well actually....
There are two aspects of OTL that are "retentive": that the bit remains true when the rung conditions are false, and that the bit remains true when "first scan" conditions occur.
OTE is wholly "non-retentive", in that the bit goes false if the rung conditions go false, AND it goes false when "first scan" conditions occur.
If A then B:=0 is truly non-retentive, as the bit remains set when both A = false AND on "first scan".
B[:=]A is truly retentive, as the bit becomes false when either A = false OR on "first scan".
B:=A is -- what? "semi-retentive" -- in that it acts like an OTE in how it responds to A (which was my point), but is immune to "first scan" resets (which is Ron's).
I would expect (though never tried it) that If A then B[:=]0 would be ¿quasi-retentive? as the B would be immune to A going false, but affected by first-scan. If so, it would be an "interesting" way of initializing a sequencer, provided one is an expert of when "first scan" applies, and when it doesn't.
I, for one, am not such an expert.