Potential due to initially uncharged induced conductor?

From Griffiths' Introduction to electrodynamics:

In this problem, Griffiths says that we will set the potential inside and on the surface of the conductor to be zero since its an equipotential surface anyway, but then he claims that, due to 'symmetry' the entire xy plane is at potential zero aswell.

I tried to make sense out of this and noticed that the potential due to the field $$\vec E = E_0 \hat z$$ in the xy plane does not change, but there's still the electric field of the induced charges on the sphere, why did he not account for it? Thanks

By symmetry the electric field lines are perpendicular to the $$xy$$ plane
As the $$xy$$ plane hits the surface of the conductor which is at zero potential so must be the $$xy$$ plane.
• @khaled014z But it will also be symmetrical about the $xy$ plane. Commented Jul 2, 2019 at 11:55
• @Farcher Can you explain why by symmetry the electric field lines are perpendicular to the $xy$ plane, please? I couldn't understand it. Commented Dec 31, 2019 at 20:00