> What do you mean by at the same time? Those two photons reach Bob at exactly the same instant?In the original idealized model of Hawking, yes. That is what the spacetime geometry says. One way of thinking about it is that, with this spacetime geometry, the path of any light ray emitted outward at the event horizon travels along that horizon into the future, but the area of the horizon decreases as the hole evaporates. At the instant of the hole's final evaporation, all of the light rays emitted anywhere along the horizon are at the same point as the final evaporation of the hole, and they all go outward from there together. This is a highly idealized model; but that's what the model says.
> We had already agreed that General Relativity should imply Bob never sees Alice reach EH, regardless of quantum effects
No, we haven't. You claimed that, and I pointed out that you were wrong. Bob never sees Alice reach the EH in the classical GR model, with no quantum effects. Quantum effects change the model. In the model including quantum effects, Bob does see Alice reach the EH. See above and my previous posts.
> there cannot be a chain of events whereby Alice evaporates into photons (emanating from EH), and those photons (originating from Alice's mass-energy) hit Bob before the photons of Alice when she was near
This is correct. And it is also what the model of a hole evaporating by Hawking radiation says. All of the light signals emitted by Alice before she reaches the EH will reach Bob before he sees the hole's final evaporation (and the light signal emitted by Alice when she crossed the EH).
> I could be terribly mistaken of course, but your arguments did not convince me
You are terribly mistaken, and you didn't read my arguments carefully enough. See above.
> the expansion of the EH as a body approximates
I don't know what you're talking about here. As a black hole evaporates by Hawking radiation, the area of its horizon decreases. It doesn't increase.