zn+1=({Δ}(2+2=≈4/4){π}zn c sum prior
zn+1=({Δ}2+2≈4∧4){π≜π≜π}zn c sum prior
zn+1≐zn{(Δn}≐(2+2≈4∧4})≐{Rxπ≐π≐π°}∪{°π≐π≐πRx}{(zn+1=H(xn)FA(z~n+1)+(1−H(xn))FB(z~n+1),),zn=(xn,yn,wnRx)}+cΣ sum prior
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Foreword: This work is 7 years in the making. 7 years of continued deduction, reaching the final step, around a month ago, defining how we...
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