I am attempting to back two claims in this problem:
I use $\textbf{Q}$ to denote minimal arithmetic for this post.
I use the term 'rudimentary sentence' to denote formulas built using only negation, conjunction, disjunction, and bounded quantifications. It has been pointed out to me that these are more formally called $\Delta_0$ sentences. Additionally, '$\exists$-rudimentary' refers to $\Sigma_1$ sentences.
A theory $T$ is 1-consistent if for all $\exists$-rudimentary sentences of the form $\exists x$ $F(x)$ , if $T\vdash \neg F(0)$, $T\vdash \neg F(1)$, $T\vdash \neg F(2)$,... then $T\not\vdash\exists x$ $F(x)$.
To show: If $T$ is 1-consistent, then $T$ is consistent.
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From this point on, I assume that $T$ is a 1-consistent theory.
I use a function now $Prv_T(x)$ to mean $\vdash_T A$ if and only if the sentence $Prv(\ulcorner A\urcorner )$ is correct under the standard interpretation.
Let $T$ now be a 1-consistent ($\Sigma_1$-sound) theory extending minimal arithmetic.
I let $G$ be such that
$T \vdash G \leftrightarrow Prv(\ulcorner G\urcorner)$.
To show: $T\not\vdash \neg G$.
I apologize in advance if this is unclear, I am not yet totally comfortable communicating this material.
Your diagonalization ansatz seems to be overkill for this purpose.
Instead simply consider that in an inconsistent theory everything is provable. So if $T$ is inconsistent, you can let $F$ be any formula whatsoever; the premise of your definition of "1-consistent" will then be satisfied but the conclusion is not. So it is impossible for a theory to be inconsistent yet 1-consistent.