If we break it down to the output stage as in your equivalent circuit, it’s definitely NOR.
As long as we agree that 0=0V and 1=Vcc (or at least >0V).
The truth table is shown for example here:
https://www.build-electronic-circuits.com/nor-gate/
The Transistor cannot be blown as the base current is limited by the emitter resistor (1k) and whatever circuit is behind it.
I realise we are actually agreeing with the same end result only relative to different inputs. What I’m saying is, if without the knowledge of the comparator’s internal circuit, the outputs of two tied together can be treated as these two outputs passing through an invisible (and non-existing) logic AND gate.
Your view point is from the signals on the bases of the two transistors (each through a resistor). Relative to these two ‘inputs’ (let’s call them B1 and B2), then the combined outputs of the two transistors is certainly NOR(B1,B2). Note that B1/B2 are only internal business of the comparators.
We are saying the same thing, because —
O1=NOT(B1), O2=NOT(B2)
AND(O1,O2)=NOT(OR(NOT(O1),NOT(O2)))
=NOT(OR(B1,B2))
=NOR(B1,B2).
On the transistor after the op amp, all after P190-2 is shown in the shaded box nearby, an LED going to GND, which is the only channel sending signal back to the main business centre of the machine. On what condition will this transistor turn on and off then?
My analysis is here (based on the behaviour of ideal op-amps):
1. When T125 (on the PSU board) is off, i.e. the motor is off (T126 is either on or off), P190-10 (connected to U1A-3) is 'subordinate' and will mirror whatever voltage on U1A-2, which will be 'driving' and is determined by the R ladder (along the dash line), and is 0.1V.
2. When T125 is on, i.e. the motor is on (T126 must be off), P190-10 (U1A-3) will be forced to what is determined by the HVDC voltage (assuming 320V) and the R ladder (R16/R17/R18), which gives 2.1V (marked as "3.8V" in the schematic). U1A-2 is also flexible and is able to mirror this voltage (from 0.1V to 2.1V) by increase of current through the last 1k R on the R ladder. This transit will present a momentarily high level on U1A-1, which turns on the transistor and in turn the LED D109 (signalling to the business centre that the motor is being turned on). As soon as the levels on U1A-2/3 are (quickly )balanced, LED D109 is off.
3. When motor continues to be on, the levels on the steps of the R ladder (along the dash line) are at different corresponding values, which will have an effect (as inputs) on the corresponding comparators. (The logic is complex, and more is to be figured out.)
Similar to the case of two comparators' outputs tied together, but slightly different, is where U3-1 and U3-2 are connected through a diode. In this case, the latter can pull the former down, but
not vice versa. In other words, whenever U3-1 is LO, U3-2 will be LO regardless of the level of U3-2 when not tied together. But the level of U3-2 will not affect U3-1 at all. (It's puzzling what the function the relative independent part, top right around U3B, is.)
My feeling about the overall working of this daughterboard circuit is like this (starting from the idle status – motor is not running). The motor on instruction comes in through phototransistor T128, on the PSU board (copied to the bottom left corner), turning on the 3F transistor, starting the charging of the C1 (which is part of the RC oscillator). Through complex log, ultimately, square wave is output through P190-6 (bottom right). This will turn on (and off) transistor T125 through transformer U146. The motor speed is able to be sensed outside of the PSU board, which can be used to adjust the intensity of light opposite of T128 (D108).