en.append(plru) # yes inverted (using bool())
print ("plru", i, en)
# boolean logic manipulation:
- # plur0 & plru1 & plur2 == ~(~plru0 | ~plru1 | ~plru2)
+ # plru0 & plru1 & plru2 == ~(~plru0 | ~plru1 | ~plru2)
m.d.comb += self.replace_en_o[i].eq(~Cat(*en).bool())
return m