-------------------------------------------------
A = A + C + BD + B'D'
-------------------------------------------------
------------------------------------------------- 1) (A + C + BD + B'D')' ; de morgan's
-------------------------------------------------
------------------------------------------------- 2) (A') (C') (B' + D') (B + D) ; distribute
-------------------------------------------------
------------------------------------------------- 3) (A') (C') (B'B + B'D + BD' + D'D) ; theorem 16: A'A = 0
-------------------------------------------------
------------------------------------------------- 4) (A') (C') (B'D + BD') ; A'B + AB' < => A xor B
-------------------------------------------------
------------------------------------------------- 5) ((A’) (C’) (B xor D))’ ; de morgan’s
-------------------------------------------------
------------------------------------------------- 6) A + C + (B xor D)’ ; final answer assuming only single - 2 input cmos gates + sinle inverters
-------------------------------------------------
A + C + BD + B'D' = 6 gates
-------------------------------------------------
------------------------------------------------- A + C + (B xor D)' = 4 gates thats a savings of 2 logic gates! the simplified XOR expression is very elegant, but i don’t have any CMOS xor gates in hand right now. so i’m sticking with the longer logic expressions which does not utilize an XOR operation. i’ve been reading and learning about analog and digital electronics during my