Relationship between independence and consistency: Why does Con$(\mathsf{ZFC})\implies\text{Con}(\mathsf{ZFC+\phi})$ imply $\phi$ can't be disproved

Solution 1:

If $\lnot \phi$ were provable in $T$ (by any proof technique whatsoever), then it would also be provable in $T + \phi$, since the validity of a proof isn't harmed by adding more axioms. Since $T + \phi$ trivially proves $\phi$, that shows that $T + \phi$ proves $\phi \land \lnot \phi$, a contradiction, so $T+\phi$ is inconsistent.

The contrapositive of this statement is that if $T + \phi$ is consistent, then $T$ does not prove $\lnot \phi$.

Solution 2:

I suspect you're overthinking this.

Suppose $T\vdash\neg\varphi$. Don't worry about how this is the case, just observe that this a fortiori means $S\vdash \neg\varphi$ for every theory $S\supseteq T$: adding axioms doesn't lose theorems. In particular, this holds for $S=T\cup\{\varphi\}$. But if $T\cup\{\varphi\}\vdash\neg\varphi$ then $T\cup\{\varphi\}\vdash\varphi\wedge\neg\varphi$ (since $S\vdash\alpha$ for every $\alpha\in S$, and $S\vdash\alpha$ and $S\vdash\beta$ implies $S\vdash\alpha\wedge\beta$), so if $T\cup\{\varphi\}\vdash\neg\varphi$ then $T\cup\{\varphi\}$ is inconsistent.