Is it possible to deduce Godel's first incompleteness theorem from Chaitin's incompleteness theorem?
Gödel's incompleteness theorem, in its modern form using Rosser's trick, only requires that (beyond being effective and sufficiently strong), the theory must be consistent. There is no requirement that the theory has to be $\omega$-consistent or meet any soundness assumption beyond simple consistency. You cannot apply Chaitin's theorem in its usual form to these theories, in general, because the usual form of Chaitin's theorem assumes more (e.g. many proofs of Chaitin's theorem assume as an extra soundness assumption that the theory only proves true statements.)
Many of the "alternate proofs" also require stronger assumptions than the standard proof of the incompleteness theorem. You have to be very careful when reading about this to check which assumptions are included in each theorem.
In the particular proof of Chaitin's theorem presented by Kritchman and Raz, however, they do not need to assume any particular soundness beyond just consistency. I am going to explain this in detail.
They do need to assume that $T$ is sufficiently strong. In particular, they assume that if $n$ and $L$ satisfy $K(n) < L$, then $T \vdash \hat K(\dot n) < \dot L$. Here $\dot n$ and $\dot L$ are terms of the form $1 + 1 + \cdots + 1$ corresponding to $n$ and $L$, and $\hat K$ is a formula of arithmetic defined directly from the definition of $K$. (Note that the set of pairs $(n,L)$ with $K(n) < L$ is recursively enumerable, so there is no real issue in assuming $T$ proves all true sentences of that form.)
Given the assumption on $T$, their proof goes as follows (rephrased in more precise terms):
Begin proof
Given $L$, we can make a program $e_L$ that does the following:
Search for any $n$ such that $T \vdash \hat K (\dot n) > \dot L$. We do this by searching through all $T$-proofs in an exhaustive manner.
Output the first such $n$ we find, if we ever find one.
Because we can code $L$ into the program as a fixed constant, using the standard coding methods, the length $|e_L|$ of $e_L$ no worse than $2\log(L) + C$ for some constant $C$. In particular, we can fix an $L$ with $|e_L| < L$. Assume such an $L$ is fixed.
For this $L$, suppose $e_L$ returns a value, $n$. Then $K(n) \leq |e_L| < L$. By our assumption that $T$ is sufficiently strong, this means $T$ proves $\hat K (\dot n) < \dot L$.
But if $e_L$ returns $n$ then $T$ also proves $\hat K(\dot n) > \dot L$. This means that if $e_L$ returns a value then $T$ is inconsistent. So, if we assume $T$ is consistent, then $e_L$ cannot return a value. This means that, for our fixed $L= L_T$, $T$ cannot prove $\hat K(\dot n) > \dot L$ for any $n$.
Thus, if we take $n = n_T$ to be such that $K(n) > L$, we have that $\hat K (\dot n) > \dot L$ is a true statement that is not provable in $T$.
End proof
This proof just given (in italics) proves:
If $T$ is a consistent, effective theory of arithmetic that proves every true formula of the form $\hat K(\dot n) < \dot L$, then there is a true statement of the form $\hat K(\dot n) > \dot L$ that is not provable in $T$.
This is almost the same as Gödel's incompleteness theorem. The only difference is that the usual proof of the incompleteness theorem gives us an explicit formula that it is true but not provable in $T$ (namely, the Rosser sentence of $T$). On the other hand, the proof in italics requires us to find $n_T$ in order to have an explicit example, and there is no algorithm to do this.
This is one motivation for what I think of as the "standard form" of Chaitin's theorem. In that form, we look instead for unprovable sentences of the form ($\dagger$): $(\exists x)[\hat K(x) > \dot L]$.
Because we can compute $L= L_T$, we can compute a specific sentence of that form for $T$. But, for the proof to work, we need to have an actual $n$ such that $T \vdash \hat K(\dot n) > \dot L$. So we have to add an additional soundness assumption to the theorem, namely that if $T$ proves a sentence of the form ($\dagger$) then there is an $n$ such that $T$ proves $\hat K(\dot n) > \dot L$.
Overall, in this version of Chaitin's theorem, we have an explicit sentence, but with a stronger soundness assumption. The proof of Gödel's incompleteness theorem using Rosser's method gives us an explicit sentence without any stronger soundness assumption.
Your confusing stems from the way many articles about Godel's incompleteness theorems are extremely imprecise. Here is a proper definition.
$\def\nn{\mathbb{N}}$
We say that a sentence $φ$ over a language $L$ is true in an $L$-structure $M$ iff $M \vDash φ$.
For convenience, when $L$ is the language of arithmetic, we say that $φ$ is true iff $\nn \vDash φ$.
Note that these definitions are only possible in a meta-system that already has a collection called $\nn$ (also known as the standard model of PA). Thus:
$\def\t#1{\text{#1}}$
$\def\con{\t{Con}}$
$\def\pa{\t{PA}}$
"$φ$ is true but unprovable" is more precisely "$\nn \vDash φ$ and $\pa \nvdash φ$".
Now there is a sentence over PA denoted by $\con(\pa)$ such that PA is consistent iff $\nn \vDash \con(\pa)$ (in other words PA is consistent iff $\con(\pa)$ is true in the standard model). It is in fact non-trivial to show that such a sentence exists, which is a crucial part of Godel's first incompleteness theorem.
The remainder of the incompleteness theorem shows that $\pa \nvdash \con(\pa)$. But the meta-system we choose always has $\nn \vDash \pa$, so $\pa$ is consistent and hence $\nn \vDash \con(\pa)$. Thus $\con(\pa)$ is the first natural example of a sentence that is true but unprovable (in the precise sense defined above).
Note that it is false that every true but unprovable sentence $φ$ can be proven by $\pa+\con(\pa)$. In particular, $\pa+\con(\pa) \nvdash \con(\pa+\con(\pa))$, even though $\nn \vDash \con(\pa+\con(\pa))$ (by essentially the same argument as above). This can be proven simply by applying Godel's proof of the incompleteness theorem to $\pa+\con(\pa)$.
Better still, we can let $\pa_0 = \pa$ and recursively let $\pa_{k+1} = \pa_k + \con(\pa_k)$ for every $k \in \nn$, and then let $\pa_ω = \bigcup_{k\in\nn} \pa_k$. Then we still have $\nn \vDash \pa_ω$, and yet $\pa_ω \nvdash \con(\pa_ω)$ even though $\pa_ω \vdash \con(\pa_k)$ for every $k \in \nn$.
Best Answer
Do you know "a fortiori"?
It's the "a maiore ad minus" argument, concluding from a general to a more special ("particular") case. https://en.m.wikipedia.org/wiki/Argumentum_a_fortiori
As in:
"All cats are mammals. In particular, all black cats are mammals."