# 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?

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.
Assume that the electric field is non-zero at any point inside the cavity. Then there has to be a field line passing through the point. Since, there are no charges inside the cavity, the field line has to go from a positive charge on the surface of the conductor to a negative charge on the surface of the conductor. Then choose $\gamma$ to be the path along the field line.
Let the electric field along the path be $E_l$. By the definition of the path $E_l> 0$ all along the path. Therefore, the line integral is non-zero, which is not possible.