*This is a revised version of a [post](https://math.stackexchange.com/q/2098832/212120) on Math.SE. It is a rather basic question (which I'd be glad to delete if the community regards as off-topic).*




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Is there a way to prove that (if consistent) $\mathsf{ZFC}$ can't prove that there exists a weakly inaccessible without first showing that $\mathsf{GCH}$ is relatively consistent? Obviously, if we can show that $\mathsf{ZFC}$ doesn't prove the *consistency* of a weakly inaccessible is much better.


It would be a little bit surprising if there are no proofs without the full logical apparatus of definability that involves $V=L$. Perhaps some wise trick using reflection might do it, but it is just a guess. 


Two weakened version of the same question are also interesting to me:


1. Showing strength of weakly Mahlo cardinals with the same restrictions;
2. Same questions, using at most the consistency of $\mathsf{CH}$.


The only really similar question on Math.SE I found is [this](https://math.stackexchange.com/q/1162740/212120), and I wasn't much too thorough in my web search because I'm pretty sure that this is just a curiosity. 


**PS.** Answers using Easton magic are not allowed!



Consistency strength of weakly inaccessibles without $\mathsf{GCH}$
Suppose $\kappa$ is weakly inaccessible. Then it is immediate $L\_\kappa$ satisfies pairing, separation, extensionality, regularity, infinity, union, and choice. To see that $L\_\kappa$ satisfies Replacement, let $h(x)=min\{\alpha|x\subseteq L\_\alpha\}$. Let $X\in L\_\kappa$, and $F(X)=\{f(x)|x\in X\}$. Let $Z=\{h(x)|x\in F(X)\}$. Such a set exists as $F(X)\subseteq L\_\kappa$. Then let $\lambda$ be the order type of $Z$, and $\eta\_{\alpha+1}=min\{\beta>\eta\_\alpha|\beta\in Z\}$ be continuous at limits, with $\eta\_0=min\,Z$. Then if $Z$ is bounded in $\kappa$, it is clear $F(X)\in L\_\kappa$. Else $\kappa=\lim\_{\alpha\to \lambda}\eta\_\alpha$, yet $\lambda<\kappa$ and $\kappa$ is regular. Contradiction.


Now it remains to show that $L\_\kappa$ satisfies power set. Let $Y\subseteq X$. Then $L\_\kappa\vDash \exists\delta>|Y|(\delta\text{ is a limit}\ \land\ Y\in L\_\delta)$. Let $L\_\kappa\vDash M\prec L\_\delta$, and $|M|=|Y|$. Then take the transitive collapse $\pi(M)$ and so there is some limit ordinal $\gamma$ such that $\pi(Y)=Y$ and $Y\in L\_\gamma$. Hence $L\_\kappa\vDash ZFC$, and so we cannot prove the consistency of "There exists a weakly inaccessible cardinal" from $ZFC$. This argument does indeed skirt very close to the argument that $V=L\rightarrow GCH$, yet does not quite prove it. The rest follows from every weakly Mahlo cardinal being weakly inaccessible.


**Edit:** If you want to avoid the $L$ construction all together, then I suggest we construct a new cumulative hierarchy. We want $W\_\alpha$ to be sufficiently large to contain pairing, separation, etc. So, we want $L\_\alpha\subseteq W\_\alpha$. We also want it satisfy the condensation lemma so we need $W\_\alpha=\cup\_{\beta<\alpha}W\_\beta^{W\_\alpha}$.


Let $W\_{\alpha+1}=\{x\subseteq W\_\alpha\mid\text{$\phi(x,p\_0...p\_n)^{W\_\alpha}$ is a definition for $x,p\_0...p\_n\in W\_\alpha$}\}$. If $x=\{y|\phi(x,p\_0...p\_n)^{W\_\alpha}\}$, then $\forall y\in W\_\alpha(y\in x\leftrightarrow \phi(x,p\_0...p\_n)^{W\_\alpha})$ is a definition for $x$, and so $L\_\alpha\subseteq W\_\alpha$. Conversely, as $W\_\alpha$ is transitive $\phi(x,p\_0...p\_n)^{{W\_\beta}^{W\_\alpha}}\leftrightarrow \phi(x,p\_0...p\_n)^{W\_\beta}$ whenever $W\_\beta\in W\_\alpha$, and so $W\_\alpha=\cup\_{\beta<\alpha}W\_\beta^{W\_\alpha}$.


From here, simply repeat the argument from before.


