Power loss into RCD snubber
I don't think that the assertion is theoretically true.
The snubber (aka diode-RC clamp) power dissipation will be the same if you halve the capacitance. The power dissipation is produced by the resistor because it needs to &quot;reset&quot; the voltage acquired by the clamp capacitor after it received leakage inductance energy during flyback. It has a certain length of time to reset that voltage hence RC must be fixed.
So, if you half the capacitance, you must double the resistance.
Given this well-known formula for capacitor energy (W): -
$$W = \dfrac{1}{2}\cdot CV^2$$
You can re-arrange for voltage: -
$$V = \sqrt{\dfrac{2\cdot W}{C}}$$
So, if you halved the capacitance, voltage would rise by \$\sqrt2\$.
The power dissipated by the resistor would be related to the peak voltage thus: -
$$P = \dfrac{V_{PK}^2}{R}$$
So if the capacitance halved, R would double and \$V_{PK}\$ would increase by \$\sqrt2\$: -
$$P = \dfrac{\sqrt2^2\cdot V_{PK}^2}{2R} = \dfrac{V_{PK}^2}{R}$$
In other words no change in dissipation.
There may be other nuances I've not considered of course but for a simple circuit, I believe the power dissipated by the clamp is unchanged when C changes.