AFAIK - no way. As you mentioned above you should use setup with 2 buffs (one hidden behaviour for detection second is the actual modifier), 1 apply behaviour periodic effect for 1st buff and 1 validator (for 1st buff this validator will be disabling it, for 2nd - deleting it).
Buff disabled by validator on themselves is still showing. Is there a way to hide it when disabled?
P.S. I know that buff disabled by other buff will be hidden. But I want to know if there is a way to hide it by itself when disabled by own validator.
AFAIK - no way. As you mentioned above you should use setup with 2 buffs (one hidden behaviour for detection second is the actual modifier), 1 apply behaviour periodic effect for 1st buff and 1 validator (for 1st buff this validator will be disabling it, for 2nd - deleting it).
And here I thought I can be lazy and get off with 1 buff D: Thanks for the help!