A) A set of wffs $\Sigma$ is inconsistent just when there is a proposition $A$ such that $\Sigma \vdash A$ and $\Sigma \vdash \neg A$.
And the question is how this relates to the textbook definition
(B) A set of wffs $\Sigma$ is inconsistent iff $\Sigma \vdash \bot$.
There's a third familiar definition
(C) A set of wffs $\Sigma$ is inconsistent iff, for all wffs $A$ (of the relevant language) $\Sigma \vdash A$.
Well, these come to just the same in a classical (and intuitionist) setting. The OP's sketched proof-idea, together with the rule that $\bot$ entails anything, can be re-used to show that (A) is equivalent to (B) and to (C). (In fancier logics, or more restricted logics these definitions can of course peel apart.)
This is incorrect: consider the situation where we have two propositional atoms $p$ and $q$, and $\Sigma=\emptyset$ (so doesn't prove anything other than tautologies). According to your definition of $u$, there is no restriction on what truth values can be assigned to each of the following sentences (each of which is undecidable in $\Sigma$):
$p$,
$q$,
$\neg(p\wedge q)$.
However, clearly we can't assign each of them "$\top$." So your approach for defining a valuation for $\Sigma$ needs to be more complicated: it needs to take into account how the various sentences, even those undecidable in $\Sigma$, interact with each other.
Note that this problem would not arise if we assumed additionally that $\Sigma$ was complete: that is, for each $\varphi$, either $\varphi\in \Sigma$ or $\neg\varphi\in\Sigma$. (Indeed, then the third clause of your definition of $u$ would be unnecessary.) So what is immediately true is:
$(*)\quad$ Any complete consistent theory has a model.
So now you need to prove:
$(**)\quad$ Any consistent theory is contained in a complete consistent theory.
It's also worth mentioning for contrast what happens in first-order logic. (In particular, this old question nicely parallels yours.)
In first-order logic, the notion of satisfiability is more complex: a model is not just a truth assignment. We still have "consistency implies satisfiability" (for the right proof system, anyways), but the proof is a bit more complicated. However, it evolves quite nicely from the proof of the completeness theorem for propositional logic; see this exposition by Bezhanishvili.
Best Answer
Every set of first-order sentences proves $\mbox{True},$ even the empty set; so, you're not negating correctly. If done correctly, you will see that 'consistent' equals 'does not prove false.' Lets just spell out the definitions to make this as clear as possible.
Definition 0. $\Sigma$ is inconsistent iff $\Sigma \vdash \mbox{False}.$
Definition 1. $\Sigma$ is consistent iff not ($\Sigma$ is inconsistent.)
Therefore, $\Sigma$ is consistent iff not ($\Sigma \vdash \mbox{False}.$)