I want to simplify the below equation but I can't get the same result with an online calculator (the result for the calculator is the correct - I check it with Logisim).
$Y = A'B'CD + A'B'C'D + ABC'D' + A'BC'D + AB'C'D'$
$Y = A'B'(C'D + CD) + A'BC'D + C'D'(AB' + AB)$
--> From Redundancy Laws: $AB + AB' = A$
$Y = A'B'D + A'BC'D + C'D'A$
--> I can simplify until here
The complete simplification is the below, but I can't find a boolean's law to use to remove the $B$.
Does anyone knows which law I have to use in order to remove the variable?
$Y = A'B'D + A'C'D + C'D'A$
$\endgroup$1 Answer
$\begingroup$Here is something you can do:
$A'B'D + A'BC'D + C'D'A =$
$A'D(B' + BC') + C'D'A =$
$A'D(B' + B)(B'+C') + C'D'A =$
$A'D1(B'+C') + C'D'A =$
$A'D(B'+C') + C'D'A =$
$A'B'D + A'C'D + C'D'A$
$\endgroup$ 1
