Why doesn't the Continuum Hypothesis show that $\text{ZFC}$ is simply consistent?

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Where is my error?

A system is simply inconsistent is every formula is provable (if $\text{A} \wedge\neg \text{A}$ were provable for some formula $\text{A}$, then by $\neg$-introduction, any formula can be proven). Taking the contraposition, we get that if there exist some formula that is unprovable, the system is simply consistent.

Since the Continuum Hypothesis is undecidable and thus unprovable in $\text{ZFC}$, shouldn't that mean that $\text{ZFC}$ is simply consistent? This also shows the other direction of Gödel's first theorem: if the system is incomplete, then it is consistent.

I know all of this can't be right because I haven't read any of this and it can't be that easy, so I must be missing something.

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The think you're missing is the conditional part of undecidability of CH: it is undecidable, provided ZFC is consistent.

Roughly, the proof goes like this: we start with an arbitrary model of ZFC (which we have if ZFC is consistent), and then the constructible universe L within is a model of ZFC+CH, while an appropriate Cohen forcing extension is a model of ZFC+$\neg$CH.

None of this makes any sense if we have no model of ZFC to begin with.