Is the following condition equivalent to a first order theory $T$ being $\omega$-consistent?
$\forall y: \big{(} \exists x \, [\operatorname {Proof}_T (x,y)] \uparrow \exists z \, [ \operatorname {Proof}_T(z,\operatorname {neg}(y))]\big{)}$
Where $\uparrow; \operatorname {Proof}_T(x,y);\operatorname {Proof}_T(x,\operatorname {neg}(y))$ are defined as usual (see here, and here).
Why I'm asking this question, is actually because I had the impression before that the above is actually the definition of ordinary consistency. But from the following Wikipedia page on $\omega$-consistency, it mentions:
A theory $T$ that interprets arithmetic is $\omega$-inconsistent if, for some property $P$ of natural numbers (defined by a formula in the language of $T$), $T$ proves $P(0), P(1), P(2)$, and so on (that is, for every standard natural number $n$, $T$ proves that $P(n) $holds), but $T$ also proves that there is some natural number $n$ such that $P(n)$ fails.
Also in the wikipedia article on the proof sketch of the first incompleteness theorem, the diagonalization, the paragraph discussing the provability of negation of $p$ (where $p$ is the Gödel sentence), it also mentions the role of the assumption of $\omega$-consistency in establishing non-provability of $\neg p$, and so in an ordinary consistent theory (that may not be $\omega$-consistent) I get it that we can have $\exists x \, [\operatorname {Proof}_T(x, \ulcorner \neg p \urcorner)] \land \exists z \, [\operatorname {Proof}_T(z, \ulcorner p \urcorner)]$, the latter being proved at some non-standard $y$.
Now, the above-mentioned consistency statement seems to prove $\omega$-consistency, since it would prevent existence of any proofs of contradicting statements coded by any naturals whether standard or not.
From all of this I also get the impression that the above statement if the naturals coding the proofs are taken to be standard, then it would be equivalent to the ordinary consistency statement.
However, I'm not sure of these, also I'm not sure if this statement is equivalent, weaker, or even stronger than $\omega$-consistency. Hence the question.