Powering the Aruino with <6.6v on the barrel + USB?

I was reading about the Arduino Uno power select subsystem, and how both USB and barrel jack power can be connected at the same time. I understand how the board switches off the USB power when about >7v is supplied to the barrel jack, using the op-amp and the mosfet to compare and switch.

What I’m trying to figure out is how the board would deal with a barrel jack voltage of, say, 6v. Just as a guess, it would barely power the LDO regulator, but still put about 4v on the 5v rail. At the same time, the voltage divider before the op-amp would make it less than the 3.3v rail it’s being compared to, allowing the 5v from USB onto the 5v rail as well.

My Question: Would that not drive the 5v rail with both sources? Would it be additive at ~9v? Would that smoke up the board? Would the LDO regulator sense 5v already on the rail and turn off? I guess there might be something that drives the VIN to ground, but I don’t see how in the schematic.

I’m new to electronics, so maybe I just don’t understand this at all. This is the first schematic I’ve tried to make sense of. I tried searching, but couldn’t find the answer. I’d appreciate any input I can get on this. Thanks!

To a certain extent, how the issue power source conflict is resolved depends on which arduino and/or which clone.
The least you can expect is a schottky diode (voltage drop c. 0.4 volts) on each power source to stop it being back fed from another. For example, you definitely don't want non-rechargeable batteries being "charged" when connected.
All the power sources share a common ground so there is not "addition" of voltages and 5 + 5 in this case equals 5.
Are you looking at the official Uno V3 schematic ?

I think so. I was reading this page. Technobyte - Core Subsystems. I don't know if the author used the official schematic, but it looks like it.