Hello again,
Yes that's right except instead of using the word 'larger' use the word 'higher' or 'more positive', the anode must be more positive than the cathode. So if we had say +1.1 volts at the anode and +1.0 volts at the cathode the diode would 'switch' to the on state.
Note that because he assumes the diode model is the perfect diode switch, when this happens we might then make the voltage at the anode equal to the voltage at the cathode, or the voltage at the cathode equal to the voltage at the anode, depending on which voltage is allowed to change. So we effectively short out the diode and analyze the circuit a second time once we know it is 'on'.
This might seem strange and it is a little bit strange, and that's now how a diode works in real life, but it's a simplified model of the diode where we allow the diode to be a short circuit when it is turned on. This simplifies the analysis. In real life we'd see a voltage drop across the diode of somewhere between 0.5v and 1v or maybe a little higher sometimes, but for these problem we dont have to be concerned about that because the lecturer is using this simplified switching model for the diode.
Note that with this diode model the anode voltage only has to be more positive by a small amount, so even if the anode was +1.001v and the cathode +1.000v, the anode is again more positive so the diode switches to the 'on' state, so we can short it out.