Vmc has to be 20% greater than Vs
Agree with most of those statements, but not that one. There's no requirement that minimum control speed be higher than stall speed - and in fact it's preferable that it be lower. Since that means you have no practical asymmetric thrust controllability limits at low speeds.
Just to clarify some of the other statments about the conditions for Vmc(a) demonstration
As far as I remeber, it is calculated considering:
- maximum take-off thrust * usually it's "maximum normally expected" or some similar statement; which means the rated max thrust isn't used, since that's considered the minimum you'll get (or take credit for). Instead some number in excess is used, to allow for engine-to-engine variation.
- airplane trimmed for take-off
- worst CG position (aft limit)
- maximum take-off weight at MSL *actually, minimum weight is the key here, since increased weight gives effectively an increased bank angle allowance. Vmc(a) increases with decreased weight, all other conditions being equal
- most critical take-off configuration but undercarriage up!
- out of ground effect
- dead engine NOT feathered unless an auto-feathering device is provided.
- maximum bank allowed 5°
- maximum foot load 150 lbs *or 180lbf in some rules/amendments, I believe)