Q:

Proof of $\bf{E}=0$ the cavity of a conductor and $q=0$ on the internal surface of the conductor

I found the following explanation of Griffiths Introduction to Electrodynamics, chapter 2.5, page 100-101.

So is it correct to state the following?
$$\oint_{\gamma} \bf{E} \cdot \bf{ds}=0 \,\,\,\,\, \forall\gamma \,\,\,\,\implies \,\,\,\,\bf{E}=0 \,\,\,\, \mathrm{everywhere \,\,\, inside \,\,\, the \,\,\, conductor}$$
Where $\gamma$ is any closed curve made as described in the text (i.e. composed of a curve joining two points of the cavity and another curve passing only inside the conductor).
Furthermore, is it possible to conclude from here that there cannot be any charge on the internal surface of the conductor? That is
$$\bf{E}=0 \,\,\,\, \mathrm{everywhere \,\,\, inside \,\,\, the \,\,\, cavity} \,\,\,\implies\,\,\,\mathrm{ q=0\,\,\,everywhere \,\,\, on \,\,\, the \,\,\, internal \,\,\,surface \,\,\, of \,\,\, the \,\,\, conductor}$$
If so, how to justify this last implication?

A:

If $$\int_{\gamma}\vec{E}.\mathrm{d}\vec{s}=0\quad \forall\gamma\\
\int_{\gamma}\vec{E}.\mathrm{d}\vec{s}=\iint_{\delta}\left(\nabla\times\vec{E}\right).\mathrm{d}\vec{A}=0\quad \forall \delta\quad \text{inside the surface}\\
\text{It means}\quad \nabla\times\vec{E}=0$$
This hints at $\vec{E}$ being a gradient of some sort, in fact we know that the electric field would be an electrostatic gradient $\vec{E}=-\nabla V$, where $V$ is the electrostatic potential function. So, we have Gauss' law which allows the field to diverge. $$\nabla.\vec{E}=\frac{\rho}{\epsilon}$$ but since inside the conductor no charge exists, hence, the charge density $\rho = 0$, thus we have $$\nabla\times\vec{E}=0\quad\text{and}\quad \nabla.\vec{E}=0$$ This implies $\vec{E}$ has to be a constant vector for sure, which is zero in this case.

