The generalized continuum hypothesis holds in L

Work inside $L$. By F1 it satisfies ZF and thinks $V=L$; by F2 it also satisfies A1. Fix an infinite internal cardinal $\kappa$, and write $\lambda =({\kappa }^{+}{)}^L$. Every $p\in {\mathcal{P}}^L(\kappa )$ is internally constructible, so the internal instance of F3 gives $p\in L_{\lambda }$. Thus inclusion is an injection ${\mathcal{P}}^L(\kappa )\hookrightarrow L_{\lambda }$.

Since $\lambda$ is an infinite ordinal, internal F4 gives $\mathrm{\mid }{L}_{\lambda }{\mathrm{\mid }}^L=\mathrm{\mid }\lambda {\mathrm{\mid }}^L=\lambda$. Hence $(2^{\kappa }{)}^L=\mathrm{\mid }{\mathcal{P}}^L(\kappa ){\mathrm{\mid }}^L\le \lambda$. On the other hand F5, applied under internal AC, gives $\kappa <(2^{\kappa }{)}^L$. By the defining minimality of the successor cardinal $\lambda$, this implies $\lambda \le (2^{\kappa }{)}^L$. Therefore $(2^{\kappa }{)}^L=\lambda =({\kappa }^{+}{)}^L$.

The argument applies to every infinite cardinal of $L$, so $L\models \mathrm{G}\mathrm{C}\mathrm{H}$, while F2 also gives $L\models \mathrm{A}\mathrm{C}$. If $V=L$, internal and ambient sets, cardinals, power sets, and successors coincide, yielding $\mathrm{A}\mathrm{C}+\mathrm{G}\mathrm{C}\mathrm{H}$ in $V$. No claim that ambient ZF alone well-orders arbitrary ambient power sets was used.